id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-5500 | Scalar Product with Inverse Unity | :$\paren {-1_R} \circ x = - x$ | Follows directly from Scalar Product with Inverse.
{{qed}} | :$\paren {-1_R} \circ x = - x$ | Follows directly from [[Scalar Product with Inverse]].
{{qed}} | Scalar Product with Inverse Unity | https://proofwiki.org/wiki/Scalar_Product_with_Inverse_Unity | https://proofwiki.org/wiki/Scalar_Product_with_Inverse_Unity | [
"Unitary Modules"
] | [] | [
"Scalar Product with Inverse"
] |
proofwiki-5501 | Scalar Product with Multiple of Unity | :$\paren {n \cdot 1_R} \circ x = n \cdot x$
that is:
:$\paren {\map {\paren {+_R}^n} {1_R} } \circ x = \map {\paren {+_G}^n} x$ | Follows directly from Scalar Product with Product.
{{qed}} | :$\paren {n \cdot 1_R} \circ x = n \cdot x$
that is:
:$\paren {\map {\paren {+_R}^n} {1_R} } \circ x = \map {\paren {+_G}^n} x$ | Follows directly from [[Scalar Product with Product]].
{{qed}} | Scalar Product with Multiple of Unity | https://proofwiki.org/wiki/Scalar_Product_with_Multiple_of_Unity | https://proofwiki.org/wiki/Scalar_Product_with_Multiple_of_Unity | [
"Unitary Modules"
] | [] | [
"Scalar Product with Product"
] |
proofwiki-5502 | Subring Module is Module/Special Case | Let $S$ be a subring of the ring $\struct {R, +, \circ}$.
Let $\circ_S$ be the restriction of $\circ$ to $S \times R$.
Then $\struct {R, +, \circ_S}_S$ is an $S$-module. | From Ring is Module over Itself, it follows that:
:$\struct {R, +, \circ}_R$ is an $R$-module.
The result follows directly from Subring Module is Module.
{{qed}} | Let $S$ be a [[Definition:Subring|subring]] of the [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$.
Let $\circ_S$ be the [[Definition:Restriction of Operation|restriction]] of $\circ$ to $S \times R$.
Then $\struct {R, +, \circ_S}_S$ is an [[Definition:Module over Ring|$S$-module]]. | From [[Ring is Module over Itself]], it follows that:
:$\struct {R, +, \circ}_R$ is an [[Definition:Module over Ring|$R$-module]].
The result follows directly from [[Subring Module is Module]].
{{qed}} | Subring Module is Module/Special Case | https://proofwiki.org/wiki/Subring_Module_is_Module/Special_Case | https://proofwiki.org/wiki/Subring_Module_is_Module/Special_Case | [
"Subring Module is Module"
] | [
"Definition:Subring",
"Definition:Ring (Abstract Algebra)",
"Definition:Restriction/Operation",
"Definition:Module over Ring"
] | [
"Ring is Module over Itself",
"Definition:Module over Ring",
"Subring Module is Module"
] |
proofwiki-5503 | Trivial Module is Module | Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $\struct {G, +_G, \ast}_R$ be the trivial $R$-module, such that:
:$\forall \lambda \in R: \forall x \in G: \lambda \ast x = e_G$
Then $\struct {G, +_G, \ast}_R$ is a module. | Checking the module axioms in turn:
:{{Module-axiom|1}}: $\quad \lambda \circ \paren {x +_G y} = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
:{{Module-axiom|2}}: $\quad \paren {\lambda +_R \mu} \circ x = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$
:{{Module-axio... | Let $\struct {G, +_G}$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G, +_G, \ast}_R$ be the [[Definition:Trivial Module|trivial $R$-module]], such that:
:$\for... | Checking the [[Axiom:Module Axioms|module axioms]] in turn:
:{{Module-axiom|1}}: $\quad \lambda \circ \paren {x +_G y} = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
:{{Module-axiom|2}}: $\quad \paren {\lambda +_R \mu} \circ x = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\m... | Trivial Module is Module | https://proofwiki.org/wiki/Trivial_Module_is_Module | https://proofwiki.org/wiki/Trivial_Module_is_Module | [
"Module Theory"
] | [
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Ring (Abstract Algebra)",
"Definition:Trivial Module",
"Definition:Module over Ring"
] | [
"Axiom:Left Module Axioms",
"Definition:Trivial Module",
"Definition:Module over Ring"
] |
proofwiki-5504 | Trivial Module is Not Unitary | Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be the trivial $R$-module, such that:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$
Then unless $R$ is a ring with unity and $G$ contains only one element, th... | By definition, for a trivial module to be unitary, $R$ needs to be a ring with unity.
For {{Module-axiom|4}} to apply, we require that:
:$\forall x \in G: 1_R \circ x = x$
But for the trivial module:
:$\forall x \in G: 1_R \circ x = e_G$
So {{Module-axiom|4}} can apply only when:
:$\forall x \in G: x = e_G$
Thus for th... | Let $\struct {G, +_G}$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G, +_G, \circ}_R$ be the [[Definition:Trivial Module|trivial $R$-module]], such that:
:$\fo... | By definition, for a [[Definition:Trivial Module|trivial module]] to be [[Definition:Unitary Module|unitary]], $R$ needs to be a [[Definition:Ring with Unity|ring with unity]].
For {{Module-axiom|4}} to apply, we require that:
:$\forall x \in G: 1_R \circ x = x$
But for the trivial module:
:$\forall x \in G: 1_R \ci... | Trivial Module is Not Unitary | https://proofwiki.org/wiki/Trivial_Module_is_Not_Unitary | https://proofwiki.org/wiki/Trivial_Module_is_Not_Unitary | [
"Module Theory"
] | [
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Ring (Abstract Algebra)",
"Definition:Trivial Module",
"Definition:Ring with Unity",
"Definition:Unitary Module over Ring"
] | [
"Definition:Trivial Module",
"Definition:Unitary Module over Ring",
"Definition:Ring with Unity",
"Definition:Unitary Module over Ring",
"Definition:Trivial Group"
] |
proofwiki-5505 | Null Module is Module | Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $G$ be the trivial group.
Let $\struct {G, +_G, \ast}_R$ be the null module.
Then $\struct {G, +_G, \ast}_R$ is a module. | Follows from the fact that $\struct {G, +_G, \ast}_R$ has to be, by definition, a trivial module:
$\ast$ can only be defined as:
:$\forall \lambda \in R: \forall x \in G: \lambda \ast x = e_G$
The result then follows from Trivial Module is Module.
{{qed}}
Category:Module Theory
0tlsqipf6e4cnq5oqkgv0z0ay7pjtty | Let $\struct {R, +_R, \circ_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $G$ be the [[Definition:Trivial Group|trivial group]].
Let $\struct {G, +_G, \ast}_R$ be the [[Definition:Null Module|null module]].
Then $\struct {G, +_G, \ast}_R$ is a [[Definition:Module over Ring|module]]. | Follows from the fact that $\struct {G, +_G, \ast}_R$ has to be, by definition, a [[Definition:Trivial Module|trivial module]]:
$\ast$ can only be defined as:
:$\forall \lambda \in R: \forall x \in G: \lambda \ast x = e_G$
The result then follows from [[Trivial Module is Module]].
{{qed}}
[[Category:Module Theory]]
... | Null Module is Module | https://proofwiki.org/wiki/Null_Module_is_Module | https://proofwiki.org/wiki/Null_Module_is_Module | [
"Module Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Trivial Group",
"Definition:Null Module",
"Definition:Module over Ring"
] | [
"Definition:Trivial Module",
"Trivial Module is Module",
"Category:Module Theory"
] |
proofwiki-5506 | Condition on Equality of Generated Sigma-Algebras | Let $X$ be a set.
Let $\GG$, $\HH$ be sets of subsets of $X$.
Suppose that:
:$\GG \subseteq \HH \subseteq \map \sigma \GG$
where $\sigma$ denotes generated $\sigma$-algebra.
Then:
:$\map \sigma \GG = \map \sigma \HH$ | From Generated Sigma-Algebra Preserves Subset, it follows that:
:$\map \sigma \GG \subseteq \map \sigma \HH$
Since $\map \sigma \GG$ is a $\sigma$-algebra containing $\HH$:
:$\map \sigma \HH \subseteq \map \sigma \GG$
from the definition of generated $\sigma$-algebra.
Hence the result, from the definition of set equali... | Let $X$ be a [[Definition:Set|set]].
Let $\GG$, $\HH$ be [[Definition:Set|sets]] of [[Definition:Subset|subsets]] of $X$.
Suppose that:
:$\GG \subseteq \HH \subseteq \map \sigma \GG$
where $\sigma$ denotes [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]].
Then:
:$\map \... | From [[Generated Sigma-Algebra Preserves Subset]], it follows that:
:$\map \sigma \GG \subseteq \map \sigma \HH$
Since $\map \sigma \GG$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]] containing $\HH$:
:$\map \sigma \HH \subseteq \map \sigma \GG$
from the definition of [[Definition:Sigma-Algebra Generated by C... | Condition on Equality of Generated Sigma-Algebras | https://proofwiki.org/wiki/Condition_on_Equality_of_Generated_Sigma-Algebras | https://proofwiki.org/wiki/Condition_on_Equality_of_Generated_Sigma-Algebras | [
"Sigma-Algebras",
"Sigma-Algebras Generated by Collection of Subsets",
"Sigma-Algebras Generated by Collection of Subsets"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Subset",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Generated Sigma-Algebra Preserves Subset",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Set Equality",
"Category:Sigma-Algebras Generated by Collection of Subsets"
] |
proofwiki-5507 | Sigma-Algebra is Monotone Class | Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Then $\Sigma$ is also a monotone class. | By definition, $\Sigma$, being a $\sigma$-algebra, is closed under countable unions.
From Sigma-Algebra Closed under Countable Intersection, it is also closed under countable intersections.
Thence, by definition, $\Sigma$ is a monotone class.
{{qed}}
Category:Sigma-Algebras
Category:Monotone Classes
fzkm77bth2qkus1e99u... | Let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on a [[Definition:Set|set]] $X$.
Then $\Sigma$ is also a [[Definition:Monotone Class|monotone class]]. | By definition, $\Sigma$, being a [[Definition:Sigma-Algebra|$\sigma$-algebra]], is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Countable Union|countable unions]].
From [[Sigma-Algebra Closed under Countable Intersection]], it is also [[Definition:Closed Algebraic Structure|closed]] under [[Defi... | Sigma-Algebra is Monotone Class | https://proofwiki.org/wiki/Sigma-Algebra_is_Monotone_Class | https://proofwiki.org/wiki/Sigma-Algebra_is_Monotone_Class | [
"Sigma-Algebras",
"Monotone Classes"
] | [
"Definition:Sigma-Algebra",
"Definition:Set",
"Definition:Monotone Class"
] | [
"Definition:Sigma-Algebra",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Set Union/Countable Union",
"Sigma-Algebra Closed under Countable Intersection",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Set Intersection/Countable Intersection",
"Defini... |
proofwiki-5508 | Linear Combination of Sequence is Linear Combination of Set | Let $G$ be an $R$-module.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of elements of $G$.
Let $b$ be an element of $G$.
Then:
:$b$ is a linear combination of the sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$
{{iff}}:
:$b$ is a linear combination of the set $\set {a_k: 1 \mathop \le ... | === Necessary Condition ===
By definition of linear combination of subset:
:Every linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is a linear combination of $\set {a_k: 1 \mathop \le k \mathop \le n}$.
{{qed|lemma}} | Let $G$ be an [[Definition:Module over Ring|$R$-module]].
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Sequence|sequence of elements]] of $G$.
Let $b$ be an [[Definition:Element|element]] of $G$.
Then:
:$b$ is a [[Definition:Linear Combination of Sequence|linear combination]] of the [[Def... | === Necessary Condition ===
By definition of [[Definition:Linear Combination of Subset|linear combination of subset]]:
:Every [[Definition:Linear Combination of Sequence|linear combination]] of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is a [[Definition:Linear Combination of Subset|linear combination]] of $\s... | Linear Combination of Sequence is Linear Combination of Set | https://proofwiki.org/wiki/Linear_Combination_of_Sequence_is_Linear_Combination_of_Set | https://proofwiki.org/wiki/Linear_Combination_of_Sequence_is_Linear_Combination_of_Set | [
"Linear Algebra"
] | [
"Definition:Module over Ring",
"Definition:Sequence",
"Definition:Element",
"Definition:Linear Combination/Sequence",
"Definition:Finite Sequence",
"Definition:Linear Combination/Subset",
"Definition:Set"
] | [
"Definition:Linear Combination/Subset",
"Definition:Linear Combination/Sequence",
"Definition:Linear Combination/Subset",
"Definition:Linear Combination/Subset",
"Definition:Linear Combination/Sequence",
"Definition:Linear Combination/Subset"
] |
proofwiki-5509 | Trace Sigma-Algebra of Generated Sigma-Algebra | Let $X$ be a Set, and let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Let $A \subseteq X$ be a subset of $X$.
Then the following equality holds:
:$A \cap \map \sigma \GG = \map \sigma {A \cap \GG}$
where
:$\map \sigma \GG$ denotes the smallest $\sigma$-algebra on $X$ that contains $\GG$
:$\map \sigma... | By definition of generated $\sigma$-algebra:
:$\GG \subseteq \map \sigma \GG$
whence from Set Intersection Preserves Subsets:
:$A \cap \GG \subseteq A \cap \map \sigma \GG$
and therefore, by definition of generated $\sigma$-algebra:
:$\map \sigma {A \cap \GG} \subseteq A \cap \map \sigma \GG$
For the reverse inclusion,... | Let $X$ be a [[Definition:Set|Set]], and let $\GG \subseteq \powerset X$ be a collection of [[Definition:Subset|subsets]] of $X$.
Let $A \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
Then the following equality holds:
:$A \cap \map \sigma \GG = \map \sigma {A \cap \GG}$
where
:$\map \sigma \GG$ denotes t... | By definition of [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]]:
:$\GG \subseteq \map \sigma \GG$
whence from [[Set Intersection Preserves Subsets]]:
:$A \cap \GG \subseteq A \cap \map \sigma \GG$
and therefore, by definition of [[Definition:Sigma-Algebra Generated by Col... | Trace Sigma-Algebra of Generated Sigma-Algebra | https://proofwiki.org/wiki/Trace_Sigma-Algebra_of_Generated_Sigma-Algebra | https://proofwiki.org/wiki/Trace_Sigma-Algebra_of_Generated_Sigma-Algebra | [
"Sigma-Algebras",
"Sigma-Algebras Generated by Collection of Subsets",
"Trace Sigma-Algebras",
"Sigma-Algebras Generated by Collection of Subsets"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Subset",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Trace Sigma-Algebra"
] | [
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Set Intersection Preserves Subsets",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Sigma-Algebra",
"Set Intersection Distributes over Set Difference",
"Intersection with Subset is Subset",
"Definition:Sigma-Alge... |
proofwiki-5510 | Intersection is Subset/General Result | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \forall T \in \mathbb S: \bigcap \mathbb S \subseteq T$
=== Family of Sets ===
In the context of a family of sets, the result can be presented as follows:
{{:Intersection is Subset/Family of Sets}} | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap \mathbb S
| c =
}}
{{eqn | ll= \leadsto
| q = \forall T \in \mathbb S
| l = x
| o = \in
| r = T
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| q = \forall T \in \mathbb S
| l = \bigcap \mathbb S
... | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \forall T \in \mathbb S: \bigcap \mathbb S \subseteq T$
=== [[Intersection is Subset/Family of Sets|Family of Sets]] ===
In the context of a [[Definition:Ind... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap \mathbb S
| c =
}}
{{eqn | ll= \leadsto
| q = \forall T \in \mathbb S
| l = x
| o = \in
| r = T
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| q = \forall T \in \mathbb S
| l = \bigcap \mathbb S
... | Intersection is Subset/General Result | https://proofwiki.org/wiki/Intersection_is_Subset/General_Result | https://proofwiki.org/wiki/Intersection_is_Subset/General_Result | [
"Set Intersection",
"Subsets"
] | [
"Definition:Set",
"Definition:Power Set",
"Intersection is Subset/Family of Sets",
"Definition:Indexing Set/Family of Subsets"
] | [
"Category:Set Intersection",
"Category:Subsets"
] |
proofwiki-5511 | Union of Relative Complements of Nested Subsets | Let $R \subseteq S \subseteq T$ be sets with the indicated inclusions.
Then:
:$\relcomp T S \cup \relcomp S R = \relcomp T R$
where $\complement$ denotes relative complement.
Phrased via Set Difference as Intersection with Relative Complement:
:$\paren {T \setminus S} \cup \paren {S \setminus R} = T \setminus R$
where ... | From Union with Set Difference:
:$T = T \setminus S \cup S$
and therefore by Set Difference is Right Distributive over Union:
:$T \setminus R = \paren {\paren {T \setminus S} \setminus R} \cup \paren {S \setminus R}$
Now, by Set Difference with Union and Union with Superset is Superset:
:$\paren {T \setminus S} \setmin... | Let $R \subseteq S \subseteq T$ be [[Definition:Set|sets]] with the indicated [[Definition:Subset|inclusions]].
Then:
:$\relcomp T S \cup \relcomp S R = \relcomp T R$
where $\complement$ denotes [[Definition:Relative Complement|relative complement]].
Phrased via [[Set Difference as Intersection with Relative Comp... | From [[Union with Set Difference]]:
:$T = T \setminus S \cup S$
and therefore by [[Set Difference is Right Distributive over Union]]:
:$T \setminus R = \paren {\paren {T \setminus S} \setminus R} \cup \paren {S \setminus R}$
Now, by [[Set Difference with Union]] and [[Union with Superset is Superset]]:
:$\paren {T... | Union of Relative Complements of Nested Subsets | https://proofwiki.org/wiki/Union_of_Relative_Complements_of_Nested_Subsets | https://proofwiki.org/wiki/Union_of_Relative_Complements_of_Nested_Subsets | [
"Relative Complement",
"Set Union"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Relative Complement",
"Set Difference as Intersection with Relative Complement",
"Definition:Set Difference"
] | [
"Union with Set Difference",
"Set Difference is Right Distributive over Union",
"Set Difference with Union",
"Union with Superset is Superset"
] |
proofwiki-5512 | Homomorphic Image of R-Module is R-Module | Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ_G}_R$ be an $R$-module.
Let $\struct {H, +_H, \circ_H}_R$ be an $R$-algebraic structure.
Let $\phi: G \to H$ be a homomorphism.
Then the homomorphic image of $\phi$ is an $R$-module. | Let us write $\phi \sqbrk G$ to denote the homomorphic image of $\phi$.
From Image of Group Homomorphism is Subgroup, $\phi \sqbrk G$ is a subgroup of $\struct {H, +_H}$.
For any $\map \phi g$ and $\map \phi {g'}$ in $\phi \sqbrk G$, we have:
{{begin-eqn}}
{{eqn | l = \map \phi g +_H \map \phi {g'}
| r = \map \ph... | Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\struct {G, +_G, \circ_G}_R$ be an [[Definition:Module over Ring|$R$-module]].
Let $\struct {H, +_H, \circ_H}_R$ be an [[Definition:R-Algebraic Structure|$R$-algebraic structure]].
Let $\phi: G \to H$ be a [[Definition:R-Algebrai... | Let us write $\phi \sqbrk G$ to denote the [[Definition:Homomorphic Image|homomorphic image]] of $\phi$.
From [[Image of Group Homomorphism is Subgroup]], $\phi \sqbrk G$ is a [[Definition:Subgroup|subgroup]] of $\struct {H, +_H}$.
For any $\map \phi g$ and $\map \phi {g'}$ in $\phi \sqbrk G$, we have:
{{begin-eqn}}... | Homomorphic Image of R-Module is R-Module | https://proofwiki.org/wiki/Homomorphic_Image_of_R-Module_is_R-Module | https://proofwiki.org/wiki/Homomorphic_Image_of_R-Module_is_R-Module | [
"Module Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Module over Ring",
"Definition:R-Algebraic Structure",
"Definition:R-Algebraic Structure Homomorphism",
"Definition:Homomorphism (Abstract Algebra)/Image",
"Definition:Module over Ring"
] | [
"Definition:Homomorphism (Abstract Algebra)/Image",
"Image of Group Homomorphism is Subgroup",
"Definition:Subgroup",
"Definition:Abelian Group",
"Definition:Module over Ring",
"Axiom:Left Module Axioms",
"Definition:Module over Ring",
"Definition:Module over Ring",
"Definition:Module over Ring",
... |
proofwiki-5513 | Evaluation Linear Transformation is Linear Transformation | Let $R$ be a commutative ring with unity.
Let $G$ be an $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the double dual of $G^*$.
Let the mapping $J: G \to G^{**}$ be the evaluation linear transformation from $G$ into $G^{**}$ defined as:
:$\forall x \in G: \map J x = x^\wedge$
where for each $x \in... | From Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual, we have that:
:$x^\wedge \in G^{**}$
Hence $x^\wedge$ a fortiori is a linear transformation.
It remains to be shown that $J: G \to G^{**}$ is a linear transformation.
That is, that the following conditions are satisfied by $J$:
:$(1)... | Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $G$ be an [[Definition:Module over Ring|$R$-module]].
Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
Let $G^{**}$ be the [[Definition:Double Dual|double dual]] of $G^*$.
Let the [[Definition:Mapping|mapp... | From [[Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual]], we have that:
:$x^\wedge \in G^{**}$
Hence $x^\wedge$ [[Definition:A Fortiori|a fortiori]] is a [[Definition:Linear Transformation|linear transformation]].
It remains to be shown that $J: G \to G^{**}$ is a [[Definition:Linea... | Evaluation Linear Transformation is Linear Transformation | https://proofwiki.org/wiki/Evaluation_Linear_Transformation_is_Linear_Transformation | https://proofwiki.org/wiki/Evaluation_Linear_Transformation_is_Linear_Transformation | [
"Linear Transformations",
"Evaluation Linear Transformations (Module Theory)"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Module over Ring",
"Definition:Algebraic Dual",
"Definition:Algebraic Dual/Double Dual",
"Definition:Mapping",
"Definition:Evaluation Linear Transformation/Module Theory",
"Definition:Linear Transformation"
] | [
"Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual",
"Definition:A Fortiori",
"Definition:Linear Transformation",
"Definition:Linear Transformation"
] |
proofwiki-5514 | Evaluation Isomorphism is Isomorphism | Let $R$ be a commutative ring with unity.
Let $G$ be a unitary $R$-module whose dimension is finite.
Then the evaluation linear transformation $J: G \to G^{**}$ is an isomorphism. | Let $\sequence {a_n}$ be an ordered basis of $G$.
Then by definition, every $x \in G$ can be written in the form:
:$\ds \sum_{k \mathop = 1}^n \lambda_k a_k$
where $\lambda_k \in R$.
From Unique Representation by Ordered Basis, this representation is unique.
From Existence of Ordered Dual Basis:
:$\sequence {\map J {a_... | Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $G$ be a [[Definition:Unitary Module|unitary $R$-module]] whose [[Definition:Dimension of Module|dimension]] is [[Definition:Finite|finite]].
Then the [[Definition:Evaluation Linear Transformation/Module Theory|evaluation linear... | Let $\sequence {a_n}$ be an [[Definition:Ordered Basis|ordered basis]] of $G$.
Then by definition, every $x \in G$ can be written in the form:
:$\ds \sum_{k \mathop = 1}^n \lambda_k a_k$
where $\lambda_k \in R$.
From [[Unique Representation by Ordered Basis]], this representation is [[Definition:Unique|unique]].
Fro... | Evaluation Isomorphism is Isomorphism | https://proofwiki.org/wiki/Evaluation_Isomorphism_is_Isomorphism | https://proofwiki.org/wiki/Evaluation_Isomorphism_is_Isomorphism | [
"Linear Transformations",
"Module Isomorphisms",
"Evaluation Linear Transformations (Module Theory)"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Unitary Module over Ring",
"Definition:Dimension of Module",
"Definition:Finite",
"Definition:Evaluation Linear Transformation/Module Theory",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism/Module Isomorphism"
] | [
"Definition:Ordered Basis",
"Unique Representation by Ordered Basis",
"Definition:Unique",
"Existence of Ordered Dual Basis",
"Definition:Ordered Dual Basis",
"Definition:Ordered Basis",
"Definition:Ordered Dual Basis",
"Unique Linear Transformation Between Modules",
"Definition:Unique",
"Definiti... |
proofwiki-5515 | Conditions for Homogeneity/Straight Line | The line $L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$ is homogeneous {{iff}} $\beta = 0$. | === Sufficient Condition ===
Let the line $L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$ be homogeneous.
Then the origin $\tuple {0, 0}$ lies on $L$.
Hence:
{{begin-eqn}}
{{eqn | l = \alpha_1 0 + \alpha_2 0
| r = \beta
| c =
}}
{{eqn | ll= \leadsto
| l = \beta
| r = 0
| c =
}}
{{end-eqn}}
{{q... | The [[Equation of Straight Line in Plane|line]] $L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$ is [[Definition:Homogeneous (Analytic Geometry)|homogeneous]] {{iff}} $\beta = 0$. | === Sufficient Condition ===
Let the [[Definition:Straight Line|line]] $L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$ be [[Definition:Homogeneous (Analytic Geometry)|homogeneous]].
Then the [[Definition:Origin|origin]] $\tuple {0, 0}$ lies on $L$.
Hence:
{{begin-eqn}}
{{eqn | l = \alpha_1 0 + \alpha_2 0
| r = \bet... | Conditions for Homogeneity/Straight Line | https://proofwiki.org/wiki/Conditions_for_Homogeneity/Straight_Line | https://proofwiki.org/wiki/Conditions_for_Homogeneity/Straight_Line | [
"Linear Algebra",
"Analytic Geometry"
] | [
"Equation of Straight Line in Plane",
"Definition:Homogeneous (Analytic Geometry)"
] | [
"Definition:Line/Straight Line",
"Definition:Homogeneous (Analytic Geometry)",
"Definition:Coordinate System/Origin",
"Definition:Homogeneous (Analytic Geometry)"
] |
proofwiki-5516 | Conditions for Homogeneity/Plane | The plane $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ is homogeneous {{iff}} $\gamma = 0$. | === Sufficient Condition ===
Let the plane $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be homogeneous.
Then the origin $\tuple {0, 0, 0}$ lies on $P$.
Hence:
{{begin-eqn}}
{{eqn | l = \alpha_1 0 + \alpha_2 0 + \alpha_3 0
| r = \gamma
| c =
}}
{{eqn | ll= \leadsto
| l = \gamma
| r ... | The [[Equation of Plane|plane]] $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ is [[Definition:Homogeneous (Analytic Geometry)|homogeneous]] {{iff}} $\gamma = 0$. | === Sufficient Condition ===
Let the plane $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be [[Definition:Homogeneous (Analytic Geometry)|homogeneous]].
Then the [[Definition:Origin|origin]] $\tuple {0, 0, 0}$ lies on $P$.
Hence:
{{begin-eqn}}
{{eqn | l = \alpha_1 0 + \alpha_2 0 + \alpha_3 0
| r = ... | Conditions for Homogeneity/Plane | https://proofwiki.org/wiki/Conditions_for_Homogeneity/Plane | https://proofwiki.org/wiki/Conditions_for_Homogeneity/Plane | [
"Linear Algebra",
"Solid Analytic Geometry"
] | [
"Equation of Plane",
"Definition:Homogeneous (Analytic Geometry)"
] | [
"Definition:Homogeneous (Analytic Geometry)",
"Definition:Coordinate System/Origin",
"Definition:Homogeneous (Analytic Geometry)"
] |
proofwiki-5517 | Zero Matrix is Identity for Hadamard Product | Let $\struct {S, \cdot}$ be a monoid whose identity is $e$.
Let $\map {\MM_S} {m, n}$ be an $m \times n$ matrix space over $S$.
Let $\mathbf e = \sqbrk e_{m n}$ be the zero matrix of $\map {\MM_S} {m, n}$.
Then $\mathbf e$ is the identity element for Hadamard product. | Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_S} {m, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A \circ \mathbf e
| r = \sqbrk a_{m n} \circ \sqbrk e_{m n}
| c = Definition of $\mathbf A$ and $\mathbf e$
}}
{{eqn | r = \sqbrk {a \cdot e}_{m n}
| c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \sqbrk a... | Let $\struct {S, \cdot}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\map {\MM_S} {m, n}$ be an [[Definition:Matrix Space|$m \times n$ matrix space]] over $S$.
Let $\mathbf e = \sqbrk e_{m n}$ be the [[Definition:Zero Matrix over General Monoid|zero matrix]] of $\map ... | Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_S} {m, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A \circ \mathbf e
| r = \sqbrk a_{m n} \circ \sqbrk e_{m n}
| c = Definition of $\mathbf A$ and $\mathbf e$
}}
{{eqn | r = \sqbrk {a \cdot e}_{m n}
| c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \sqbrk... | Zero Matrix is Identity for Hadamard Product | https://proofwiki.org/wiki/Zero_Matrix_is_Identity_for_Hadamard_Product | https://proofwiki.org/wiki/Zero_Matrix_is_Identity_for_Hadamard_Product | [
"Hadamard Product",
"Zero Matrix"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Matrix Space",
"Definition:Zero Matrix/General Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Hadamard Product"
] | [] |
proofwiki-5518 | Sum of Ideals is Ideal/General Result | Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.
Then:
:$J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.
where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product with respect to $\struct{R, +}$. | Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.
$\map P 1$ is true, as this just says $J_1$ is an ideal of $R$. | Let $J_1, J_2, \ldots, J_n$ be [[Definition:Ideal of Ring|ideals]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$.
Then:
:$J = J_1 + J_2 + \cdots + J_n$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
where $J_1 + J_2 + \cdots + J_n$ is as defined in [[Definition:Subset Product|subset produ... | Let $J_1, J_2, \ldots, J_n$ be [[Definition:Ideal of Ring|ideals]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$.
Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$J_1 + J_2 + \cdots + J_n$ i... | Sum of Ideals is Ideal/General Result | https://proofwiki.org/wiki/Sum_of_Ideals_is_Ideal/General_Result | https://proofwiki.org/wiki/Sum_of_Ideals_is_Ideal/General_Result | [
"Ideal Theory"
] | [
"Definition:Ideal of Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring",
"Definition:Subset Product"
] | [
"Definition:Ideal of Ring",
"Definition:Ring (Abstract Algebra)",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",... |
proofwiki-5519 | Integer is Expressible as Product of Primes | Let $n$ be an integer such that $n > 1$.
Then $n$ can be expressed as the product of one or more primes. | {{AimForCont}} this supposition is false.
Let $m$ be the smallest integer which can not be expressed as the product of primes.
As a prime number is trivially a product of primes, $m$ can not itself be prime.
Hence:
:$\exists r, s \in \Z: 1 < r < m, 1 < s < m: m = r s$
As $m$ is our least counterexample, both $r$ and $s... | Let $n$ be an [[Definition:Integer|integer]] such that $n > 1$.
Then $n$ can be expressed as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]]. | {{AimForCont}} this supposition is false.
Let $m$ be the smallest [[Definition:Integer|integer]] which can not be expressed as the [[Definition:Integer Multiplication|product]] of [[Definition:Prime Number|primes]].
As a [[Definition:Prime Number|prime number]] is trivially a [[Definition:Integer Multiplication|produ... | Integer is Expressible as Product of Primes/Proof 1 | https://proofwiki.org/wiki/Integer_is_Expressible_as_Product_of_Primes | https://proofwiki.org/wiki/Integer_is_Expressible_as_Product_of_Primes/Proof_1 | [
"Integer is Expressible as Product of Primes",
"Fundamental Theorem of Arithmetic",
"Prime Decompositions",
"Prime Numbers",
"Factorization"
] | [
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Prime Number"
] | [
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Multiplication/Integers",
"Definition:Prime Number",
"Definition:Prime Number",
"Principle of Least Counterexample",
"Definition:Multiplication/Integers",
"Definition:Prime... |
proofwiki-5520 | Integer is Expressible as Product of Primes | Let $n$ be an integer such that $n > 1$.
Then $n$ can be expressed as the product of one or more primes. | If $n$ is prime, the result is immediate.
Let $n$ be composite.
Then by Composite Number has Two Divisors Less Than It:
:$\exists r, s \in \Z: n = r s, 1 < r < n, 1 < s < n$
This being the case, the set $S_1 = \set {d: d \divides n, 1 < d < n}$ is nonempty, and bounded below by $1$.
By Set of Integers Bounded Below by ... | Let $n$ be an [[Definition:Integer|integer]] such that $n > 1$.
Then $n$ can be expressed as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]]. | If $n$ is [[Definition:Prime Number|prime]], the result is immediate.
Let $n$ be [[Definition:Composite Number|composite]].
Then by [[Composite Number has Two Divisors Less Than It]]:
:$\exists r, s \in \Z: n = r s, 1 < r < n, 1 < s < n$
This being the case, the set $S_1 = \set {d: d \divides n, 1 < d < n}$ is [[De... | Integer is Expressible as Product of Primes/Proof 2 | https://proofwiki.org/wiki/Integer_is_Expressible_as_Product_of_Primes | https://proofwiki.org/wiki/Integer_is_Expressible_as_Product_of_Primes/Proof_2 | [
"Integer is Expressible as Product of Primes",
"Fundamental Theorem of Arithmetic",
"Prime Decompositions",
"Prime Numbers",
"Factorization"
] | [
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Prime Number"
] | [
"Definition:Prime Number",
"Definition:Composite Number",
"Composite Number has Two Divisors Less Than It",
"Definition:Non-Empty Set",
"Definition:Bounded Below Set",
"Set of Integers Bounded Below by Integer has Smallest Element",
"Definition:Smallest Element",
"Definition:Composite Number",
"Comp... |
proofwiki-5521 | Integer is Expressible as Product of Primes | Let $n$ be an integer such that $n > 1$.
Then $n$ can be expressed as the product of one or more primes. | The proof proceeds by induction.
For all $n \in \N_{> 1}$, let $\map P n$ be the proposition:
:$n$ can be expressed as a product of prime numbers.
First note that if $n$ is prime, the result is immediate.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$n$ can be expressed as a product of prime numbers.
As $2$... | Let $n$ be an [[Definition:Integer|integer]] such that $n > 1$.
Then $n$ can be expressed as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{> 1}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$n$ can be expressed as a product of [[Definition:Prime Number|prime numbers]].
First note that if $n$ is [[Definition:Prime Number|prime]], the result is ... | Integer is Expressible as Product of Primes/Proof 3 | https://proofwiki.org/wiki/Integer_is_Expressible_as_Product_of_Primes | https://proofwiki.org/wiki/Integer_is_Expressible_as_Product_of_Primes/Proof_3 | [
"Integer is Expressible as Product of Primes",
"Fundamental Theorem of Arithmetic",
"Prime Decompositions",
"Prime Numbers",
"Factorization"
] | [
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Prime Number"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Prime Number",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Definition:Prime Number",
"Definition:Prime Nu... |
proofwiki-5522 | Expression for Integer as Product of Primes is Unique | Let $n$ be an integer such that $n > 1$.
Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear. | {{AimForCont}} the supposition false.
That is, suppose there is at least one positive integer that can be expressed in more than one way as a product of primes.
Let the smallest of these be $m$.
Thus:
:$m = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s$
where all of $p_1, \ldots p_r, q_1, \ldots q_s$ are prime.
By definition... | Let $n$ be an [[Definition:Integer|integer]] such that $n > 1$.
Then the expression for $n$ as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]] is [[Definition:Unique|unique]] up to the order in which they appear. | {{AimForCont}} the supposition false.
That is, suppose there is at least one [[Definition:Positive Integer|positive integer]] that can be expressed in more than one way as a product of [[Definition:Prime Number|primes]].
Let the smallest of these be $m$.
Thus:
:$m = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s$
where all... | Expression for Integer as Product of Primes is Unique/Proof 1 | https://proofwiki.org/wiki/Expression_for_Integer_as_Product_of_Primes_is_Unique | https://proofwiki.org/wiki/Expression_for_Integer_as_Product_of_Primes_is_Unique/Proof_1 | [
"Expression for Integer as Product of Primes is Unique",
"Fundamental Theorem of Arithmetic",
"Prime Decompositions",
"Prime Numbers",
"Factorization"
] | [
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Prime Number",
"Definition:Unique"
] | [
"Definition:Positive/Integer",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Positive/Integer",
"Prime not Divisor implies Coprime",
"Definition:Divisor (Algebra)/Integer",
"Euclid's Lemma for ... |
proofwiki-5523 | Expression for Integer as Product of Primes is Unique | Let $n$ be an integer such that $n > 1$.
Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear. | {{AimForCont}} $n$ has two prime factorizations:
:$n = p_1 p_2 \dots p_r = q_1 q_2 \dots q_s$
where $r \le s$ and each $p_i$ and $q_j$ is prime with $p_1 \le p_2 \le \dots \le p_r$ and $q_1 \le q_2 \le \dots \le q_s$.
Since $p_1 \divides q_1 q_2 \dots q_s$, it follows from Euclid's Lemma for Prime Divisors that $p_1 =... | Let $n$ be an [[Definition:Integer|integer]] such that $n > 1$.
Then the expression for $n$ as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]] is [[Definition:Unique|unique]] up to the order in which they appear. | {{AimForCont}} $n$ has two [[Definition:Prime Decomposition|prime factorizations]]:
:$n = p_1 p_2 \dots p_r = q_1 q_2 \dots q_s$
where $r \le s$ and each $p_i$ and $q_j$ is prime with $p_1 \le p_2 \le \dots \le p_r$ and $q_1 \le q_2 \le \dots \le q_s$.
Since $p_1 \divides q_1 q_2 \dots q_s$, it follows from [[Euclid'... | Expression for Integer as Product of Primes is Unique/Proof 2 | https://proofwiki.org/wiki/Expression_for_Integer_as_Product_of_Primes_is_Unique | https://proofwiki.org/wiki/Expression_for_Integer_as_Product_of_Primes_is_Unique/Proof_2 | [
"Expression for Integer as Product of Primes is Unique",
"Fundamental Theorem of Arithmetic",
"Prime Decompositions",
"Prime Numbers",
"Factorization"
] | [
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Prime Number",
"Definition:Unique"
] | [
"Definition:Prime Decomposition",
"Euclid's Lemma for Prime Divisors",
"Euclid's Lemma for Prime Divisors",
"Definition:Common Divisor/Integers",
"Definition:Common Divisor/Integers",
"Divisors of One",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number",
"Proof by Contradiction",
"D... |
proofwiki-5524 | Expression for Integer as Product of Primes is Unique | Let $n$ be an integer such that $n > 1$.
Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear. | The proof proceeds by strong induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:the prime decomposition for $n$ is unique up to order of presentation.
Note that it has been established in Integer is Expressible as Product of Primes that $n$ does in fact have at least $1$ prime decomposition.
===... | Let $n$ be an [[Definition:Integer|integer]] such that $n > 1$.
Then the expression for $n$ as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]] is [[Definition:Unique|unique]] up to the order in which they appear. | The proof proceeds by [[Second Principle of Mathematical Induction|strong induction]].
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:the [[Definition:Prime Decomposition|prime decomposition]] for $n$ is [[Definition:Unique|unique]] up to order of presentation.
Note that i... | Expression for Integer as Product of Primes is Unique/Proof 3 | https://proofwiki.org/wiki/Expression_for_Integer_as_Product_of_Primes_is_Unique | https://proofwiki.org/wiki/Expression_for_Integer_as_Product_of_Primes_is_Unique/Proof_3 | [
"Expression for Integer as Product of Primes is Unique",
"Fundamental Theorem of Arithmetic",
"Prime Decompositions",
"Prime Numbers",
"Factorization"
] | [
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Prime Number",
"Definition:Unique"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Prime Decomposition",
"Definition:Unique",
"Integer is Expressible as Product of Primes",
"Definition:Prime Decomposition",
"Second Principle of Mathematical Induction",
"Second Principle of Mathematical Induction",
... |
proofwiki-5525 | Modulo Addition is Closed/Integers | Let $m \in \Z$ be an integer.
Then addition modulo $m$ on the set of integers modulo $m$ is closed:
:$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m +_m \eqclass y m \in \Z_m$. | From the definition of addition modulo $m$, we have:
:$\eqclass x m +_m \eqclass y m = \eqclass {x + y} m$
By the Division Theorem:
:$x + y = q m + r$ where $0 \le r < m$
Therefore for all $0 \le r < m$:
:$\eqclass {x + y} m = \eqclass r m$.
Therefore from the definition of integers modulo $m$:
:$\eqclass {x + y} m \in... | Let $m \in \Z$ be an [[Definition:Integer|integer]].
Then [[Definition:Modulo Addition|addition modulo $m$]] on the [[Definition:Integers Modulo m|set of integers modulo $m$]] is [[Definition:Closed Algebraic Structure|closed]]:
:$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m +_m \eqclass y m \in \Z_m$. | From the definition of [[Definition:Modulo Addition|addition modulo $m$]], we have:
:$\eqclass x m +_m \eqclass y m = \eqclass {x + y} m$
By the [[Division Theorem]]:
:$x + y = q m + r$ where $0 \le r < m$
Therefore for all $0 \le r < m$:
:$\eqclass {x + y} m = \eqclass r m$.
Therefore from the definition of [[Defin... | Modulo Addition is Closed/Integers | https://proofwiki.org/wiki/Modulo_Addition_is_Closed/Integers | https://proofwiki.org/wiki/Modulo_Addition_is_Closed/Integers | [
"Modulo Addition"
] | [
"Definition:Integer",
"Definition:Modulo Addition",
"Definition:Integers Modulo m",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Modulo Addition",
"Division Theorem",
"Definition:Integers Modulo m"
] |
proofwiki-5526 | Modulo Addition is Closed/Real Numbers | Let $z \in \R$ be a real number.
Then addition modulo $z$ on the set of residue classes modulo $z$ is closed:
:$\forall \eqclass x z, \eqclass y z \in \R_z: \eqclass x z +_z \eqclass y z \in \R_z$. | From the definition of addition modulo $z$, we have:
:$\eqclass x z +_z \eqclass y z = \eqclass {x + y} z$
As $x, y \in R$, we have that $x + y \in \R$ as Real Addition is Closed.
Hence by definition of congruence, $\eqclass {x + y} z \in \R_z$.
{{qed}} | Let $z \in \R$ be a [[Definition:Real Number|real number]].
Then [[Definition:Modulo Addition|addition modulo $z$]] on the [[Definition:Set of Residue Classes|set of residue classes modulo $z$]] is [[Definition:Closed Algebraic Structure|closed]]:
:$\forall \eqclass x z, \eqclass y z \in \R_z: \eqclass x z +_z \eqcla... | From the definition of [[Definition:Modulo Addition|addition modulo $z$]], we have:
:$\eqclass x z +_z \eqclass y z = \eqclass {x + y} z$
As $x, y \in R$, we have that $x + y \in \R$ as [[Real Addition is Closed]].
Hence by definition of [[Definition:Congruence (Number Theory)|congruence]], $\eqclass {x + y} z \in \R... | Modulo Addition is Closed/Real Numbers | https://proofwiki.org/wiki/Modulo_Addition_is_Closed/Real_Numbers | https://proofwiki.org/wiki/Modulo_Addition_is_Closed/Real_Numbers | [
"Modulo Addition"
] | [
"Definition:Real Number",
"Definition:Modulo Addition",
"Definition:Set of Residue Classes",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Modulo Addition",
"Real Addition is Closed",
"Definition:Congruence (Number Theory)"
] |
proofwiki-5527 | Field Homomorphism Preserves Subfields | Let $\struct {F_1, +_1, \circ_1}$ and $\struct {F_2, +_2, \circ_2}$ be fields.
Let $\phi: F_1 \to F_2$ be a field homomorphism such that $\phi$ is not the trivial homomorphism.
If $K$ is a subfield of $F_1$, then $\phi \sqbrk K$ is a subfield of $F_2$. | First note that if $\phi$ is the trivial homomorphism then $\phi \sqbrk K = 0_{F_2}$ and so is not a field.
Since $K$ is a field, we have that:
:$0_{F_1} \in K$
:$1_{F_1} \in K$
and so by Ring Homomorphism Preserves Zero and Field Homomorphism Preserves Unity:
:$\map \phi {0_{F_1} } = 0_{F_2} \in \phi \sqbrk K$
:$\map ... | Let $\struct {F_1, +_1, \circ_1}$ and $\struct {F_2, +_2, \circ_2}$ be [[Definition:Field (Abstract Algebra)|fields]].
Let $\phi: F_1 \to F_2$ be a [[Definition:Field Homomorphism|field homomorphism]] such that $\phi$ is not the [[Definition:Trivial Homomorphism|trivial homomorphism]].
If $K$ is a [[Definition:Subf... | First note that if $\phi$ is the [[Definition:Trivial Homomorphism|trivial homomorphism]] then $\phi \sqbrk K = 0_{F_2}$ and so is not a [[Definition:Field (Abstract Algebra)|field]].
Since $K$ is a [[Definition:Field (Abstract Algebra)|field]], we have that:
:$0_{F_1} \in K$
:$1_{F_1} \in K$
and so by [[Ring Homomorp... | Field Homomorphism Preserves Subfields | https://proofwiki.org/wiki/Field_Homomorphism_Preserves_Subfields | https://proofwiki.org/wiki/Field_Homomorphism_Preserves_Subfields | [
"Field Homomorphisms",
"Subfields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Homomorphism",
"Definition:Zero Homomorphism",
"Definition:Subfield",
"Definition:Subfield"
] | [
"Definition:Zero Homomorphism",
"Definition:Field (Abstract Algebra)",
"Definition:Field (Abstract Algebra)",
"Ring Homomorphism Preserves Zero",
"Field Homomorphism Preserves Unity",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Group Homomorphism Preserves Subgroups",
"Definition... |
proofwiki-5528 | Set is Subset of Union/General Result | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$ | Let $x \in T$ for some $T \in \mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \bigcup \mathbb S
| c = {{Defof|Set Union}}
}}
{{eqn | ll= \leadsto
| l = T
| o = \subseteq
| r = \bigcup \mat... | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$ | Let $x \in T$ for some $T \in \mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \bigcup \mathbb S
| c = {{Defof|Set Union}}
}}
{{eqn | ll= \leadsto
| l = T
| o = \subseteq
| r = \bigcup \m... | Set is Subset of Union/General Result | https://proofwiki.org/wiki/Set_is_Subset_of_Union/General_Result | https://proofwiki.org/wiki/Set_is_Subset_of_Union/General_Result | [
"Set Union",
"Subsets"
] | [
"Definition:Set",
"Definition:Power Set"
] | [
"Category:Set Union",
"Category:Subsets"
] |
proofwiki-5529 | Intersection Distributes over Union/General Result | Let $S$ and $T$ be sets.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
:$\ds S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ | === Intersection Subset of Union ===
Let $\ds x \in S \cap \bigcup \mathbb T$.
We need to show that $\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ and then by definition of subset we will have shown that $\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$.
So, w... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\powerset T$ be the [[Definition:Power Set|power set]] of $T$.
Let $\mathbb T$ be a [[Definition:Subset|subset]] of $\powerset T$.
Then:
:$\ds S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ | === Intersection Subset of Union ===
Let $\ds x \in S \cap \bigcup \mathbb T$.
We need to show that $\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ and then by definition of [[Definition:Subset|subset]] we will have shown that $\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \... | Intersection Distributes over Union/General Result | https://proofwiki.org/wiki/Intersection_Distributes_over_Union/General_Result | https://proofwiki.org/wiki/Intersection_Distributes_over_Union/General_Result | [
"Intersection Distributes over Union"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset"
] | [
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Set Union",
"Definition:Subset",
"Definition:Set Union",
"Definition:Set Intersection",
"Definition:Set Union",
"Definition:Set Intersection"
] |
proofwiki-5530 | Union Distributes over Intersection/General Result | Let $S$ and $T$ be sets.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
:$\ds S \cup \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ | === Union Subset of Intersection ===
Let $\ds x \in S \cup \bigcap \mathbb T$.
We need to show that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and then by definition of subset we will have shown that:
:$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.
S... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\powerset T$ be the [[Definition:Power Set|power set]] of $T$.
Let $\mathbb T$ be a [[Definition:Subset|subset]] of $\powerset T$.
Then:
:$\ds S \cup \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ | === Union Subset of Intersection ===
Let $\ds x \in S \cup \bigcap \mathbb T$.
We need to show that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and then by definition of [[Definition:Subset|subset]] we will have shown that:
:$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb... | Union Distributes over Intersection/General Result/Proof | https://proofwiki.org/wiki/Union_Distributes_over_Intersection/General_Result | https://proofwiki.org/wiki/Union_Distributes_over_Intersection/General_Result/Proof | [
"Union Distributes over Intersection"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset"
] | [
"Definition:Subset",
"Definition:Set Union",
"Definition:Set Union",
"Definition:Set Intersection",
"Definition:Set Union",
"Proof by Cases",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Set Intersection",
"Definition:Set Union",
"Definition:Set Union",
"Proof by Cases",
"... |
proofwiki-5531 | Scalar Multiple of Simple Function is Simple Function | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function, and let $\lambda \in \R$.
Then the pointwise scalar multiple $\lambda f: X \to \R$ of $f$ is also a simple function. | Let $\Img f$ denote the image of $f$.
Let $\Img {\lambda f}$ denote the image of $\lambda f$.
Consider the surjection $l_\lambda: \Img f \to \Img {\lambda f}$ defined by:
:$\map {l_\lambda} {\map f x} := \lambda \map f x$
By Measurable Function is Simple Function iff Finite Image Set, $\card {\Img f}$ is finite.
Hence ... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \R$ be a [[Definition:Simple Function|simple function]], and let $\lambda \in \R$.
Then the [[Definition:Pointwise Scalar Multiplication of Real-Valued Functions|pointwise scalar multiple]] $\lambda f: X \to \R$ of $f$ is ... | Let $\Img f$ denote the [[Definition:Image of Mapping|image]] of $f$.
Let $\Img {\lambda f}$ denote the [[Definition:Image of Mapping|image]] of $\lambda f$.
Consider the [[Definition:Surjection|surjection]] $l_\lambda: \Img f \to \Img {\lambda f}$ defined by:
:$\map {l_\lambda} {\map f x} := \lambda \map f x$
By [... | Scalar Multiple of Simple Function is Simple Function | https://proofwiki.org/wiki/Scalar_Multiple_of_Simple_Function_is_Simple_Function | https://proofwiki.org/wiki/Scalar_Multiple_of_Simple_Function_is_Simple_Function | [
"Simple Functions"
] | [
"Definition:Measurable Space",
"Definition:Simple Function",
"Definition:Pointwise Scalar Multiplication of Mappings/Real-Valued Functions",
"Definition:Simple Function"
] | [
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Surjection",
"Measurable Function is Simple Function iff Finite Image Set",
"Definition:Finite Set",
"Cardinality of Codomain of Surjection",
"Definition:Finite Set",
"Measurable Function is S... |
proofwiki-5532 | Image of Composite Mapping | Let $f: S \to T$ and $g: R \to S$ be mappings.
Then:
:$\Img {f \circ g} = f \sqbrk {\Img g}$
where:
: $f \circ g$ is the composition of $g$ and $f$
: $\operatorname{Img}$ denotes image
: $f \sqbrk \cdot$ denotes taking image of a subset under $f$. | By definition of image:
:$\Img {f \circ g} = \set {t \in T: \exists r \in R: \map {\paren {f \circ g} } r = t}$
and by definition of the image of a subset:
:$f \sqbrk {\Img g} = \set {t \in T: \exists s \in \Img g: \map f s = t}$
which, expanding what it means that $s \in \Img g$, equals:
:$f \sqbrk {\Img g} = \set {t ... | Let $f: S \to T$ and $g: R \to S$ be [[Definition:Mapping|mappings]].
Then:
:$\Img {f \circ g} = f \sqbrk {\Img g}$
where:
: $f \circ g$ is the [[Definition:Composite Mapping|composition]] of $g$ and $f$
: $\operatorname{Img}$ denotes [[Definition:Image of Mapping|image]]
: $f \sqbrk \cdot$ denotes taking [[Definit... | By definition of [[Definition:Image of Mapping|image]]:
:$\Img {f \circ g} = \set {t \in T: \exists r \in R: \map {\paren {f \circ g} } r = t}$
and by definition of the [[Definition:Image of Subset under Mapping|image of a subset]]:
:$f \sqbrk {\Img g} = \set {t \in T: \exists s \in \Img g: \map f s = t}$
which, ex... | Image of Composite Mapping | https://proofwiki.org/wiki/Image_of_Composite_Mapping | https://proofwiki.org/wiki/Image_of_Composite_Mapping | [
"Composite Mappings"
] | [
"Definition:Mapping",
"Definition:Composition of Mappings",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Image (Set Theory)/Mapping/Subset"
] | [
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Non-Empty Set",
"Null Relation is Mapping iff Domain is Empty Set",
"Definition:Empty Mapping",
"Image of Empty Set is Empty Set"
] |
proofwiki-5533 | Scalar Multiple of Integrable Function is Integrable Function | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function, and let $\lambda \in \R$.
Then $\lambda f: X \to \overline \R$, the pointwise $\lambda$-multiple of $f$, is also $\mu$-integrable.
That is, the space of integrable functions $\LL^1_{\overline \R}$ is closed un... | {{proof wanted|standard machinery}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R$ be a [[Definition:Measure-Integrable Function|$\mu$-integrable function]], and let $\lambda \in \R$.
Then $\lambda f: X \to \overline \R$, the [[Definition:Pointwise Scalar Multiplication of Extended Real-Valu... | {{proof wanted|standard machinery}} | Scalar Multiple of Integrable Function is Integrable Function | https://proofwiki.org/wiki/Scalar_Multiple_of_Integrable_Function_is_Integrable_Function | https://proofwiki.org/wiki/Scalar_Multiple_of_Integrable_Function_is_Integrable_Function | [
"Measure-Integrable Functions"
] | [
"Definition:Measure Space",
"Definition:Integrable Function/Measure Space",
"Definition:Pointwise Scalar Multiplication of Extended Real-Valued Functions",
"Definition:Integrable Function/Measure Space",
"Definition:Space of Integrable Functions",
"Definition:Closure (Abstract Algebra)",
"Definition:Poi... | [] |
proofwiki-5534 | Idempotent Elements for Integer Multiplication | There are exactly two integers which are idempotent with respect to multiplication:
:$0 \times 0 = 0$
:$1 \times 1 = 1$ | The integers $\struct {\Z, +, \times}$ form an integral domain.
By definition of integral domain, therefore, the integers form a ring with no (proper) zero divisors.
The result follows from Idempotent Elements of Ring with No Proper Zero Divisors.
{{qed}} | There are exactly two [[Definition:Integer|integers]] which are [[Definition:Idempotent Element|idempotent]] with respect to [[Definition:Integer Multiplication|multiplication]]:
:$0 \times 0 = 0$
:$1 \times 1 = 1$ | The [[Definition:Integer|integers]] $\struct {\Z, +, \times}$ [[Integers form Integral Domain|form an integral domain]].
By definition of [[Definition:Integral Domain|integral domain]], therefore, the [[Definition:Integer|integers]] form a [[Definition:Ring (Abstract Algebra)|ring]] with no [[Definition:Proper Zero Di... | Idempotent Elements for Integer Multiplication | https://proofwiki.org/wiki/Idempotent_Elements_for_Integer_Multiplication | https://proofwiki.org/wiki/Idempotent_Elements_for_Integer_Multiplication | [
"Integer Multiplication"
] | [
"Definition:Integer",
"Definition:Idempotence/Element",
"Definition:Multiplication/Integers"
] | [
"Definition:Integer",
"Integers form Integral Domain",
"Definition:Integral Domain",
"Definition:Integer",
"Definition:Ring (Abstract Algebra)",
"Definition:Proper Zero Divisor",
"Idempotent Elements of Ring with No Proper Zero Divisors"
] |
proofwiki-5535 | Left Distributive Law for Natural Numbers | The operation of multiplication is left distributive over addition on the set of natural numbers $\N$:
:$\forall x, y, n \in \N_{> 0}: n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$ | Let us cast the proposition in the form:
:$\forall a, b, n \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall a, b \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$
=== Basis for the I... | The operation of [[Definition:Natural Number Multiplication|multiplication]] is [[Definition:Left Distributive Operation|left distributive]] over [[Definition:Natural Number Addition|addition]] on the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N$:
:$\forall x, y, n \in \N_{> 0}: n \times... | Let us cast the proposition in the form:
:$\forall a, b, n \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\forall a, b \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \t... | Left Distributive Law for Natural Numbers | https://proofwiki.org/wiki/Left_Distributive_Law_for_Natural_Numbers | https://proofwiki.org/wiki/Left_Distributive_Law_for_Natural_Numbers | [
"Natural Numbers/1-Based"
] | [
"Definition:Multiplication/Natural Numbers",
"Definition:Distributive Operation/Left",
"Definition:Addition/Natural Numbers",
"Definition:Set",
"Definition:Natural Numbers"
] | [
"Definition:Proposition",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Axiom:Axiomatization... |
proofwiki-5536 | Right Distributive Law for Natural Numbers | The operation of multiplication is right distributive over addition on the set of natural numbers $\N_{> 0}$:
:$\forall x, y, n \in \N_{> 0}: \paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$ | For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \paren {a + b} \times 1
| r = a + b
| c = Axiom $\text A$
}}
{{eqn ... | The operation of [[Definition:Natural Number Multiplication|multiplication]] is [[Definition:Right Distributive Operation|right distributive]] over [[Definition:Natural Number Addition|addition]] on the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N_{> 0}$:
:$\forall x, y, n \in \N_{> 0}: ... | For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \paren {a + b} \times 1
| r = a + b
... | Right Distributive Law for Natural Numbers | https://proofwiki.org/wiki/Right_Distributive_Law_for_Natural_Numbers | https://proofwiki.org/wiki/Right_Distributive_Law_for_Natural_Numbers | [
"Natural Numbers/1-Based"
] | [
"Definition:Multiplication/Natural Numbers",
"Definition:Distributive Operation/Right",
"Definition:Addition/Natural Numbers",
"Definition:Set",
"Definition:Natural Numbers"
] | [
"Definition:Proposition",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Right Distributive L... |
proofwiki-5537 | Standard Machinery | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^1_{\overline \R} } \mu$ be the space of $\mu$-integrable functions.
Let $\map P {f_1, \ldots, f_n}$ be a proposition, where the variables $f_i$ denote $\mu$-measurable functions $f_i: X \to \overline \R$.
Let every occurrence of an $f_i$ be of the form:
... | {{proof wanted}}
Category:Proof Techniques
Category:Measure Theory
lf6pm67jgqh6a126f0ew3uujbhppymn | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\map {\LL^1_{\overline \R} } \mu$ be the [[Definition:Space of Integrable Functions|space of $\mu$-integrable functions]].
Let $\map P {f_1, \ldots, f_n}$ be a [[Definition:Proposition|proposition]], where the [[Definition:Variable|v... | {{proof wanted}}
[[Category:Proof Techniques]]
[[Category:Measure Theory]]
lf6pm67jgqh6a126f0ew3uujbhppymn | Standard Machinery | https://proofwiki.org/wiki/Standard_Machinery | https://proofwiki.org/wiki/Standard_Machinery | [
"Proof Techniques",
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Space of Integrable Functions",
"Definition:Proposition",
"Definition:Variable",
"Definition:Measurable Function",
"Definition:Indexing Set",
"Definition:Multilinear Mapping",
"Definition:Set",
"Definition:Characteristic Function (Set Theory)/Set",
"Definiti... | [
"Category:Proof Techniques",
"Category:Measure Theory"
] |
proofwiki-5538 | Third Principle of Mathematical Induction | Let $\map P n$ be a propositional function depending on $n \in \N$.
If:
:$(1): \quad \map P n$ is true for all $n \le d$ for some $d \in \N$
:$(2): \quad \forall m \in \N: \paren {\forall k \in \N, m \le k < m + d: \map P k} \implies \map P {m + d}$
then $\map P n$ is true for all $n \in \N$. | Let $A = \set {n \in \N: \map P n}$.
We show that $A$ is an inductive set.
By $(1)$:
:$\forall 1 \le i \le d: i \in A$
Let:
:$\forall x \ge d: \set {1, 2, \dotsc, x} \subset A$
Then by definition of $A$:
:$\forall k \in \N: x - \paren {d - 1} \le k < x + 1: \map P k$
Thus $\map P {x + 1} \implies x + 1 \in A$
Thus $A$ ... | Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N$.
If:
:$(1): \quad \map P n$ is true for all $n \le d$ for some $d \in \N$
:$(2): \quad \forall m \in \N: \paren {\forall k \in \N, m \le k < m + d: \map P k} \implies \map P {m + d}$
then $\map P n$ is true for a... | Let $A = \set {n \in \N: \map P n}$.
We show that $A$ is an [[Definition:Inductive Set as Subset of Real Numbers|inductive set]].
By $(1)$:
:$\forall 1 \le i \le d: i \in A$
Let:
:$\forall x \ge d: \set {1, 2, \dotsc, x} \subset A$
Then by definition of $A$:
:$\forall k \in \N: x - \paren {d - 1} \le k < x + 1: \ma... | Third Principle of Mathematical Induction | https://proofwiki.org/wiki/Third_Principle_of_Mathematical_Induction | https://proofwiki.org/wiki/Third_Principle_of_Mathematical_Induction | [
"Number Theory",
"Named Theorems",
"Mathematical Induction",
"Proof Techniques"
] | [
"Definition:Propositional Function"
] | [
"Definition:Inductive Set/Subset of Real Numbers"
] |
proofwiki-5539 | Cross-Relation is Equivalence Relation | {{Cross-Relation Context Definition}}
Then $\boxtimes$ is an equivalence relation on $\struct {S_1 \times S_2, \oplus}$. | === Reflexivity ===
:$\forall \tuple {x_1, y_1} \in S_1 \times S_2: x_1 \circ y_1 = x_1 \circ y_1 \implies \tuple {x_1, y_1} \boxtimes \tuple {x_1, y_1}$
So $\boxtimes$ is a reflexive relation.
{{qed|lemma}} | {{Cross-Relation Context Definition}}
Then $\boxtimes$ is an [[Definition:Equivalence Relation|equivalence relation]] on $\struct {S_1 \times S_2, \oplus}$. | === Reflexivity ===
:$\forall \tuple {x_1, y_1} \in S_1 \times S_2: x_1 \circ y_1 = x_1 \circ y_1 \implies \tuple {x_1, y_1} \boxtimes \tuple {x_1, y_1}$
So $\boxtimes$ is a [[Definition:Reflexive Relation|reflexive relation]].
{{qed|lemma}} | Cross-Relation is Equivalence Relation | https://proofwiki.org/wiki/Cross-Relation_is_Equivalence_Relation | https://proofwiki.org/wiki/Cross-Relation_is_Equivalence_Relation | [
"Cross-Relations"
] | [
"Definition:Equivalence Relation"
] | [
"Definition:Reflexive Relation"
] |
proofwiki-5540 | Cross-Relation on Natural Numbers is Equivalence Relation | Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
The relation $\boxtimes$ defined on $\N \times \N$ by:
:$\tuple {x_1... | $\boxtimes$ is an instance of a cross-relation.
We also have that Natural Number Addition is Commutative.
The result therefore follows from Cross-Relation is Equivalence Relation.
{{qed}} | Let $\struct {\N, +}$ be the [[Natural Numbers under Addition form Commutative Semigroup|semigroup of natural numbers under addition]].
Let $\struct {\N \times \N, \oplus}$ be the [[Definition:External Direct Product|(external) direct product]] of $\struct {\N, +}$ with itself, where $\oplus$ is the [[Definition:Opera... | $\boxtimes$ is an instance of a [[Definition:Cross-Relation|cross-relation]].
We also have that [[Natural Number Addition is Commutative]].
The result therefore follows from [[Cross-Relation is Equivalence Relation]].
{{qed}} | Cross-Relation on Natural Numbers is Equivalence Relation | https://proofwiki.org/wiki/Cross-Relation_on_Natural_Numbers_is_Equivalence_Relation | https://proofwiki.org/wiki/Cross-Relation_on_Natural_Numbers_is_Equivalence_Relation | [
"Natural Numbers",
"Examples of Equivalence Relations",
"Cross-Relations"
] | [
"Natural Numbers under Addition form Commutative Semigroup",
"Definition:External Direct Product",
"Definition:Operation Induced by Direct Product",
"Definition:Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Cross-Relation",
"Natural Number Addition is Commutative",
"Cross-Relation is Equivalence Relation"
] |
proofwiki-5541 | Equivalence Classes of Cross-Relation on Natural Numbers | Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:
... | We have that $\struct {\N, +}$ is a commutative semigroup from Natural Number Addition is Commutative.
We also have the result Cross-Relation on Natural Numbers is Equivalence Relation.
The result follows from the definition of equivalence class.
{{qed}} | Let $\struct {\N, +}$ be the [[Natural Numbers under Addition form Commutative Semigroup|semigroup of natural numbers under addition]].
Let $\struct {\N \times \N, \oplus}$ be the [[Definition:External Direct Product|(external) direct product]] of $\struct {\N, +}$ with itself, where $\oplus$ is the [[Definition:Opera... | We have that $\struct {\N, +}$ is a [[Definition:Commutative Semigroup|commutative semigroup]] from [[Natural Number Addition is Commutative]].
We also have the result [[Cross-Relation on Natural Numbers is Equivalence Relation]].
The result follows from the definition of [[Definition:Equivalence Class|equivalence cl... | Equivalence Classes of Cross-Relation on Natural Numbers | https://proofwiki.org/wiki/Equivalence_Classes_of_Cross-Relation_on_Natural_Numbers | https://proofwiki.org/wiki/Equivalence_Classes_of_Cross-Relation_on_Natural_Numbers | [
"Natural Numbers",
"Cartesian Product",
"Examples of Equivalence Classes"
] | [
"Natural Numbers under Addition form Commutative Semigroup",
"Definition:External Direct Product",
"Definition:Operation Induced by Direct Product",
"Definition:Cross-Relation",
"Definition:Equivalence Class",
"Definition:Equivalence Class"
] | [
"Definition:Commutative Semigroup",
"Natural Number Addition is Commutative",
"Cross-Relation on Natural Numbers is Equivalence Relation",
"Definition:Equivalence Class"
] |
proofwiki-5542 | Integer Addition is Well-Defined | Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:
... | Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.
Then:
{{begin-eqn}}
{{eqn | l = \eqclass {a_1, b_1} {}
| r = \e... | Let $\struct {\N, +}$ be the [[Natural Numbers under Addition form Commutative Semigroup|semigroup of natural numbers under addition]].
Let $\struct {\N \times \N, \oplus}$ be the [[Definition:External Direct Product|(external) direct product]] of $\struct {\N, +}$ with itself, where $\oplus$ is the [[Definition:Opera... | Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be [[Definition:Equivalence Class|$\boxtimes$-equivalence classes]] such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.
Then:
{{begin-eqn}}
{{eqn | l = \e... | Integer Addition is Well-Defined/Proof 1 | https://proofwiki.org/wiki/Integer_Addition_is_Well-Defined | https://proofwiki.org/wiki/Integer_Addition_is_Well-Defined/Proof_1 | [
"Integer Addition",
"Integer Addition is Well-Defined"
] | [
"Natural Numbers under Addition form Commutative Semigroup",
"Definition:External Direct Product",
"Definition:Operation Induced by Direct Product",
"Definition:Cross-Relation",
"Equivalence Classes of Cross-Relation on Natural Numbers",
"Definition:Quotient Set",
"Definition:Equivalence Class",
"Defi... | [
"Definition:Equivalence Class",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Commutative/Operation",
"Definition:Associative Operation"
] |
proofwiki-5543 | Multiplication of Cross-Relation Equivalence Classes on Natural Numbers is Well-Defined | Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:
... | Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.
Then:
{{begin-eqn}}
{{eqn | l = \eqclass {a_1, b_1} {}
| r = \e... | Let $\struct {\N, +}$ be the [[Natural Numbers under Addition form Commutative Semigroup|semigroup of natural numbers under addition]].
Let $\struct {\N \times \N, \oplus}$ be the [[Definition:External Direct Product|(external) direct product]] of $\struct {\N, +}$ with itself, where $\oplus$ is the [[Definition:Opera... | Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be [[Definition:Equivalence Class|$\boxtimes$-equivalence classes]] such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.
Then:
{{begin-eqn}}
{{eqn | l = \e... | Multiplication of Cross-Relation Equivalence Classes on Natural Numbers is Well-Defined | https://proofwiki.org/wiki/Multiplication_of_Cross-Relation_Equivalence_Classes_on_Natural_Numbers_is_Well-Defined | https://proofwiki.org/wiki/Multiplication_of_Cross-Relation_Equivalence_Classes_on_Natural_Numbers_is_Well-Defined | [
"Integers",
"Multiplication",
"Examples of Well-Defined Mappings"
] | [
"Natural Numbers under Addition form Commutative Semigroup",
"Definition:External Direct Product",
"Definition:Operation Induced by Direct Product",
"Definition:Cross-Relation",
"Equivalence Classes of Cross-Relation on Natural Numbers",
"Definition:Operation/Binary Operation",
"Definition:Equivalence C... | [
"Definition:Equivalence Class",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Multiplication/Integers",
"Definition:Multiplication/Integers",
"Definition:Well-Defined/Operation"
] |
proofwiki-5544 | Addition of Cross-Relation Equivalence Classes on Natural Numbers is Cancellable | Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:
... | {{begin-eqn}}
{{eqn | l = \eqclass {a, b} {} \oplus \eqclass {c_1, d_1} {}
| r = \eqclass {a, b} {} \oplus \eqclass {c_2, d_2} {}
| c =
}}
{{eqn | ll= \leadsto
| l = \eqclass {b, a} {} \oplus \paren {\eqclass {a, b} {} \oplus \eqclass {c_1, d_1} {} }
| r = \eqclass {b, a} {} \oplus \paren {\eqc... | Let $\struct {\N, +}$ be the [[Natural Numbers under Addition form Commutative Semigroup|semigroup of natural numbers under addition]].
Let $\struct {\N \times \N, \oplus}$ be the [[Definition:External Direct Product|(external) direct product]] of $\struct {\N, +}$ with itself, where $\oplus$ is the [[Definition:Opera... | {{begin-eqn}}
{{eqn | l = \eqclass {a, b} {} \oplus \eqclass {c_1, d_1} {}
| r = \eqclass {a, b} {} \oplus \eqclass {c_2, d_2} {}
| c =
}}
{{eqn | ll= \leadsto
| l = \eqclass {b, a} {} \oplus \paren {\eqclass {a, b} {} \oplus \eqclass {c_1, d_1} {} }
| r = \eqclass {b, a} {} \oplus \paren {\eqc... | Addition of Cross-Relation Equivalence Classes on Natural Numbers is Cancellable | https://proofwiki.org/wiki/Addition_of_Cross-Relation_Equivalence_Classes_on_Natural_Numbers_is_Cancellable | https://proofwiki.org/wiki/Addition_of_Cross-Relation_Equivalence_Classes_on_Natural_Numbers_is_Cancellable | [
"Natural Number Addition",
"Cross-Relations",
"Equivalence Relations"
] | [
"Natural Numbers under Addition form Commutative Semigroup",
"Definition:External Direct Product",
"Definition:Operation Induced by Direct Product",
"Definition:Cross-Relation",
"Equivalence Classes of Cross-Relation on Natural Numbers",
"Definition:Quotient Set",
"Definition:Equivalence Class",
"Defi... | [
"Integer Addition is Associative",
"Identity for Addition of Cross-Relation Equivalence Classes on Natural Numbers"
] |
proofwiki-5545 | Cross-Relation Equivalence Classes on Natural Numbers are Cancellable for Addition | Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:
... | Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.
Then:
{{begin-eqn}}
{{eqn | l = \eqclass {a_1, b_1} {}
| r = \e... | Let $\struct {\N, +}$ be the [[Natural Numbers under Addition form Commutative Semigroup|semigroup of natural numbers under addition]].
Let $\struct {\N \times \N, \oplus}$ be the [[Definition:External Direct Product|(external) direct product]] of $\struct {\N, +}$ with itself, where $\oplus$ is the [[Definition:Opera... | Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be [[Definition:Equivalence Class|$\boxtimes$-equivalence classes]] such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.
Then:
{{begin-eqn}}
{{eqn | l = \e... | Cross-Relation Equivalence Classes on Natural Numbers are Cancellable for Addition | https://proofwiki.org/wiki/Cross-Relation_Equivalence_Classes_on_Natural_Numbers_are_Cancellable_for_Addition | https://proofwiki.org/wiki/Cross-Relation_Equivalence_Classes_on_Natural_Numbers_are_Cancellable_for_Addition | [
"Natural Number Addition",
"Cartesian Product",
"Equivalence Relations"
] | [
"Natural Numbers under Addition form Commutative Semigroup",
"Definition:External Direct Product",
"Definition:Operation Induced by Direct Product",
"Definition:Cross-Relation",
"Equivalence Classes of Cross-Relation on Natural Numbers",
"Definition:Quotient Set",
"Definition:Equivalence Class",
"Defi... | [
"Definition:Equivalence Class",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Commutative/Operation",
"Definition:Associative Operation"
] |
proofwiki-5546 | Cross-Relation is Congruence Relation | {{Cross-Relation Context Definition}}
The cross-relation $\boxtimes$ is a congruence relation on $\struct {S_1 \times S_2, \oplus}$. | From Cross-Relation is Equivalence Relation we have that $\boxtimes$ is an equivalence relation.
We now need to show that:
{{begin-eqn}}
{{eqn | l = \tuple {x_1, y_1}
| o = \boxtimes
| r = \tuple {x_2, y_2}
| c =
}}
{{eqn | lo= \land
| l = \tuple {u_1, v_1}
| o = \boxtimes
| r = \tu... | {{Cross-Relation Context Definition}}
The [[Definition:Cross-Relation|cross-relation]] $\boxtimes$ is a [[Definition:Congruence Relation|congruence relation]] on $\struct {S_1 \times S_2, \oplus}$. | From [[Cross-Relation is Equivalence Relation]] we have that $\boxtimes$ is an [[Definition:Equivalence Relation|equivalence relation]].
We now need to show that:
{{begin-eqn}}
{{eqn | l = \tuple {x_1, y_1}
| o = \boxtimes
| r = \tuple {x_2, y_2}
| c =
}}
{{eqn | lo= \land
| l = \tuple {u_1,... | Cross-Relation is Congruence Relation | https://proofwiki.org/wiki/Cross-Relation_is_Congruence_Relation | https://proofwiki.org/wiki/Cross-Relation_is_Congruence_Relation | [
"Cross-Relations"
] | [
"Definition:Cross-Relation",
"Definition:Congruence Relation"
] | [
"Cross-Relation is Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Commutative/Operation",
"Definition:Commutative/Operation",
"Definition:Commutative/Operation",
"Definition:Congruence Relation"
] |
proofwiki-5547 | Elements of Cross-Relation Equivalence Class | {{Cross-Relation Context Definition}}
Let $\eqclass {\tuple {x, y} } \boxtimes$ be the $\boxtimes$-equivalence class of $\tuple {x, y}$, where $\tuple {x, y} \in S_1 \times S_2$.
Then:
$\forall x, y \in S_1, a, b \in S_2:$
:$(1): \quad \eqclass {\tuple {x \circ a, a} } \boxtimes = \eqclass {\tuple {y \circ b, b} } \box... | {{begin-eqn}}
{{eqn | n = 1
| l = \eqclass {\tuple {x \circ a, a} } \boxtimes
| r = \eqclass {\tuple {y \circ b, b} } \boxtimes
| c =
}}
{{eqn | l = \tuple {x \circ a, a}
| o = \boxtimes
| r = \tuple {y \circ b, b}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \circ a \circ b
... | {{Cross-Relation Context Definition}}
Let $\eqclass {\tuple {x, y} } \boxtimes$ be the [[Definition:Equivalence Class|$\boxtimes$-equivalence class of $\tuple {x, y}$]], where $\tuple {x, y} \in S_1 \times S_2$.
Then:
$\forall x, y \in S_1, a, b \in S_2:$
:$(1): \quad \eqclass {\tuple {x \circ a, a} } \boxtimes = ... | {{begin-eqn}}
{{eqn | n = 1
| l = \eqclass {\tuple {x \circ a, a} } \boxtimes
| r = \eqclass {\tuple {y \circ b, b} } \boxtimes
| c =
}}
{{eqn | l = \tuple {x \circ a, a}
| o = \boxtimes
| r = \tuple {y \circ b, b}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \circ a \circ b
... | Elements of Cross-Relation Equivalence Class | https://proofwiki.org/wiki/Elements_of_Cross-Relation_Equivalence_Class | https://proofwiki.org/wiki/Elements_of_Cross-Relation_Equivalence_Class | [
"Cross-Relations"
] | [
"Definition:Equivalence Class"
] | [
"Definition:Commutative/Operation",
"Definition:Cancellable Element",
"Definition:Associative Operation"
] |
proofwiki-5548 | Equivalence Class of Equal Elements of Cross-Relation | {{Cross-Relation Context Definition}}
Then:
:$\forall c, d \in S_1 \cap S_2: \tuple {c, c} \boxtimes \tuple {d, d}$ | Note that in order for $\tuple {c, c}$ and $\tuple {d, d}$ to be defined, $c$ and $d$ must be in both $S_1$ and $S_2$.
Hence the restriction given:
:$\forall c, d \in S_1 \cap S_2$
Then:
{{begin-eqn}}
{{eqn | q = \forall c, d \in S_1 \cap S_2
| l = c \circ d
| r = d \circ c
| c = Commutativity of $\ci... | {{Cross-Relation Context Definition}}
Then:
:$\forall c, d \in S_1 \cap S_2: \tuple {c, c} \boxtimes \tuple {d, d}$ | Note that in order for $\tuple {c, c}$ and $\tuple {d, d}$ to be defined, $c$ and $d$ must be in both $S_1$ and $S_2$.
Hence the restriction given:
:$\forall c, d \in S_1 \cap S_2$
Then:
{{begin-eqn}}
{{eqn | q = \forall c, d \in S_1 \cap S_2
| l = c \circ d
| r = d \circ c
| c = [[Definition:Commu... | Equivalence Class of Equal Elements of Cross-Relation | https://proofwiki.org/wiki/Equivalence_Class_of_Equal_Elements_of_Cross-Relation | https://proofwiki.org/wiki/Equivalence_Class_of_Equal_Elements_of_Cross-Relation | [
"Cross-Relations",
"Examples of Equivalence Classes"
] | [] | [
"Definition:Commutative/Operation"
] |
proofwiki-5549 | Length of Concatenation | Let $S$ and $T$ be words, and let $ST$ be their concatenation.
Then:
:$\size {S T}\ = \size S + \size T$
where $\size S$ denotes the length of $S$. | Because of the unique readability of $ST$, we can determine for each symbol $s$ that is part of $S T$, whether:
:$s$ is part of $S$
:$s$ is part of $T$
and furthermore, precisely one of these options occurs.
There are $\size S$ symbols in $S$, and $\size T$ symbols in $T$.
In total, then, $S T$ is seen to consist of $\... | Let $S$ and $T$ be [[Definition:Word (Formal Systems)|words]], and let $ST$ be their [[Definition:Concatenation (Formal Systems)|concatenation]].
Then:
:$\size {S T}\ = \size S + \size T$
where $\size S$ denotes the [[Definition:Length of String|length]] of $S$. | Because of the [[Definition:Unique Readability|unique readability]] of $ST$, we can determine for each [[Definition:Symbol|symbol]] $s$ that is part of $S T$, whether:
:$s$ is part of $S$
:$s$ is part of $T$
and furthermore, precisely one of these options occurs.
There are $\size S$ [[Definition:Symbol|symbols]] in ... | Length of Concatenation | https://proofwiki.org/wiki/Length_of_Concatenation | https://proofwiki.org/wiki/Length_of_Concatenation | [
"Collations"
] | [
"Definition:Word (Formal Systems)",
"Definition:Concatenation (Formal Systems)",
"Definition:Length of String"
] | [
"Definition:Collation/Unique Readability",
"Definition:Symbol",
"Definition:Symbol",
"Definition:Symbol",
"Definition:Symbol",
"Category:Collations"
] |
proofwiki-5550 | Preimage of Serial Relation is Domain | Let $\RR$ be a serial relation on $S$.
Then the preimage of $\RR$ is $S$ (the domain of $\RR$). | {{begin-eqn}}
{{eqn | l = S
| o = \supseteq
| r = \Preimg \RR
| c = {{Defof|Preimage of Relation}}
}}
{{eqn | q = \forall s \in S: \exists t \in S
| l = \tuple {s, t}
| o = \in
| r = \RR
| c = {{Defof|Serial Relation}}
}}
{{eqn | ll= \leadsto
| q = \forall s \in S
|... | Let $\RR$ be a [[Definition:Serial Relation|serial relation]] on $S$.
Then the [[Definition:Preimage of Relation|preimage]] of $\RR$ is $S$ (the [[Definition:Domain of Relation|domain]] of $\RR$). | {{begin-eqn}}
{{eqn | l = S
| o = \supseteq
| r = \Preimg \RR
| c = {{Defof|Preimage of Relation}}
}}
{{eqn | q = \forall s \in S: \exists t \in S
| l = \tuple {s, t}
| o = \in
| r = \RR
| c = {{Defof|Serial Relation}}
}}
{{eqn | ll= \leadsto
| q = \forall s \in S
|... | Preimage of Serial Relation is Domain | https://proofwiki.org/wiki/Preimage_of_Serial_Relation_is_Domain | https://proofwiki.org/wiki/Preimage_of_Serial_Relation_is_Domain | [
"Serial Relations"
] | [
"Definition:Serial Relation",
"Definition:Preimage/Relation/Relation",
"Definition:Domain (Set Theory)/Relation"
] | [
"Category:Serial Relations"
] |
proofwiki-5551 | Relation is Connected and Reflexive iff Total | Let $S$ be a set.
Let $\RR$ be a relation on $S$.
Then:
:$\RR$ is both a connected relation and a reflexive relation
{{iff}}:
:$\RR$ is a total relation. | === Necessary Condition ===
Let $\RR$ be a relation on $S$ which is both connected and reflexive.
Let $\tuple {a, b} \in S \times S$.
Suppose $a = b$.
Then as $\RR$ is reflexive, $\tuple {a, b} \in \RR$.
Suppose $a \ne b$.
Then as $\RR$ is connected, $\tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$.
That is:
:$\foral... | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be a [[Definition:Relation|relation]] on $S$.
Then:
:$\RR$ is both a [[Definition:Connected Relation|connected relation]] and a [[Definition:Reflexive Relation|reflexive relation]]
{{iff}}:
:$\RR$ is a [[Definition:Total Relation|total relation]]. | === Necessary Condition ===
Let $\RR$ be a [[Definition:Relation|relation]] on $S$ which is both [[Definition:Connected Relation|connected]] and [[Definition:Reflexive Relation|reflexive]].
Let $\tuple {a, b} \in S \times S$.
Suppose $a = b$.
Then as $\RR$ is [[Definition:Reflexive Relation|reflexive]], $\tuple {a,... | Relation is Connected and Reflexive iff Total | https://proofwiki.org/wiki/Relation_is_Connected_and_Reflexive_iff_Total | https://proofwiki.org/wiki/Relation_is_Connected_and_Reflexive_iff_Total | [
"Total Relations",
"Reflexive Relations",
"Connected Relations"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Connected Relation",
"Definition:Reflexive Relation",
"Definition:Total Relation"
] | [
"Definition:Relation",
"Definition:Connected Relation",
"Definition:Reflexive Relation",
"Definition:Reflexive Relation",
"Definition:Connected Relation",
"Definition:Total Relation",
"Definition:Total Relation",
"Definition:Reflexive Relation",
"Definition:Connected Relation"
] |
proofwiki-5552 | Total Ordering is Total Relation | Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a total ordering.
Then $\RR$ is a total relation. | By definition of total ordering:
:$\RR$ is a reflexive relation on the strength of being an ordering
:$\RR$ is a connected relation on the strength of being a total ordering.
{{qed}}
Category:Total Orderings
Category:Total Relations
ks8n1nv30vxvjpdtoogegijgsgy65gf | Let $S$ be a [[Definition:Set|set]].
Let $\RR \subseteq S \times S$ be a [[Definition:Total Ordering|total ordering]].
Then $\RR$ is a [[Definition:Total Relation|total relation]]. | By definition of [[Definition:Total Ordering|total ordering]]:
:$\RR$ is a [[Definition:Reflexive Relation|reflexive relation]] on the strength of being an [[Definition:Ordering|ordering]]
:$\RR$ is a [[Definition:Connected Relation|connected relation]] on the strength of being a [[Definition:Total Ordering|total orde... | Total Ordering is Total Relation | https://proofwiki.org/wiki/Total_Ordering_is_Total_Relation | https://proofwiki.org/wiki/Total_Ordering_is_Total_Relation | [
"Total Orderings",
"Total Relations"
] | [
"Definition:Set",
"Definition:Total Ordering",
"Definition:Total Relation"
] | [
"Definition:Total Ordering",
"Definition:Reflexive Relation",
"Definition:Ordering",
"Definition:Connected Relation",
"Definition:Total Ordering",
"Category:Total Orderings",
"Category:Total Relations"
] |
proofwiki-5553 | Natural Number Addition Commutativity with Successor/Proof 2 | Let $\N$ be the natural numbers.
Then:
:$\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$ | Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Proof by induction:
From Axiomatization of $1$-Based Natural Numbers, we have by definition that:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
| l = m + 0
| r = m
}}
{{eqn | l = \paren {m + n}^+
| r = m + n^+
}}
{{end-eq... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Then:
:$\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$ | Using the [[Axiom:Axiomatization of 1-Based Natural Numbers|following axioms]]:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Proof by [[Principle of Mathematical Induction|induction]]:
From [[Axiom:Axiomatization of 1-Based Natural Numbers|Axiomatization of $1$-Based Natural Numbers]], we have by definition ... | Natural Number Addition Commutativity with Successor/Proof 2 | https://proofwiki.org/wiki/Natural_Number_Addition_Commutativity_with_Successor/Proof_2 | https://proofwiki.org/wiki/Natural_Number_Addition_Commutativity_with_Successor/Proof_2 | [
"Natural Number Addition Commutativity with Successor"
] | [
"Definition:Natural Numbers"
] | [
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Principle of Mathematical Induction",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Natural Number Addition is Assoc... |
proofwiki-5554 | Natural Number Commutes with 1 under Addition | Let $n \in \N_{> 0}$ be a natural number.
Then $n$ commutes with $1$ under the operation of addition:
:$\forall n \in \N_{> 0}: n + 1 = 1 + n$ | Using the axiomatization:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$n + 1 = 1 + n$ | Let $n \in \N_{> 0}$ be a [[Definition:Natural Number|natural number]].
Then $n$ [[Definition:Commute|commutes]] with $1$ under the [[Definition:Binary Operation|operation]] of [[Definition:Natural Number Addition|addition]]:
:$\forall n \in \N_{> 0}: n + 1 = 1 + n$ | Using the [[Axiom:Axiomatization of 1-Based Natural Numbers|axiomatization]]:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$n + 1 = 1 + n$ | Natural Number Commutes with 1 under Addition | https://proofwiki.org/wiki/Natural_Number_Commutes_with_1_under_Addition | https://proofwiki.org/wiki/Natural_Number_Commutes_with_1_under_Addition | [
"Natural Numbers/1-Based",
"Natural Number Addition"
] | [
"Definition:Natural Numbers",
"Definition:Commutative/Elements",
"Definition:Operation/Binary Operation",
"Definition:Addition/Natural Numbers"
] | [
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Proposition",
"Axiom:Axiomatization of 1-Based Natural Numbers"
] |
proofwiki-5555 | Natural Number Addition is Commutative/Proof 3 | The operation of addition on the set of natural numbers $\N_{> 0}$ is commutative:
:$\forall x, y \in \N_{> 0}: x + y = y + x$ | Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Let $x \in \N_{> 0}$ be arbitrary.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$x + n = n + x$
=== Basis for the Induction ===
From Natural Number Commutes with 1 under Addition, we have that:
:$\forall x \in \N_{> 0}: x ... | The operation of [[Definition:Natural Number Addition|addition]] on the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N_{> 0}$ is [[Definition:Commutative Operation|commutative]]:
:$\forall x, y \in \N_{> 0}: x + y = y + x$ | Using the [[Axiom:Axiomatization of 1-Based Natural Numbers|following axioms]]:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Let $x \in \N_{> 0}$ be arbitrary.
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$x + n = n + x$
=== Basis for the Induction ===
From [[Nat... | Natural Number Addition is Commutative/Proof 3 | https://proofwiki.org/wiki/Natural_Number_Addition_is_Commutative/Proof_3 | https://proofwiki.org/wiki/Natural_Number_Addition_is_Commutative/Proof_3 | [
"Natural Numbers/1-Based",
"Natural Number Addition is Commutative"
] | [
"Definition:Addition/Natural Numbers",
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Commutative/Operation"
] | [
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Proposition",
"Natural Number Commutes with 1 under Addition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Natural Number Addition is Associative",
"Natural Number Addition is Comm... |
proofwiki-5556 | Ordering on Natural Numbers is Trichotomy | Let $\N$ be the natural numbers.
Let $<$ be the (strict) ordering on $\N$.
Then exactly one of the following is true:
:$(1): \quad a = b$
:$(2): \quad a > b$
:$(3): \quad a < b$
That is, $<$ is a trichotomy on $\N$. | Applying the definition of $<$, the theorem becomes:
Exactly one of the following is true:
:$(1): \quad a = b$
:$(2): \quad \exists n \in \N_{>0} : b + n = a$
:$(3): \quad \exists n \in \N_{>0} : a + n = b$
We will use the principle of Mathematical Induction.
Let $P \left({a}\right)$ be the proposition that exactly one... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $<$ be the [[Definition:Ordering on Natural Numbers|(strict) ordering on $\N$]].
Then exactly one of the following is true:
:$(1): \quad a = b$
:$(2): \quad a > b$
:$(3): \quad a < b$
That is, $<$ is a [[Definition:Trichotomy|trichotomy]] on $\N$. | Applying the definition of $<$, the theorem becomes:
Exactly one of the following is true:
:$(1): \quad a = b$
:$(2): \quad \exists n \in \N_{>0} : b + n = a$
:$(3): \quad \exists n \in \N_{>0} : a + n = b$
We will use the [[Principle of Mathematical Induction|principle of Mathematical Induction]].
Let $P \left({a... | Ordering on Natural Numbers is Trichotomy | https://proofwiki.org/wiki/Ordering_on_Natural_Numbers_is_Trichotomy | https://proofwiki.org/wiki/Ordering_on_Natural_Numbers_is_Trichotomy | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Ordering on Natural Numbers",
"Definition:Trichotomy"
] | [
"Principle of Mathematical Induction"
] |
proofwiki-5557 | Ordering on 1-Based Natural Numbers is Transitive | Let $\N_{> 0}$ be the $1$-based natural numbers.
Let $<$ be the strict ordering on $\N_{>0}$.
Then $<$ is a transitive relation. | Let $a < b$ and $b < c$.
Then:
{{begin-eqn}}
{{eqn | q = \exists x \in \N_{> 0}
| l = a + x
| r = b
| c =
}}
{{eqn | ll= \land
| q = \exists y \in \N_{> 0}
| l = b + y
| r = c
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {a + x} + y
| r = c
| c = substituting ... | Let $\N_{> 0}$ be the [[Definition:1-Based Natural Numbers|$1$-based natural numbers]].
Let $<$ be the [[Definition:Ordering on 1-Based Natural Numbers|strict ordering on $\N_{>0}$]].
Then $<$ is a [[Definition:Transitive Relation|transitive relation]]. | Let $a < b$ and $b < c$.
Then:
{{begin-eqn}}
{{eqn | q = \exists x \in \N_{> 0}
| l = a + x
| r = b
| c =
}}
{{eqn | ll= \land
| q = \exists y \in \N_{> 0}
| l = b + y
| r = c
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {a + x} + y
| r = c
| c = substitutin... | Ordering on 1-Based Natural Numbers is Transitive | https://proofwiki.org/wiki/Ordering_on_1-Based_Natural_Numbers_is_Transitive | https://proofwiki.org/wiki/Ordering_on_1-Based_Natural_Numbers_is_Transitive | [
"Natural Numbers/1-Based"
] | [
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Ordering on Natural Numbers/1-Based",
"Definition:Transitive Relation"
] | [
"Natural Number Addition is Associative"
] |
proofwiki-5558 | Trichotomy is Antireflexive | Let $\RR$ be a trichotomy.
Then $\RR$ is an antireflexive relation. | Let $\RR$ be a trichotomy on a set $S$.
Let $x \in S$.
By definition of a trichotomy, for all $a, b \in S$, either:
:$a \mathrel \RR b$
:$a = b$
:$b \mathrel \RR a$
As $x = x$ it follows directly that $x \not < x$.
Hence the result by definition of antireflexive relation.
{{qed}}
Category:Antireflexive Relations
Catego... | Let $\RR$ be a [[Definition:Trichotomy|trichotomy]].
Then $\RR$ is an [[Definition:Antireflexive Relation|antireflexive relation]]. | Let $\RR$ be a [[Definition:Trichotomy|trichotomy]] on a [[Definition:Set|set]] $S$.
Let $x \in S$.
By definition of a [[Definition:Trichotomy|trichotomy]], for all $a, b \in S$, either:
:$a \mathrel \RR b$
:$a = b$
:$b \mathrel \RR a$
As $x = x$ it follows directly that $x \not < x$.
Hence the result by definitio... | Trichotomy is Antireflexive | https://proofwiki.org/wiki/Trichotomy_is_Antireflexive | https://proofwiki.org/wiki/Trichotomy_is_Antireflexive | [
"Antireflexive Relations",
"Trichotomies"
] | [
"Definition:Trichotomy",
"Definition:Antireflexive Relation"
] | [
"Definition:Trichotomy",
"Definition:Set",
"Definition:Trichotomy",
"Definition:Antireflexive Relation",
"Category:Antireflexive Relations",
"Category:Trichotomies"
] |
proofwiki-5559 | Ordering on 1-Based Natural Numbers is Total Ordering | Let $\N_{> 0}$ be the $1$-based natural numbers.
Let $<$ be the strict ordering on $\N_{>0}$.
Then $<$ is a (strict) total ordering. | From Ordering on $1$-Based Natural Numbers is Trichotomy we have that $<$ is trichotomy.
From Ordering on $1$-Based Natural Numbers is Transitive we have that $<$ is transitive.
From Trichotomy is Antireflexive it follows that $<$ is antireflexive.
It follows by definition that $<$ is a strict ordering.
By the trichoto... | Let $\N_{> 0}$ be the [[Definition:1-Based Natural Numbers|$1$-based natural numbers]].
Let $<$ be the [[Definition:Ordering on 1-Based Natural Numbers|strict ordering on $\N_{>0}$]].
Then $<$ is a [[Definition:Strict Total Ordering|(strict) total ordering]]. | From [[Ordering on 1-Based Natural Numbers is Trichotomy|Ordering on $1$-Based Natural Numbers is Trichotomy]] we have that $<$ is [[Definition:Trichotomy|trichotomy]].
From [[Ordering on 1-Based Natural Numbers is Transitive|Ordering on $1$-Based Natural Numbers is Transitive]] we have that $<$ is [[Definition:Transi... | Ordering on 1-Based Natural Numbers is Total Ordering | https://proofwiki.org/wiki/Ordering_on_1-Based_Natural_Numbers_is_Total_Ordering | https://proofwiki.org/wiki/Ordering_on_1-Based_Natural_Numbers_is_Total_Ordering | [
"Natural Numbers/1-Based"
] | [
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Ordering on Natural Numbers/1-Based",
"Definition:Strict Total Ordering"
] | [
"Ordering on 1-Based Natural Numbers is Trichotomy",
"Definition:Trichotomy",
"Ordering on 1-Based Natural Numbers is Transitive",
"Definition:Transitive Relation",
"Trichotomy is Antireflexive",
"Definition:Antireflexive Relation",
"Definition:Strict Ordering",
"Trichotomy Law (Ordering)",
"Definit... |
proofwiki-5560 | Left Cancellable Commutative Operation is Right Cancellable | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be left cancellable and also commutative.
Then $\circ$ is also right cancellable. | Let $\circ$ be both left cancellable and commutative on a set $S$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ c
| r = b \circ c
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ a
| r = c \circ b
| c = $\circ$ is Commutative
}}
{{eqn | ll= \leadsto
| l = a
| r = b
| c = $\circ$ i... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\circ$ be [[Definition:Left Cancellable Operation|left cancellable]] and also [[Definition:Commutative Operation|commutative]].
Then $\circ$ is also [[Definition:Right Cancellable Operation|right cancellabl... | Let $\circ$ be both [[Definition:Left Cancellable Operation|left cancellable]] and [[Definition:Commutative Operation|commutative]] on a [[Definition:Set|set]] $S$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ c
| r = b \circ c
| c =
}}
{{eqn | ll= \leadsto
| l = c \circ a
| r = c \circ b
| ... | Left Cancellable Commutative Operation is Right Cancellable | https://proofwiki.org/wiki/Left_Cancellable_Commutative_Operation_is_Right_Cancellable | https://proofwiki.org/wiki/Left_Cancellable_Commutative_Operation_is_Right_Cancellable | [
"Commutativity",
"Cancellability"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Left Cancellable Operation",
"Definition:Commutative/Operation",
"Definition:Right Cancellable Operation"
] | [
"Definition:Left Cancellable Operation",
"Definition:Commutative/Operation",
"Definition:Set",
"Definition:Commutative/Operation",
"Definition:Left Cancellable Operation"
] |
proofwiki-5561 | Right Cancellable Commutative Operation is Left Cancellable | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be right cancellable and also commutative.
Then $\circ$ is also left cancellable. | Let $\circ$ be both right cancellable and commutative on a set $S$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
| r = a \circ c
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ a
| r = c \circ a
| c = $\circ$ is Commutative
}}
{{eqn | ll= \leadsto
| l = b
| r = c
| c = $\circ$ ... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\circ$ be [[Definition:Right Cancellable Operation|right cancellable]] and also [[Definition:Commutative Operation|commutative]].
Then $\circ$ is also [[Definition:Left Cancellable Operation|left cancellabl... | Let $\circ$ be both [[Definition:Right Cancellable Operation|right cancellable]] and [[Definition:Commutative Operation|commutative]] on a [[Definition:Set|set]] $S$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
| r = a \circ c
| c =
}}
{{eqn | ll= \leadsto
| l = b \circ a
| r = c \circ a
... | Right Cancellable Commutative Operation is Left Cancellable | https://proofwiki.org/wiki/Right_Cancellable_Commutative_Operation_is_Left_Cancellable | https://proofwiki.org/wiki/Right_Cancellable_Commutative_Operation_is_Left_Cancellable | [
"Commutativity",
"Cancellability"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Right Cancellable Operation",
"Definition:Commutative/Operation",
"Definition:Left Cancellable Operation"
] | [
"Definition:Right Cancellable Operation",
"Definition:Commutative/Operation",
"Definition:Set",
"Definition:Commutative/Operation",
"Definition:Right Cancellable Operation"
] |
proofwiki-5562 | Natural Number Addition is Cancellable | Let $\N$ be the natural numbers.
Let $+$ be addition on $\N$.
Then:
:$\forall a, b, c \in \N: a + c = b + c \implies a = b$
:$\forall a, b, c \in \N: a + b = a + c \implies b = c$
That is, $+$ is cancellable on $\N$. | Consider the natural numbers $\N$ defined as a naturally ordered semigroup $\struct {\N, +, \le}$.
By {{NOSAxiom|2}}, every element of $\struct {\N, +}$ is cancellable.
{{qed}} | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $+$ be [[Definition:Natural Number Addition|addition]] on $\N$.
Then:
:$\forall a, b, c \in \N: a + c = b + c \implies a = b$
:$\forall a, b, c \in \N: a + b = a + c \implies b = c$
That is, $+$ is [[Definition:Cancellable Operation|cancellable]] ... | Consider the [[Definition:Natural Numbers|natural numbers]] $\N$ defined as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]] $\struct {\N, +, \le}$.
By {{NOSAxiom|2}}, every element of $\struct {\N, +}$ is [[Definition:Cancellable Element|cancellable]].
{{qed}} | Natural Number Addition is Cancellable/Proof 1 | https://proofwiki.org/wiki/Natural_Number_Addition_is_Cancellable | https://proofwiki.org/wiki/Natural_Number_Addition_is_Cancellable/Proof_1 | [
"Natural Number Addition",
"Examples of Cancellable Operations",
"Natural Number Addition is Cancellable"
] | [
"Definition:Natural Numbers",
"Definition:Addition/Natural Numbers",
"Definition:Cancellable Operation"
] | [
"Definition:Natural Numbers",
"Definition:Naturally Ordered Semigroup",
"Definition:Cancellable Element"
] |
proofwiki-5563 | Natural Number Addition is Cancellable | Let $\N$ be the natural numbers.
Let $+$ be addition on $\N$.
Then:
:$\forall a, b, c \in \N: a + c = b + c \implies a = b$
:$\forall a, b, c \in \N: a + b = a + c \implies b = c$
That is, $+$ is cancellable on $\N$. | By Natural Number Addition is Commutative, we only need to prove the first statement.
Proof by induction.
Consider the natural numbers $\N$ defined in terms of Peano's Axioms.
From the definition of addition in Peano structure, we have that:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
| l = m + 0
| r = m
... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $+$ be [[Definition:Natural Number Addition|addition]] on $\N$.
Then:
:$\forall a, b, c \in \N: a + c = b + c \implies a = b$
:$\forall a, b, c \in \N: a + b = a + c \implies b = c$
That is, $+$ is [[Definition:Cancellable Operation|cancellable]] ... | By [[Natural Number Addition is Commutative/Proof 2|Natural Number Addition is Commutative]], we only need to prove the first statement.
Proof by [[Principle of Mathematical Induction for Peano Structure|induction]].
Consider the [[Definition:Natural Numbers|natural numbers]] $\N$ defined in terms of [[Axiom:Peano's... | Natural Number Addition is Cancellable/Proof 2 | https://proofwiki.org/wiki/Natural_Number_Addition_is_Cancellable | https://proofwiki.org/wiki/Natural_Number_Addition_is_Cancellable/Proof_2 | [
"Natural Number Addition",
"Examples of Cancellable Operations",
"Natural Number Addition is Cancellable"
] | [
"Definition:Natural Numbers",
"Definition:Addition/Natural Numbers",
"Definition:Cancellable Operation"
] | [
"Natural Number Addition is Commutative/Proof 2",
"Principle of Mathematical Induction/Peano Structure",
"Definition:Natural Numbers",
"Axiom:Peano's Axioms",
"Definition:Addition/Peano Structure",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"De... |
proofwiki-5564 | Uniqueness of Measures | Let $\struct {X, \Sigma}$ be a measurable space.
Let $\GG \subseteq \powerset X$ be a generator for $\Sigma$; that is, $\Sigma = \map \sigma \GG$.
Suppose that $\GG$ satisfies the following conditions:
:$(1):\quad \forall G, H \in \GG: G \cap H \in \GG$
:$(2):\quad$ There exists an exhausting sequence $\sequence {G_n}_... | Define, for all $n \in \N$, $\DD_n$ by:
:$\DD_n := \set {E \in \Sigma: \map \mu {G_n \cap E} = \map \nu {G_n \cap E} }$
Let us show that $\DD_n$ is a Dynkin system.
By Intersection with Subset is Subset, $G_n \cap X = G_n$, whence $(3)$ implies that $X \in \DD_n$.
Now, let $D \in \DD_n$. Then:
{{begin-eqn}}
{{eqn | l =... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\GG \subseteq \powerset X$ be a [[Definition:Sigma-Algebra Generated by Collection of Subsets/Generator|generator]] for $\Sigma$; that is, $\Sigma = \map \sigma \GG$.
Suppose that $\GG$ satisfies the following conditions:
:$(1):\qu... | Define, for all $n \in \N$, $\DD_n$ by:
:$\DD_n := \set {E \in \Sigma: \map \mu {G_n \cap E} = \map \nu {G_n \cap E} }$
Let us show that $\DD_n$ is a [[Definition:Dynkin System|Dynkin system]].
By [[Intersection with Subset is Subset]], $G_n \cap X = G_n$, whence $(3)$ implies that $X \in \DD_n$.
Now, let $D \in \... | Uniqueness of Measures/Proof 1 | https://proofwiki.org/wiki/Uniqueness_of_Measures | https://proofwiki.org/wiki/Uniqueness_of_Measures/Proof_1 | [
"Measure Theory",
"Measures",
"Uniqueness of Measures",
"Measures",
"Uniqueness of Measures"
] | [
"Definition:Measurable Space",
"Definition:Sigma-Algebra Generated by Collection of Subsets/Generator",
"Definition:Exhausting Sequence of Sets",
"Definition:Measure (Measure Theory)",
"Definition:Finite Extended Real Number"
] | [
"Definition:Dynkin System",
"Intersection with Subset is Subset",
"Intersection with Set Difference is Set Difference with Intersection",
"Set Difference and Intersection form Partition",
"Measure is Finitely Additive Function",
"Definition:Sequence",
"Definition:Pairwise Disjoint",
"Intersection Dist... |
proofwiki-5565 | Uniqueness of Measures | Let $\struct {X, \Sigma}$ be a measurable space.
Let $\GG \subseteq \powerset X$ be a generator for $\Sigma$; that is, $\Sigma = \map \sigma \GG$.
Suppose that $\GG$ satisfies the following conditions:
:$(1):\quad \forall G, H \in \GG: G \cap H \in \GG$
:$(2):\quad$ There exists an exhausting sequence $\sequence {G_n}_... | Define the set:
:$\AA = \set {T \in \Sigma: \forall G \in \GG: \map \mu {G \cap T} = \map \nu {G \cap T} }$
By Intersection with Subset is Subset, it follows that:
:$\forall G \in \GG: G \cap X = G$
Therefore, {{hypothesis}} $(3)$:
:$X \in \AA$
Now, define the set:
:$\Sigma' = \set {S \in \Sigma: \forall T \in \AA: \ma... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\GG \subseteq \powerset X$ be a [[Definition:Sigma-Algebra Generated by Collection of Subsets/Generator|generator]] for $\Sigma$; that is, $\Sigma = \map \sigma \GG$.
Suppose that $\GG$ satisfies the following conditions:
:$(1):\qu... | Define the [[Definition:Set|set]]:
:$\AA = \set {T \in \Sigma: \forall G \in \GG: \map \mu {G \cap T} = \map \nu {G \cap T} }$
By [[Intersection with Subset is Subset]], it follows that:
:$\forall G \in \GG: G \cap X = G$
Therefore, {{hypothesis}} $(3)$:
:$X \in \AA$
Now, define the [[Definition:Set|set]]:
:$\Sigma... | Uniqueness of Measures/Proof 2 | https://proofwiki.org/wiki/Uniqueness_of_Measures | https://proofwiki.org/wiki/Uniqueness_of_Measures/Proof_2 | [
"Measure Theory",
"Measures",
"Uniqueness of Measures",
"Measures",
"Uniqueness of Measures"
] | [
"Definition:Measurable Space",
"Definition:Sigma-Algebra Generated by Collection of Subsets/Generator",
"Definition:Exhausting Sequence of Sets",
"Definition:Measure (Measure Theory)",
"Definition:Finite Extended Real Number"
] | [
"Definition:Set",
"Intersection with Subset is Subset",
"Definition:Set",
"Definition:Restriction/Mapping",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Set Equality",
"Definition:Sigma-Algebra",
"Intersection is Associative",
"Definition:By Hypothesis",
"Intersection... |
proofwiki-5566 | Natural Number Multiplication is Cancellable | Let $\N$ be the natural numbers.
Let $\times$ be multiplication on $\N$.
Then:
:$\forall x, y, z \in \N_{>0}: x \times z = y \times z \implies x = y$
:$\forall x, y, z \in \N_{>0}: x \times y = x \times z \implies y = z$
That is, $\times$ is cancellable on $\N_{>0}$. | By Natural Number Multiplication is Commutative, proving one of the assertions suffices.
From Ordering on Natural Numbers is Compatible with Multiplication it follows that:
:$\forall x, y, z \in \N: x < y \iff x \times z < y \times z$
Interchanging $x$ and $y$, we obtain:
:$\forall x, y, z \in \N: y < x \iff y \times z... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $\times$ be [[Definition:Natural Number Multiplication|multiplication]] on $\N$.
Then:
:$\forall x, y, z \in \N_{>0}: x \times z = y \times z \implies x = y$
:$\forall x, y, z \in \N_{>0}: x \times y = x \times z \implies y = z$
That is, $\times$ ... | By [[Natural Number Multiplication is Commutative]], proving one of the assertions suffices.
From [[Ordering on Natural Numbers is Compatible with Multiplication]] it follows that:
:$\forall x, y, z \in \N: x < y \iff x \times z < y \times z$
Interchanging $x$ and $y$, we obtain:
:$\forall x, y, z \in \N: y < x \i... | Natural Number Multiplication is Cancellable | https://proofwiki.org/wiki/Natural_Number_Multiplication_is_Cancellable | https://proofwiki.org/wiki/Natural_Number_Multiplication_is_Cancellable | [
"Examples of Cancellable Operations",
"Natural Number Multiplication"
] | [
"Definition:Natural Numbers",
"Definition:Multiplication/Natural Numbers",
"Definition:Cancellable Operation"
] | [
"Natural Number Multiplication is Commutative",
"Ordering on Natural Numbers is Compatible with Multiplication",
"Ordering on Natural Numbers is Trichotomy",
"Definition:Cancellable Operation"
] |
proofwiki-5567 | Natural Number Addition is Cancellable for Ordering | Let $\N$ be the natural numbers.
Let $<$ be the strict ordering on $\N$.
Let $+$ be addition on $\N$.
Then:
:$\forall a, b, c \in \N_{>0}: a + c < b + c \implies a < b$
:$\forall a, b, c \in \N_{>0}: a + b < a + c \implies b < c$
That is, $+$ is cancellable on $\N$ for $<$. | {{ProofWanted}}
Category:Natural Number Addition
Category:Examples of Cancellable Operations
nbgpynwtclm58j7no3ognibx6pfc6nq | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $<$ be the [[Definition:Ordering on Natural Numbers|strict ordering on $\N$]].
Let $+$ be [[Definition:Natural Number Addition|addition]] on $\N$.
Then:
:$\forall a, b, c \in \N_{>0}: a + c < b + c \implies a < b$
:$\forall a, b, c \in \N_{>0}: a +... | {{ProofWanted}}
[[Category:Natural Number Addition]]
[[Category:Examples of Cancellable Operations]]
nbgpynwtclm58j7no3ognibx6pfc6nq | Natural Number Addition is Cancellable for Ordering | https://proofwiki.org/wiki/Natural_Number_Addition_is_Cancellable_for_Ordering | https://proofwiki.org/wiki/Natural_Number_Addition_is_Cancellable_for_Ordering | [
"Natural Number Addition",
"Examples of Cancellable Operations"
] | [
"Definition:Natural Numbers",
"Definition:Ordering on Natural Numbers",
"Definition:Addition/Natural Numbers",
"Definition:Cancellable Operation"
] | [
"Category:Natural Number Addition",
"Category:Examples of Cancellable Operations"
] |
proofwiki-5568 | Natural Number Multiplication is Cancellable for Ordering | Let $\N$ be the natural numbers.
Let $\times$ be multiplication on $\N$.
Let $<$ be the strict ordering on $\N$.
Then:
:$\forall a, b, c \in \N: a \times c < b \times c \implies a < b$
:$\forall a, b, c \in \N: a \times b < a \times c \implies b < c$
That is, $\times$ is cancellable on $\N$ for $<$. | {{proof wanted}}
Category:Examples of Cancellable Operations
Category:Natural Number Multiplication
4jy5waacgramzyozgw5mrge19x723xf | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $\times$ be [[Definition:Natural Number Multiplication|multiplication]] on $\N$.
Let $<$ be the [[Definition:Ordering on Natural Numbers|strict ordering on $\N$]].
Then:
:$\forall a, b, c \in \N: a \times c < b \times c \implies a < b$
:$\forall a,... | {{proof wanted}}
[[Category:Examples of Cancellable Operations]]
[[Category:Natural Number Multiplication]]
4jy5waacgramzyozgw5mrge19x723xf | Natural Number Multiplication is Cancellable for Ordering | https://proofwiki.org/wiki/Natural_Number_Multiplication_is_Cancellable_for_Ordering | https://proofwiki.org/wiki/Natural_Number_Multiplication_is_Cancellable_for_Ordering | [
"Examples of Cancellable Operations",
"Natural Number Multiplication"
] | [
"Definition:Natural Numbers",
"Definition:Multiplication/Natural Numbers",
"Definition:Ordering on Natural Numbers",
"Definition:Cancellable Operation"
] | [
"Category:Examples of Cancellable Operations",
"Category:Natural Number Multiplication"
] |
proofwiki-5569 | Successor to Natural Number | Let $\N_{> 0}$ be the 1-based natural numbers:
:$\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$
Let $<$ be the ordering on $\N_{> 0}$:
:$\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$
Let $a \in \N_{>0}$.
Then there exists no natural number $n$ such that $a < n < a + 1$. | Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Suppose that $\exists n \in \N_{>0}: a < n < a + 1$.
Then by the definition of ordering on natural numbers:
{{begin-eqn}}
{{eqn | l = a + x
| r = n
| c = Definition of Ordering on Natural Numbers: $a < n$
}}
{{eqn | l = n + y
... | Let $\N_{> 0}$ be the [[Definition:Natural Numbers|1-based natural numbers]]:
:$\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$
Let $<$ be the [[Definition:Ordering on 1-Based Natural Numbers|ordering on $\N_{> 0}$]]:
:$\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$
Let $a \in \N_{>0}$.
Then th... | Using the [[Axiom:Axiomatization of 1-Based Natural Numbers|following axioms]]:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Suppose that $\exists n \in \N_{>0}: a < n < a + 1$.
Then by the definition of [[Definition:Ordering on Natural Numbers|ordering on natural numbers]]:
{{begin-eqn}}
{{eqn | l = a + x
... | Successor to Natural Number | https://proofwiki.org/wiki/Successor_to_Natural_Number | https://proofwiki.org/wiki/Successor_to_Natural_Number | [
"Natural Numbers/1-Based"
] | [
"Definition:Natural Numbers",
"Definition:Ordering on Natural Numbers/1-Based",
"Definition:Natural Numbers"
] | [
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Ordering on Natural Numbers",
"Definition:Ordering on Natural Numbers",
"Definition:Ordering on Natural Numbers",
"Natural Number Addition is Associative",
"Addition on 1-Based Natural Numbers is Cancellable",
"Axiom:Axiomatization of 1-Base... |
proofwiki-5570 | Natural Number is Not Equal to Successor | Let $\N_{> 0}$ be the $1$-based natural numbers:
:$\N_{> 0} = \set {1, 2, 3, \ldots}$
Then:
:$\forall n \in \N_{> 0}: n \ne n + 1$ | Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$n \ne n + 1$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$1 \ne 1 + 1$
From axiom $E$:
:$E: \quad \forall a, b \in \N_{> 0}: \text{ exactly 1 of... | Let $\N_{> 0}$ be the [[Definition:Natural Numbers|$1$-based natural numbers]]:
:$\N_{> 0} = \set {1, 2, 3, \ldots}$
Then:
:$\forall n \in \N_{> 0}: n \ne n + 1$ | Using the [[Axiom:Axiomatization of 1-Based Natural Numbers|following axioms]]:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$n \ne n + 1$
=== Basis for the Ind... | Natural Number is Not Equal to Successor/Proof 1 | https://proofwiki.org/wiki/Natural_Number_is_Not_Equal_to_Successor | https://proofwiki.org/wiki/Natural_Number_is_Not_Equal_to_Successor/Proof_1 | [
"Natural Numbers/1-Based"
] | [
"Definition:Natural Numbers"
] | [
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Natural Number is Not Equal to S... |
proofwiki-5571 | Biconditional is Commutative | ==== Formulation 1 ====
{{:Biconditional is Commutative/Formulation 1}}
==== Formulation 2 ====
{{:Biconditional is Commutative/Formulation 2}} | {{BeginTableau|p \iff q \vdash q \iff p}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{BiconditionalIntro|4|1|q \iff p|3|2}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|q \iff p \vdash p \iff q}}
{{Premise|1|q \iff p}}
{{BiconditionalElimin... | ==== [[Biconditional is Commutative/Formulation 1|Formulation 1]] ====
{{:Biconditional is Commutative/Formulation 1}}
==== [[Biconditional is Commutative/Formulation 2|Formulation 2]] ====
{{:Biconditional is Commutative/Formulation 2}} | {{BeginTableau|p \iff q \vdash q \iff p}}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{BiconditionalIntro|4|1|q \iff p|3|2}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|q \iff p \vdash p \iff q}}
{{Premise|1|q \iff p}}
{{BiconditionalElim... | Biconditional is Commutative/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Biconditional_is_Commutative | https://proofwiki.org/wiki/Biconditional_is_Commutative/Formulation_1/Proof_1 | [
"Biconditional"
] | [
"Biconditional is Commutative/Formulation 1",
"Biconditional is Commutative/Formulation 2"
] | [] |
proofwiki-5572 | Biconditional is Commutative | ==== Formulation 1 ====
{{:Biconditional is Commutative/Formulation 1}}
==== Formulation 2 ====
{{:Biconditional is Commutative/Formulation 2}} | {{BeginTableau|p \iff q \vdash q \iff p}}
{{Premise|1|p \iff q}}
{{SequentIntro|2|1|\paren {p \implies q} \land \paren {q \implies p}|1|Rule of Material Equivalence}}
{{Commutation|3|1|\paren {q \implies p} \land \paren {p \implies q}|2|Conjunction}}
{{SequentIntro|4|1|q \iff p|3|Rule of Material Equivalence}}
{{EndTab... | ==== [[Biconditional is Commutative/Formulation 1|Formulation 1]] ====
{{:Biconditional is Commutative/Formulation 1}}
==== [[Biconditional is Commutative/Formulation 2|Formulation 2]] ====
{{:Biconditional is Commutative/Formulation 2}} | {{BeginTableau|p \iff q \vdash q \iff p}}
{{Premise|1|p \iff q}}
{{SequentIntro|2|1|\paren {p \implies q} \land \paren {q \implies p}|1|[[Rule of Material Equivalence]]}}
{{Commutation|3|1|\paren {q \implies p} \land \paren {p \implies q}|2|Conjunction}}
{{SequentIntro|4|1|q \iff p|3|[[Rule of Material Equivalence]]}}
... | Biconditional is Commutative/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Biconditional_is_Commutative | https://proofwiki.org/wiki/Biconditional_is_Commutative/Formulation_1/Proof_2 | [
"Biconditional"
] | [
"Biconditional is Commutative/Formulation 1",
"Biconditional is Commutative/Formulation 2"
] | [
"Rule of Material Equivalence",
"Rule of Material Equivalence",
"Rule of Material Equivalence",
"Rule of Material Equivalence"
] |
proofwiki-5573 | Biconditional is Commutative | ==== Formulation 1 ====
{{:Biconditional is Commutative/Formulation 1}}
==== Formulation 2 ====
{{:Biconditional is Commutative/Formulation 2}} | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccc|} \hline
p & \iff & q & q & \iff & p \\
\hline
\F & \T & \F & \F & \T & \F \\
\F & \F & \T & \T & \F & \F \\
\T & \F & \F & \F & \F & \T \\
\T & ... | ==== [[Biconditional is Commutative/Formulation 1|Formulation 1]] ====
{{:Biconditional is Commutative/Formulation 1}}
==== [[Biconditional is Commutative/Formulation 2|Formulation 2]] ====
{{:Biconditional is Commutative/Formulation 2}} | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|ccc||ccc|} \hline
p & \iff & ... | Biconditional is Commutative/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Biconditional_is_Commutative | https://proofwiki.org/wiki/Biconditional_is_Commutative/Formulation_1/Proof_by_Truth_Table | [
"Biconditional"
] | [
"Biconditional is Commutative/Formulation 1",
"Biconditional is Commutative/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-5574 | Biconditional is Reflexive | :$\vdash p \iff p$ | {{BeginTableau|\vdash p \iff p}}
{{TheoremIntro|1|p \implies p|Law of Identity: Formulation 2}}
{{BiconditionalIntro|2||p \iff p|1|1}}
{{EndTableau|qed}} | :$\vdash p \iff p$ | {{BeginTableau|\vdash p \iff p}}
{{TheoremIntro|1|p \implies p|[[Law of Identity/Formulation 2|Law of Identity: Formulation 2]]}}
{{BiconditionalIntro|2||p \iff p|1|1}}
{{EndTableau|qed}} | Biconditional is Reflexive/Proof 1 | https://proofwiki.org/wiki/Biconditional_is_Reflexive | https://proofwiki.org/wiki/Biconditional_is_Reflexive/Proof_1 | [
"Biconditional is Reflexive",
"Biconditional"
] | [] | [
"Law of Identity/Formulation 2"
] |
proofwiki-5575 | Biconditional is Reflexive | :$\vdash p \iff p$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective match for both boolean interpretations.
:<nowiki>$\begin {array} {|ccc|} \hline
p & \iff & p \\
\hline
\F & \T & \F \\
\T & \T & \T \\
\hline
\end {array}$</nowiki>
{{qed}} | :$\vdash p \iff p$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] match for both [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|ccc|} \hline
p & \i... | Biconditional is Reflexive/Proof by Truth Table | https://proofwiki.org/wiki/Biconditional_is_Reflexive | https://proofwiki.org/wiki/Biconditional_is_Reflexive/Proof_by_Truth_Table | [
"Biconditional is Reflexive",
"Biconditional"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-5576 | Biconditional is Transitive | The biconditional operator is transitive: | {{BeginTableau|p \iff q, q \iff r \vdash p \iff r}}
{{Premise|1|p \iff q}}
{{Premise|2|q \iff r}}
{{BiconditionalElimination|3|1|p \implies q|1|1}}
{{BiconditionalElimination|4|2|q \implies r|2|1}}
{{SequentIntro|5|1, 2|p \implies r|1, 2|Hypothetical Syllogism: Formulation 1}}
{{BiconditionalElimination|6|1|q \implies ... | The [[Definition:Biconditional|biconditional operator]] is [[Definition:Transitive Relation|transitive]]: | {{BeginTableau|p \iff q, q \iff r \vdash p \iff r}}
{{Premise|1|p \iff q}}
{{Premise|2|q \iff r}}
{{BiconditionalElimination|3|1|p \implies q|1|1}}
{{BiconditionalElimination|4|2|q \implies r|2|1}}
{{SequentIntro|5|1, 2|p \implies r|1, 2|[[Hypothetical Syllogism/Formulation 1|Hypothetical Syllogism: Formulation 1]]}}
{... | Biconditional is Transitive/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Biconditional_is_Transitive | https://proofwiki.org/wiki/Biconditional_is_Transitive/Formulation_1/Proof_1 | [
"Biconditional"
] | [
"Definition:Biconditional",
"Definition:Transitive Relation"
] | [
"Hypothetical Syllogism/Formulation 1",
"Hypothetical Syllogism/Formulation 1"
] |
proofwiki-5577 | Biconditional is Transitive | The biconditional operator is transitive: | We apply the Method of Truth Tables.
As can be seen for all boolean interpretations by inspection, where the truth values under the main connective on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$:
:<nowiki>$\begin {array} {|ccccccc||ccc|} \hline
(p & \iff & q) & \land & (q & \iff & r) & p & \iff & ... | The [[Definition:Biconditional|biconditional operator]] is [[Definition:Transitive Relation|transitive]]: | We apply the [[Method of Truth Tables]].
As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] on the {{LHS}} is $T$, that under the one on the ... | Biconditional is Transitive/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Biconditional_is_Transitive | https://proofwiki.org/wiki/Biconditional_is_Transitive/Formulation_1/Proof_by_Truth_Table | [
"Biconditional"
] | [
"Definition:Biconditional",
"Definition:Transitive Relation"
] | [
"Method of Truth Tables",
"Definition:Boolean Interpretation",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic"
] |
proofwiki-5578 | Ordinals are Totally Ordered | The ordinals are totally ordered. | Follows directly from Relation between Two Ordinals and Subset Relation is Ordering.
{{qed}}
Category:Ordinals
7j5gt4exlrw1dxv82z8s62oow1hw3bv | The [[Definition:Ordinal|ordinals]] are [[Definition:Total Ordering|totally ordered]]. | Follows directly from [[Relation between Two Ordinals]] and [[Subset Relation is Ordering]].
{{qed}}
[[Category:Ordinals]]
7j5gt4exlrw1dxv82z8s62oow1hw3bv | Ordinals are Totally Ordered | https://proofwiki.org/wiki/Ordinals_are_Totally_Ordered | https://proofwiki.org/wiki/Ordinals_are_Totally_Ordered | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Total Ordering"
] | [
"Relation between Two Ordinals",
"Subset Relation is Ordering",
"Category:Ordinals"
] |
proofwiki-5579 | Successor Set of Transitive Set is Transitive | Let $S$ be a transitive set.
Then its successor set $S^+ = S \cup \set S$ is also transitive. | Recall that $S$ is '''transitive''' {{iff}}:
:$x \in S \implies x \subseteq S$
Hence:
{{begin-eqn}}
{{eqn | q = \forall x \in S^+
| l = x
| o = \in
| r = S
| c =
}}
{{eqn | lo= \lor
| l = x
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \subseteq
| r = ... | Let $S$ be a [[Definition:Transitive Set|transitive set]].
Then its [[Definition:Successor Set|successor set]] $S^+ = S \cup \set S$ is also [[Definition:Transitive Set|transitive]]. | Recall that $S$ is '''[[Definition:Transitive Set|transitive]]''' {{iff}}:
:$x \in S \implies x \subseteq S$
Hence:
{{begin-eqn}}
{{eqn | q = \forall x \in S^+
| l = x
| o = \in
| r = S
| c =
}}
{{eqn | lo= \lor
| l = x
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = x
... | Successor Set of Transitive Set is Transitive | https://proofwiki.org/wiki/Successor_Set_of_Transitive_Set_is_Transitive | https://proofwiki.org/wiki/Successor_Set_of_Transitive_Set_is_Transitive | [
"Successor Mapping",
"Transitive Classes"
] | [
"Definition:Transitive Class",
"Definition:Successor Mapping/Successor Set",
"Definition:Transitive Class"
] | [
"Definition:Transitive Class",
"Set is Subset of Itself",
"Proof by Cases",
"Set is Subset of Union"
] |
proofwiki-5580 | Successor Set of Ordinal is Ordinal | Let $\On$ denote the class of all ordinals.
Let $\alpha \in \On$ be an ordinal.
Then its successor set $\alpha^+ = \alpha \cup \set \alpha$ is also an ordinal. | We have the result that Class of All Ordinals is Minimally Superinductive over Successor Mapping.
Hence $\On$ is {{afortiori}} a superinductive class {{WRT}} the successor mapping.
Hence, by definition of superinductive class:
:$\On$ is closed under the successor mapping.
That is:
:$\forall \alpha \in \On: \alpha^+ \in... | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $\alpha \in \On$ be an [[Definition:Ordinal|ordinal]].
Then its [[Definition:Successor Set|successor set]] $\alpha^+ = \alpha \cup \set \alpha$ is also an [[Definition:Ordinal|ordinal]]. | We have the result that [[Class of All Ordinals is Minimally Superinductive over Successor Mapping]].
Hence $\On$ is {{afortiori}} a [[Definition:Superinductive Class|superinductive class]] {{WRT}} the [[Definition:Successor Mapping|successor mapping]].
Hence, by definition of [[Definition:Superinductive Class|superi... | Successor Set of Ordinal is Ordinal/Proof 1 | https://proofwiki.org/wiki/Successor_Set_of_Ordinal_is_Ordinal | https://proofwiki.org/wiki/Successor_Set_of_Ordinal_is_Ordinal/Proof_1 | [
"Successor Set of Ordinal is Ordinal",
"Ordinals",
"Successor Mapping"
] | [
"Definition:Class of All Ordinals",
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:Ordinal"
] | [
"Class of All Ordinals is Minimally Superinductive over Successor Mapping",
"Definition:Superinductive Class",
"Definition:Successor Mapping",
"Definition:Superinductive Class",
"Definition:Closed under Mapping/Class Theory",
"Definition:Successor Mapping"
] |
proofwiki-5581 | Successor Set of Ordinal is Ordinal | Let $\On$ denote the class of all ordinals.
Let $\alpha \in \On$ be an ordinal.
Then its successor set $\alpha^+ = \alpha \cup \set \alpha$ is also an ordinal. | {{explain|Would be insightful to make the definition of ordinal being used explicit.}}
From Ordinal is Transitive, it follows by Successor Set of Transitive Set is Transitive that $\alpha^+$ is transitive.
We now have to show that $\alpha^+$ is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_{\... | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $\alpha \in \On$ be an [[Definition:Ordinal|ordinal]].
Then its [[Definition:Successor Set|successor set]] $\alpha^+ = \alpha \cup \set \alpha$ is also an [[Definition:Ordinal|ordinal]]. | {{explain|Would be insightful to make the definition of ordinal being used explicit.}}
From [[Ordinal is Transitive]], it follows by [[Successor Set of Transitive Set is Transitive]] that $\alpha^+$ is [[Definition:Transitive Set|transitive]].
We now have to show that $\alpha^+$ is [[Definition:Strict Well-Ordering|st... | Successor Set of Ordinal is Ordinal/Proof 2 | https://proofwiki.org/wiki/Successor_Set_of_Ordinal_is_Ordinal | https://proofwiki.org/wiki/Successor_Set_of_Ordinal_is_Ordinal/Proof_2 | [
"Successor Set of Ordinal is Ordinal",
"Ordinals",
"Successor Mapping"
] | [
"Definition:Class of All Ordinals",
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:Ordinal"
] | [
"Ordinal is Transitive",
"Successor Set of Transitive Set is Transitive",
"Definition:Transitive Class",
"Definition:Strict Well-Ordering",
"Definition:Epsilon Relation/Restriction",
"Definition:Subset",
"Definition:Non-Empty Set",
"Intersection with Subset is Subset",
"Intersection Distributes over... |
proofwiki-5582 | Ordinals are Well-Ordered | The ordinals are well-ordered. | Recall that the Ordinals are Totally Ordered.
Let $A$ be a non-empty set of ordinals.
Let $\alpha \in A$.
Let $B = \alpha^+ \cap A$, where $\alpha^+$ denotes the successor set of $\alpha$.
Recall that $\alpha^+$ is an ordinal.
Note that $\alpha \in B$, so $B$ is non-empty.
By Intersection is Subset, $B \subseteq \alpha... | The [[Definition:Ordinal|ordinals]] are [[Definition:Well-Ordering|well-ordered]]. | Recall that the [[Ordinals are Totally Ordered]].
Let $A$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] of [[Definition:Ordinal|ordinals]].
Let $\alpha \in A$.
Let $B = \alpha^+ \cap A$, where $\alpha^+$ denotes the [[Definition:Successor Set|successor set]] of $\alpha$.
Recall that [[Successor... | Ordinals are Well-Ordered/Proof 1 | https://proofwiki.org/wiki/Ordinals_are_Well-Ordered | https://proofwiki.org/wiki/Ordinals_are_Well-Ordered/Proof_1 | [
"Ordinals",
"Ordinals are Well-Ordered"
] | [
"Definition:Ordinal",
"Definition:Well-Ordering"
] | [
"Ordinals are Totally Ordered",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set",
"Successor Set of Ordinal is Ordinal",
"Definition:Empty Set",
"Intersection is Subset",
"Definition:Existential Quantifier",
"Definition:Smallest Eleme... |
proofwiki-5583 | Ordinals are Well-Ordered | The ordinals are well-ordered. | By Ordinals are Totally Ordered, the ordinals are totally ordered by $\subseteq$:
{{begin-eqn}}
{{eqn | l = X
| o = \subsetneqq
| r = Y
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists a \in Y
| l = X
| r = Y_a
| c = where $Y_a$ denotes the initial segment of $Y$ determined ... | The [[Definition:Ordinal|ordinals]] are [[Definition:Well-Ordering|well-ordered]]. | By [[Ordinals are Totally Ordered]], the [[Definition:Ordinal|ordinals]] are [[Definition:Total Ordering|totally ordered]] by $\subseteq$:
{{begin-eqn}}
{{eqn | l = X
| o = \subsetneqq
| r = Y
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \exists a \in Y
| l = X
| r = Y_a
| c = w... | Ordinals are Well-Ordered/Proof 2 | https://proofwiki.org/wiki/Ordinals_are_Well-Ordered | https://proofwiki.org/wiki/Ordinals_are_Well-Ordered/Proof_2 | [
"Ordinals",
"Ordinals are Well-Ordered"
] | [
"Definition:Ordinal",
"Definition:Well-Ordering"
] | [
"Ordinals are Totally Ordered",
"Definition:Ordinal",
"Definition:Total Ordering",
"Definition:Initial Segment",
"Definition:Strict Total Ordering",
"Definition:Ordinal",
"Definition:Strict Total Ordering",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Well-Ordering",
"Definition:Seque... |
proofwiki-5584 | Set Intersection Preserves Subsets/Corollary/Proof 1 | Let $A, B, S$ be sets.
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$ | Let $A \subseteq B$, and let $S$ be any set.
From Set Intersection Preserves Subsets:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
:$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From Set is Subset of Itself, $S \... | Let $A, B, S$ be [[Definition:Set|sets]].
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$ | Let $A \subseteq B$, and let $S$ be any set.
From [[Set Intersection Preserves Subsets]]:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
:$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From [[Set is Subset of It... | Set Intersection Preserves Subsets/Corollary/Proof 1 | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Corollary/Proof_1 | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Corollary/Proof_1 | [
"Set Intersection Preserves Subsets"
] | [
"Definition:Set"
] | [
"Set Intersection Preserves Subsets",
"Set is Subset of Itself"
] |
proofwiki-5585 | Set Intersection Preserves Subsets/Corollary | Let $A, B, S$ be sets.
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$ | Let $A \subseteq B$, and let $S$ be any set.
From Set Intersection Preserves Subsets:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
:$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From Set is Subset of Itself, $S \... | Let $A, B, S$ be [[Definition:Set|sets]].
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$ | Let $A \subseteq B$, and let $S$ be any set.
From [[Set Intersection Preserves Subsets]]:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
:$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From [[Set is Subset of It... | Set Intersection Preserves Subsets/Corollary/Proof 1 | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Corollary | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Corollary/Proof_1 | [
"Set Intersection Preserves Subsets"
] | [
"Definition:Set"
] | [
"Set Intersection Preserves Subsets",
"Set is Subset of Itself"
] |
proofwiki-5586 | Set Intersection Preserves Subsets/Corollary | Let $A, B, S$ be sets.
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$ | Recall the Factor Principles, themselves a corollary of the Praeclarum Theorema:
:$\paren {p \implies q} \vdash \paren {p \land r} \implies \paren {q \land r}$
This is applied as:
{{begin-eqn}}
{{eqn | o =
| r = A \subseteq B
| c =
}}
{{eqn | o = \leadsto
| r = \paren {x \in A \implies x \in B}
... | Let $A, B, S$ be [[Definition:Set|sets]].
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$ | Recall the [[Factor Principles]], themselves a [[Definition:Corollary|corollary]] of the [[Praeclarum Theorema]]:
:$\paren {p \implies q} \vdash \paren {p \land r} \implies \paren {q \land r}$
This is applied as:
{{begin-eqn}}
{{eqn | o =
| r = A \subseteq B
| c =
}}
{{eqn | o = \leadsto
| r = \p... | Set Intersection Preserves Subsets/Corollary/Proof 2 | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Corollary | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Corollary/Proof_2 | [
"Set Intersection Preserves Subsets"
] | [
"Definition:Set"
] | [
"Factor Principles",
"Definition:Corollary",
"Praeclarum Theorema",
"Factor Principles"
] |
proofwiki-5587 | Set Intersection Preserves Subsets/Families of Sets/Corollary | Let $I$ be an indexing set.
Let $\family {B_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $A$ be a set such that $A \subseteq B_\alpha$ for all $\alpha \in I$.
Then:
:$\ds A \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$ | For each $\alpha \in I$, define $A_\alpha := A$.
Then by Set Intersection is Idempotent, it follows that:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha = A$
Since $A \subseteq B_\alpha$ for all $\alpha \in I$, the premises of Set Intersection Preserves Subsets are satisfied.
Applying this theorem gives:
:$\ds A = \bigc... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {B_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family of subsets]] of a [[Definition:Set|set]] $S$.
Let $A$ be a [[Definition:Set|set]] such that $A \subseteq B_\alpha$ for all $\alpha \in I$.
Then:
:$\ds... | For each $\alpha \in I$, define $A_\alpha := A$.
Then by [[Set Intersection is Idempotent/Indexed Family|Set Intersection is Idempotent]], it follows that:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha = A$
Since $A \subseteq B_\alpha$ for all $\alpha \in I$, the premises of [[Set Intersection Preserves Subsets/Fam... | Set Intersection Preserves Subsets/Families of Sets/Corollary | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Families_of_Sets/Corollary | https://proofwiki.org/wiki/Set_Intersection_Preserves_Subsets/Families_of_Sets/Corollary | [
"Set Intersection Preserves Subsets",
"Indexed Families"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Set",
"Definition:Set"
] | [
"Set Intersection is Idempotent/Indexed Family",
"Set Intersection Preserves Subsets/Families of Sets",
"Category:Set Intersection Preserves Subsets",
"Category:Indexed Families"
] |
proofwiki-5588 | Intersection is Largest Subset/General Result | Let $T$ be a set.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
:$\paren {\forall X \in \mathbb T: S \subseteq X} \iff S \subseteq \bigcap \mathbb T$
=== Family of Sets ===
In the context of a family of sets, the result can be presented as follows:
{{:Intersection is Lar... | Let $\mathbb T \subseteq \powerset T$.
Suppose that $\forall X \in \mathbb T: S \subseteq X$.
Consider any $x \in S$.
By definition of subset, it follows that:
:$\forall X \in \mathbb T: x \in X$
Thus it follows from definition of set intersection that:
:$x \in \bigcap \mathbb T$
Thus by definition of subset, it follow... | Let $T$ be a [[Definition:Set|set]].
Let $\powerset T$ be the [[Definition:Power Set|power set]] of $T$.
Let $\mathbb T$ be a [[Definition:Subset|subset]] of $\powerset T$.
Then:
:$\paren {\forall X \in \mathbb T: S \subseteq X} \iff S \subseteq \bigcap \mathbb T$
=== [[Intersection is Largest Subset/Family of Se... | Let $\mathbb T \subseteq \powerset T$.
Suppose that $\forall X \in \mathbb T: S \subseteq X$.
Consider any $x \in S$.
By definition of [[Definition:Subset|subset]], it follows that:
:$\forall X \in \mathbb T: x \in X$
Thus it follows from definition of [[Definition:Set Intersection|set intersection]] that:
:$x \in ... | Intersection is Largest Subset/General Result | https://proofwiki.org/wiki/Intersection_is_Largest_Subset/General_Result | https://proofwiki.org/wiki/Intersection_is_Largest_Subset/General_Result | [
"Set Intersection",
"Subsets"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset",
"Intersection is Largest Subset/Family of Sets",
"Definition:Indexing Set/Family of Subsets"
] | [
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Subset",
"Intersection is Subset/General Result",
"Subset Relation is Transitive"
] |
proofwiki-5589 | Union is Smallest Superset/General Result | Let $S$ and $T$ be sets.
Let $\powerset S$ denote the power set of $S$.
Let $\mathbb S$ be a subset of $\powerset S$.
Then:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
=== Family of Sets ===
In the context of a family of sets, the result can be presented as follows:
{{:Unio... | Let $\mathbb S \subseteq \powerset S$.
By Union of Subsets is Subset: Subset of Power Set:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
{{qed|lemma}}
Now suppose that $\ds \bigcup \mathbb S \subseteq T$.
Consider any $X \in \mathbb S$ and take any $x \in X$.
From Set is ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$.
Let $\mathbb S$ be a [[Definition:Subset|subset]] of $\powerset S$.
Then:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
=== [[Union is Smallest Superset/F... | Let $\mathbb S \subseteq \powerset S$.
By [[Union of Subsets is Subset/Subset of Power Set|Union of Subsets is Subset: Subset of Power Set]]:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
{{qed|lemma}}
Now suppose that $\ds \bigcup \mathbb S \subseteq T$.
Consider an... | Union is Smallest Superset/General Result | https://proofwiki.org/wiki/Union_is_Smallest_Superset/General_Result | https://proofwiki.org/wiki/Union_is_Smallest_Superset/General_Result | [
"Set Union",
"Subsets"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset",
"Union is Smallest Superset/Family of Sets",
"Definition:Indexing Set/Family of Subsets"
] | [
"Union of Subsets is Subset/Subset of Power Set",
"Set is Subset of Union/General Result"
] |
proofwiki-5590 | Mapping is Involution iff Bijective and Symmetric | Let $S$ be a set.
Let $f: S \to S$ be a mapping on $S$.
Then $f$ is an involution {{iff}} $f$ is both a bijection and a symmetric relation. | By definition an involution on $S$ is a mapping such that:
:$\forall x \in S: \map f {\map f x} = x$ | Let $S$ be a [[Definition:Set|set]].
Let $f: S \to S$ be a [[Definition:Mapping|mapping]] on $S$.
Then $f$ is an [[Definition:Involution (Mapping)|involution]] {{iff}} $f$ is both a [[Definition:Bijection|bijection]] and a [[Definition:Symmetric Relation|symmetric relation]]. | By definition an [[Definition:Involution (Mapping)|involution on $S$]] is a [[Definition:Mapping|mapping]] such that:
:$\forall x \in S: \map f {\map f x} = x$ | Mapping is Involution iff Bijective and Symmetric | https://proofwiki.org/wiki/Mapping_is_Involution_iff_Bijective_and_Symmetric | https://proofwiki.org/wiki/Mapping_is_Involution_iff_Bijective_and_Symmetric | [
"Involutions",
"Bijections",
"Symmetric Relations"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Involution (Mapping)",
"Definition:Bijection",
"Definition:Symmetric Relation"
] | [
"Definition:Involution (Mapping)",
"Definition:Mapping",
"Definition:Involution (Mapping)",
"Definition:Involution (Mapping)",
"Definition:Mapping",
"Definition:Involution (Mapping)"
] |
proofwiki-5591 | Cardinality of Set Union/Corollary | Let $S_1, S_2, \ldots, S_n$ be finite sets which are pairwise disjoint.
Then:
:$\ds \card {\bigcup_{i \mathop = 1}^n S_i} = \sum_{i \mathop = 1}^n \card {S_i}$
Specifically:
:$\card {S_1 \cup S_2} = \card {S_1} + \card {S_2}$ | As $S_1, S_2, \ldots, S_n$ are pairwise disjoint, their intersections are all empty.
The Cardinality of Set Union holds, but from Cardinality of Empty Set, all the terms apart from the first vanish.
{{qed}} | Let $S_1, S_2, \ldots, S_n$ be [[Definition:Finite Set|finite sets]] which are [[Definition:Disjoint Sets|pairwise disjoint]].
Then:
:$\ds \card {\bigcup_{i \mathop = 1}^n S_i} = \sum_{i \mathop = 1}^n \card {S_i}$
Specifically:
:$\card {S_1 \cup S_2} = \card {S_1} + \card {S_2}$ | As $S_1, S_2, \ldots, S_n$ are [[Definition:Disjoint Sets|pairwise disjoint]], their [[Definition:Set Intersection|intersections]] are all [[Definition:Empty Set|empty]].
The [[Cardinality of Set Union]] holds, but from [[Cardinality of Empty Set]], all the terms apart from the first vanish.
{{qed}} | Cardinality of Set Union/Corollary | https://proofwiki.org/wiki/Cardinality_of_Set_Union/Corollary | https://proofwiki.org/wiki/Cardinality_of_Set_Union/Corollary | [
"Cardinality of Set Union"
] | [
"Definition:Finite Set",
"Definition:Disjoint Sets"
] | [
"Definition:Disjoint Sets",
"Definition:Set Intersection",
"Definition:Empty Set",
"Cardinality of Set Union",
"Cardinality of Empty Set"
] |
proofwiki-5592 | Cartesian Product of Preimage with Image of Relation is Correspondence | Let $\RR \subseteq S \times T$ be a relation.
Then the restriction of $\RR$ to $\Preimg \RR \times \Img \RR$ is a correspondence. | By the definition of a correspondence it will be shown that $\RR$ is both left-total and right-total.
$\RR$ is left-total {{iff}}:
:$\forall x \in S: \exists y \in T: x \mathrel \RR y$
By the definition of the pre-image of $\RR$:
:$\Preimg \RR = \set {x \in S: \exists y \in T: x \mathrel \RR y}$
Therefore $\RR$ is left... | Let $\RR \subseteq S \times T$ be a [[Definition:relation|relation]].
Then the [[Definition:Restriction of Relation|restriction]] of $\RR$ to $\Preimg \RR \times \Img \RR$ is a [[Definition:Correspondence|correspondence]]. | By the definition of a [[Definition:Correspondence|correspondence]] it will be shown that $\RR$ is both [[Definition:Left-Total Relation|left-total]] and [[Definition:Right-Total Relation|right-total]].
$\RR$ is [[Definition:Left-Total Relation|left-total]] {{iff}}:
:$\forall x \in S: \exists y \in T: x \mathrel \RR... | Cartesian Product of Preimage with Image of Relation is Correspondence | https://proofwiki.org/wiki/Cartesian_Product_of_Preimage_with_Image_of_Relation_is_Correspondence | https://proofwiki.org/wiki/Cartesian_Product_of_Preimage_with_Image_of_Relation_is_Correspondence | [
"Relation Theory"
] | [
"Definition:relation",
"Definition:Restriction/Relation",
"Definition:Correspondence"
] | [
"Definition:Correspondence",
"Definition:Left-Total Relation",
"Definition:Right-Total Relation",
"Definition:Left-Total Relation",
"Definition:Preimage/Relation/Relation",
"Definition:Left-Total Relation",
"Definition:Right-Total Relation",
"Definition:Image (Set Theory)/Relation/Relation",
"Defini... |
proofwiki-5593 | Image of Intersection under Relation/General Result | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$ | {{begin-eqn}}
{{eqn | q = \forall X \in \mathbb S
| l = \bigcap \mathbb S
| o = \subseteq
| r = X
| c = Intersection is Subset: General Result
}}
{{eqn | ll= \leadsto
| q = \forall X \in \mathbb S
| l = \RR \sqbrk {\bigcap \mathbb S}
| o = \subseteq
| r = \RR \sqbrk X
... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sq... | {{begin-eqn}}
{{eqn | q = \forall X \in \mathbb S
| l = \bigcap \mathbb S
| o = \subseteq
| r = X
| c = [[Intersection is Subset/General Result|Intersection is Subset: General Result]]
}}
{{eqn | ll= \leadsto
| q = \forall X \in \mathbb S
| l = \RR \sqbrk {\bigcap \mathbb S}
| ... | Image of Intersection under Relation/General Result | https://proofwiki.org/wiki/Image_of_Intersection_under_Relation/General_Result | https://proofwiki.org/wiki/Image_of_Intersection_under_Relation/General_Result | [
"Image of Intersection under Relation"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Power Set"
] | [
"Intersection is Subset/General Result",
"Image of Subset under Relation is Subset of Image",
"Intersection is Largest Subset/General Result",
"Category:Image of Intersection under Relation"
] |
proofwiki-5594 | Image of Intersection under One-to-Many Relation/General Result | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\powerset S$ be the power set of $S$.
Then:
:$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
{{iff}} $\RR$ is one-to-many. | === Sufficient Condition ===
Suppose:
:$\ds \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
where $\mathbb S$ is ''any'' subset of $\powerset S$.
Then by definition of $\mathbb S$:
:$\forall S_1, S_2 \in \mathbb S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
and... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Then:
:$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$ ... | === Sufficient Condition ===
Suppose:
:$\ds \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
where $\mathbb S$ is ''any'' [[Definition:Subset|subset]] of $\powerset S$.
Then by definition of $\mathbb S$:
:$\forall S_1, S_2 \in \mathbb S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \c... | Image of Intersection under One-to-Many Relation/General Result | https://proofwiki.org/wiki/Image_of_Intersection_under_One-to-Many_Relation/General_Result | https://proofwiki.org/wiki/Image_of_Intersection_under_One-to-Many_Relation/General_Result | [
"Image of Intersection under One-to-Many Relation"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Power Set",
"Definition:One-to-Many Relation"
] | [
"Definition:Subset",
"Image of Intersection under One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:One-to-Many Relation",
"Definition:Subset"
] |
proofwiki-5595 | Image of Union under Relation/General Result | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \RR \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} \RR \sqbrk X$ | {{begin-eqn}}
{{eqn | l = t
| o = \in
| r = \RR \sqbrk {\bigcup \mathbb S}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists s \in \bigcup \mathbb S
| l = t
| o = \in
| r = \map \RR s
| c = Image of Subset under Relation equals Union of Images of Elements
}}
{{eqn | ll= \leadstoandf... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \RR \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} \RR \sqbrk X$ | {{begin-eqn}}
{{eqn | l = t
| o = \in
| r = \RR \sqbrk {\bigcup \mathbb S}
}}
{{eqn | ll= \leadstoandfrom
| q = \exists s \in \bigcup \mathbb S
| l = t
| o = \in
| r = \map \RR s
| c = [[Image of Subset under Relation equals Union of Images of Elements]]
}}
{{eqn | ll= \leadsto... | Image of Union under Relation/General Result | https://proofwiki.org/wiki/Image_of_Union_under_Relation/General_Result | https://proofwiki.org/wiki/Image_of_Union_under_Relation/General_Result | [
"Image of Union under Relation"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Power Set"
] | [
"Image of Subset under Relation equals Union of Images of Elements",
"Category:Image of Union under Relation"
] |
proofwiki-5596 | Image of Union under Mapping/General Result | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds f \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} f \sqbrk X$
This can be expressed in the language and notation of direct image mappings as:
:$\ds \forall... | As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation: General Result:
:$\ds \RR \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} \RR \sqbrk X$
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds f \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} f \sqbrk X$
This can be express... | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Image of Union under Relation/General Result|Image of Union under Relation: General Result]]:
:$\ds \RR \sqbrk {\bigcup \mathbb S} = \bigcup_{X \mathop \in \mathbb S} \RR \sqbrk X$
{{qed}} | Image of Union under Mapping/General Result | https://proofwiki.org/wiki/Image_of_Union_under_Mapping/General_Result | https://proofwiki.org/wiki/Image_of_Union_under_Mapping/General_Result | [
"Image of Union under Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Power Set",
"Definition:Direct Image Mapping"
] | [
"Definition:Mapping",
"Definition:Relation",
"Image of Union under Relation/General Result"
] |
proofwiki-5597 | Preimage of Union under Mapping/General Result | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds f^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} f^{-1} \sqbrk X$ | As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation: General Result:
:$\ds \RR^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $\powerset T$ be the [[Definition:Power Set|power set]] of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds f^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} f^{-1} \sqbrk X$ | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Preimage of Union under Relation/General Result|Preimage of Union under Relation: General Result]]:
:$\ds \RR^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$
{{qed}} | Preimage of Union under Mapping/General Result | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping/General_Result | https://proofwiki.org/wiki/Preimage_of_Union_under_Mapping/General_Result | [
"Preimage of Union under Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Power Set"
] | [
"Definition:Mapping",
"Definition:Relation",
"Preimage of Union under Relation/General Result"
] |
proofwiki-5598 | Preimage of Union under Relation/General Result | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds \RR^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} \RR^{-1} \sqbrk X$
where $\RR^{-1} \sqbrk X$ denotes the preimage of $X$ under $\RR... | We have that $\RR^{-1}$ is a relation.
The result follows from Image of Union under Relation: General Result.
{{qed}}
Category:Preimage of Union under Relation
16qvgywv4ku7oc1lyi9d44gxd95zlxw | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]].
Let $\powerset T$ be the [[Definition:Power Set|power set]] of $T$.
Let $\mathbb T \subseteq \powerset T$.
Then:
:$\ds \RR^{-1} \sqbrk {\bigcup \mathbb T} = \bigcup_{X \mathop \in \mathbb T} \RR^{-1} \... | We have that $\RR^{-1}$ is a [[Definition:Relation|relation]].
The result follows from [[Image of Union under Relation/General Result|Image of Union under Relation: General Result]].
{{qed}}
[[Category:Preimage of Union under Relation]]
16qvgywv4ku7oc1lyi9d44gxd95zlxw | Preimage of Union under Relation/General Result | https://proofwiki.org/wiki/Preimage_of_Union_under_Relation/General_Result | https://proofwiki.org/wiki/Preimage_of_Union_under_Relation/General_Result | [
"Preimage of Union under Relation"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Power Set",
"Definition:Preimage/Relation/Subset"
] | [
"Definition:Relation",
"Image of Union under Relation/General Result",
"Category:Preimage of Union under Relation"
] |
proofwiki-5599 | Image of Intersection under Mapping/General Result | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds f \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$ | As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation: General Result:
:$\ds \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds f \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$ | As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], we can apply [[Image of Intersection under Relation/General Result|Image of Intersection under Relation: General Result]]:
:$\ds \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
{{qed}} | Image of Intersection under Mapping/General Result | https://proofwiki.org/wiki/Image_of_Intersection_under_Mapping/General_Result | https://proofwiki.org/wiki/Image_of_Intersection_under_Mapping/General_Result | [
"Image of Intersection under Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Power Set"
] | [
"Definition:Mapping",
"Definition:Relation",
"Image of Intersection under Relation/General Result"
] |
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