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-2800 | Bijection iff Left and Right Cancellable | Let $f$ be a mapping.
Then $f$ is a bijection {{iff}} $f$ is both left cancellable and right cancellable. | Follows directly from:
:Injection iff Left Cancellable: $f$ is an injection {{iff}} $f$ is left cancellable
:Surjection iff Right Cancellable: $f$ is a surjection {{iff}} $f$ is right cancellable.
{{qed}} | Let $f$ be a [[Definition:Mapping|mapping]].
Then $f$ is a [[Definition:Bijection|bijection]] {{iff}} $f$ is both [[Definition:Left Cancellable Mapping|left cancellable]] and [[Definition:Right Cancellable Mapping|right cancellable]]. | Follows directly from:
:[[Injection iff Left Cancellable]]: $f$ is an [[Definition:Injection|injection]] {{iff}} $f$ is [[Definition:Left Cancellable Mapping|left cancellable]]
:[[Surjection iff Right Cancellable]]: $f$ is a [[Definition:Surjection|surjection]] {{iff}} $f$ is [[Definition:Right Cancellable Mapping|righ... | Bijection iff Left and Right Cancellable | https://proofwiki.org/wiki/Bijection_iff_Left_and_Right_Cancellable | https://proofwiki.org/wiki/Bijection_iff_Left_and_Right_Cancellable | [
"Bijections"
] | [
"Definition:Mapping",
"Definition:Bijection",
"Definition:Left Cancellable Mapping",
"Definition:Right Cancellable Mapping"
] | [
"Injection iff Left Cancellable",
"Definition:Injection",
"Definition:Left Cancellable Mapping",
"Surjection iff Right Cancellable",
"Definition:Surjection",
"Definition:Right Cancellable Mapping"
] |
proofwiki-2801 | Relation Induced by Mapping is Equivalence Relation | Let $f: S \to T$ be a mapping.
Let $\RR_f \subseteq S \times S$ be the relation induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Then $\RR_f$ is an equivalence relation. | We need to show that $\RR_f$ is an equivalence relation.
Checking in turn each of the criteria for equivalence: | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $\RR_f \subseteq S \times S$ be the [[Definition:Equivalence Relation Induced by Mapping|relation induced by $f$]]:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Then $\RR_f$ is an [[Definition:Equivalence Relation|equivalence relation]]. | We need to show that $\RR_f$ is an [[Definition:Equivalence Relation|equivalence relation]].
Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Relation Induced by Mapping is Equivalence Relation | https://proofwiki.org/wiki/Relation_Induced_by_Mapping_is_Equivalence_Relation | https://proofwiki.org/wiki/Relation_Induced_by_Mapping_is_Equivalence_Relation | [
"Mapping Theory",
"Equivalence Relations",
"Examples of Equivalence Relations"
] | [
"Definition:Mapping",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-2802 | Canonical Injection is Monomorphism | Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be algebraic structures with identities $e_1, e_2$ respectively.
The canonical injections:
:$\inj_1: \struct {S_1, \circ_1} \to \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}: \forall x \in S_1: \map {\inj_1} x = \tuple {x, e_2}$
:$\inj_2: \struct {S_2, \c... | From Canonical Injection is Injection we have that the canonical injections are in fact injective.
It remains to prove the morphism property.
Let $x, y \in \struct {S_1, \circ_1}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\inj_1} {x \circ_1 y}
| r = \tuple {x \circ_1 y, e_2}
}}
{{eqn | r = \tuple {x \circ_1 y, e_2 \... | Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]] with [[Definition:Identity Element|identities]] $e_1, e_2$ respectively.
The [[Definition:Canonical Injection (Abstract Algebra)|canonical injections]]:
:$\inj_1: \struct {S_1, \cir... | From [[Canonical Injection is Injection]] we have that the [[Definition:Canonical Injection (Abstract Algebra)|canonical injections]] are in fact [[Definition:Injection|injective]].
It remains to prove the [[Definition:Morphism Property|morphism property]].
Let $x, y \in \struct {S_1, \circ_1}$.
Then:
{{begin-eqn}... | Canonical Injection is Monomorphism | https://proofwiki.org/wiki/Canonical_Injection_is_Monomorphism | https://proofwiki.org/wiki/Canonical_Injection_is_Monomorphism | [
"Canonical Injection is Monomorphism",
"Canonical Injections",
"Monomorphisms (Abstract Algebra)"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Monomorphism (Abstract Algebra)"
] | [
"Canonical Injection is Injection",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Injection",
"Definition:Morphism Property",
"Definition:Morphism Property",
"Definition:Injection",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Monomorphism (Abstract Algebra)"
] |
proofwiki-2803 | Order Embedding is Injection | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be an order embedding.
That is:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
Then $\phi$ is an injection. | Suppose $\phi: S \to T$ is a mapping such that:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
Then, for all $x, y \in S$:
{{begin-eqn}}
{{eqn | l = \map \phi x
| r = \map \phi y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x \preceq_2 \map \phi y
| o = \land
... | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $\phi: S \to T$ be an [[Definition:Order Embedding/Definition 1|order embedding]].
That is:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
Then $\phi$ is an [[Definition:Injection|... | Suppose $\phi: S \to T$ is a [[Definition:Mapping|mapping]] such that:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
Then, for all $x, y \in S$:
{{begin-eqn}}
{{eqn | l = \map \phi x
| r = \map \phi y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi x \preceq_2 \map \phi ... | Order Embedding is Injection | https://proofwiki.org/wiki/Order_Embedding_is_Injection | https://proofwiki.org/wiki/Order_Embedding_is_Injection | [
"Order Embeddings",
"Injections"
] | [
"Definition:Ordered Set",
"Definition:Order Embedding/Definition 1",
"Definition:Injection"
] | [
"Definition:Mapping",
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Ordering",
"Definition:Antisymmetric Relation",
"Definition:Injection"
] |
proofwiki-2804 | Order Isomorphism is Surjective Order Embedding | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $f: S \to T$ be a mapping.
Then $f$ is an order isomorphism {{iff}}:
:$(1): \quad f$ is a surjection
:$(2): \quad \forall x, y \in S: x \preceq_1 y \iff \map f x \preceq_2 \map f y$
That is, {{iff}} $f$ is an order embedding which is also a ... | === Necessary Condition ===
Suppose $f$ is an order isomorphism.
Then by definition $f$ is a bijection and so a surjection.
Also by definition, $f$ is increasing, and so:
:$\forall x, y \in S: x \preceq_1 y \implies \map f x \preceq_2 \map f y$
Also by definition $f^{-1}$ is also a bijection which is increasing, and so... | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then $f$ is an [[Definition:Order Isomorphism|order isomorphism]] {{iff}}:
:$(1): \quad f$ is a [[Definition:Surjection|surjection]]
:$(2): \quad \forall x, y \i... | === Necessary Condition ===
Suppose $f$ is an [[Definition:Order Isomorphism|order isomorphism]].
Then by definition $f$ is a [[Definition:Bijection|bijection]] and so a [[Definition:Surjection|surjection]].
Also by definition, $f$ is [[Definition:Increasing|increasing]], and so:
:$\forall x, y \in S: x \preceq_1 y ... | Order Isomorphism is Surjective Order Embedding | https://proofwiki.org/wiki/Order_Isomorphism_is_Surjective_Order_Embedding | https://proofwiki.org/wiki/Order_Isomorphism_is_Surjective_Order_Embedding | [
"Order Isomorphisms",
"Order Embeddings",
"Surjections"
] | [
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Order Isomorphism",
"Definition:Surjection",
"Definition:Order Embedding",
"Definition:Surjection"
] | [
"Definition:Order Isomorphism",
"Definition:Bijection",
"Definition:Surjection",
"Definition:Increasing",
"Definition:Bijection",
"Definition:Increasing",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Orde... |
proofwiki-2805 | Infimum of Power Set | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the relation $\subseteq$.
(From Subset Relation on Power Set is Partial Ordering, this is an ordered set.)
Then the infimum of $\struct {\powerset S, \subseteq}$ is... | Follows directly from Empty Set is Subset of All Sets and the definition of infimum.
{{qed}}
Category:Power Set
Category:Empty Set
Category:Order Theory
mxzl51u3522bax9ahoq1hxuit81hyvj | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\struct {\powerset S, \subseteq}$ be the [[Definition:Relational Structure|relational structure]] defined on $\powerset S$ by the [[Definition:Relation|relation]] $\subseteq$.
(From [[Subset Relation on Po... | Follows directly from [[Empty Set is Subset of All Sets]] and the definition of [[Definition:Infimum of Set|infimum]].
{{qed}}
[[Category:Power Set]]
[[Category:Empty Set]]
[[Category:Order Theory]]
mxzl51u3522bax9ahoq1hxuit81hyvj | Infimum of Power Set | https://proofwiki.org/wiki/Infimum_of_Power_Set | https://proofwiki.org/wiki/Infimum_of_Power_Set | [
"Power Set",
"Empty Set",
"Order Theory"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Relational Structure",
"Definition:Relation",
"Subset Relation on Power Set is Partial Ordering",
"Definition:Ordered Set",
"Definition:Infimum of Set",
"Definition:Empty Set"
] | [
"Empty Set is Subset of All Sets",
"Definition:Infimum of Set",
"Category:Power Set",
"Category:Empty Set",
"Category:Order Theory"
] |
proofwiki-2806 | Supremum of Power Set | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the relation $\subseteq$.
(From Subset Relation on Power Set is Partial Ordering, this is an ordered set.)
Then the supremum of $\struct {\powerset S, \subseteq}$ i... | By the definition of the power set:
:$\forall X \in \powerset S: X \subseteq S$
The result then follows from the definition of supremum.
{{qed}}
Category:Power Set
Category:Order Theory
n7tk2sqicrbpkrz6e4rw2pl32muxe4c | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\struct {\powerset S, \subseteq}$ be the [[Definition:Relational Structure|relational structure]] defined on $\powerset S$ by the [[Definition:Relation|relation]] $\subseteq$.
(From [[Subset Relation on Po... | By the definition of the [[Definition:Power Set|power set]]:
:$\forall X \in \powerset S: X \subseteq S$
The result then follows from the definition of [[Definition:Supremum of Set|supremum]].
{{qed}}
[[Category:Power Set]]
[[Category:Order Theory]]
n7tk2sqicrbpkrz6e4rw2pl32muxe4c | Supremum of Power Set | https://proofwiki.org/wiki/Supremum_of_Power_Set | https://proofwiki.org/wiki/Supremum_of_Power_Set | [
"Power Set",
"Order Theory"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Relational Structure",
"Definition:Relation",
"Subset Relation on Power Set is Partial Ordering",
"Definition:Ordered Set",
"Definition:Supremum of Set"
] | [
"Definition:Power Set",
"Definition:Supremum of Set",
"Category:Power Set",
"Category:Order Theory"
] |
proofwiki-2807 | Diagonal Relation is Ordering and Equivalence | Let $\struct {S, \Delta_S}$ be a relational structure where $\Delta_S$ is the diagonal relation, defined as:
:$\forall x, y \in S: \paren {x, y} \in \Delta_S \iff x = y$
Then $\Delta_S$ is the '''only''' relation on $S$ which is both an equivalence and an ordering. | From Trivial Ordering is Universally Compatible and Diagonal Relation is Equivalence we know that the diagonal relation possesses these properties.
We now need to show that it is the '''only''' relation on $S$ which possesses these properties.
Both an equivalence and an ordering are reflexive and transitive.
Also:
:An ... | Let $\struct {S, \Delta_S}$ be a [[Definition:Relational Structure|relational structure]] where $\Delta_S$ is the [[Definition:Diagonal Relation|diagonal relation]], defined as:
:$\forall x, y \in S: \paren {x, y} \in \Delta_S \iff x = y$
Then $\Delta_S$ is the '''only''' [[Definition:Relation|relation]] on $S$ whic... | From [[Trivial Ordering is Universally Compatible]] and [[Diagonal Relation is Equivalence]] we know that the [[Definition:Diagonal Relation|diagonal relation]] possesses these properties.
We now need to show that it is the '''only''' [[Definition:Relation|relation]] on $S$ which possesses these properties.
Both an ... | Diagonal Relation is Ordering and Equivalence | https://proofwiki.org/wiki/Diagonal_Relation_is_Ordering_and_Equivalence | https://proofwiki.org/wiki/Diagonal_Relation_is_Ordering_and_Equivalence | [
"Examples of Equivalence Relations",
"Order Theory"
] | [
"Definition:Relational Structure",
"Definition:Diagonal Relation",
"Definition:Relation",
"Definition:Equivalence Relation",
"Definition:Ordering"
] | [
"Trivial Ordering is Universally Compatible",
"Diagonal Relation is Equivalence",
"Definition:Diagonal Relation",
"Definition:Relation",
"Definition:Equivalence Relation",
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Transitive Relation",
"Definition:Equivalence Relation",
"... |
proofwiki-2808 | Cantor-Bernstein-Schröder Theorem/Proof 3 | Let $S$ and $T$ be sets, such that:
:$\exists f: S \to T$ such that $f$ is an injection
:$\exists g: T \to S$ such that $g$ is an injection.
Then there exists a bijection from $S$ to $T$. | Let $S, T$ be sets.
Let $\powerset S, \powerset T$ denote their power sets.
Let $f: S \to T$ and $g: T \to S$ be injections that we know to exist between $S$ and $T$.
Consider the relative complements of elements of $\powerset S$ and $\powerset T$ as mappings:
:$\complement_S: \powerset S \to \powerset S: \forall X \in... | Let $S$ and $T$ be [[Definition:Set|sets]], such that:
:$\exists f: S \to T$ such that $f$ is an [[Definition:Injection|injection]]
:$\exists g: T \to S$ such that $g$ is an [[Definition:Injection|injection]].
Then there exists a [[Definition:Bijection|bijection]] from $S$ to $T$. | Let $S, T$ be [[Definition:Set|sets]].
Let $\powerset S, \powerset T$ denote their [[Definition:Power Set|power sets]].
Let $f: S \to T$ and $g: T \to S$ be [[Definition:Injection|injections]] that we know to exist between $S$ and $T$.
Consider the [[Definition:Relative Complement|relative complements]] of elements... | Cantor-Bernstein-Schröder Theorem/Proof 3 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_3 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_3 | [
"Cantor-Bernstein-Schröder Theorem"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Injection",
"Definition:Bijection"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Injection",
"Definition:Relative Complement",
"Definition:Mapping",
"Definition:Relative Complement",
"Definition:Direct Image Mapping/Mapping",
"Definition:Mapping",
"Definition:Composition of Mappings",
"Image of Subset under Mapping is Subse... |
proofwiki-2809 | Cantor-Bernstein-Schröder Theorem/Proof 2 | Let $S$ and $T$ be sets, such that:
:$S \preccurlyeq T$, that is: $T$ dominates $S$
:$T \preccurlyeq S$, that is: $S$ dominates $T$.
Then:
:$S \sim T$
that is, $S$ is equivalent to $T$. | Suppose $S \preccurlyeq T$ and $T \preccurlyeq S$.
By definition, we have that there exist injections $f: S \to T$ and $g: T \to S$.
We are going to try to build a sequence $t_1, s_1, t_2, s_2, t_3 \ldots$ as follows.
Consider any $t_1 \in T$.
By Law of Excluded Middle there are two choices:
{{begin-eqn}}
{{eqn | q = \... | Let $S$ and $T$ be [[Definition:Set|sets]], such that:
:$S \preccurlyeq T$, that is: $T$ [[Definition:Dominate (Set Theory)|dominates]] $S$
:$T \preccurlyeq S$, that is: $S$ [[Definition:Dominate (Set Theory)|dominates]] $T$.
Then:
:$S \sim T$
that is, $S$ is [[Definition:Set Equivalence|equivalent]] to $T$. | Suppose $S \preccurlyeq T$ and $T \preccurlyeq S$.
By [[Definition:Dominate (Set Theory)|definition]], we have that there exist [[Definition:Injection|injections]] $f: S \to T$ and $g: T \to S$.
We are going to try to build a [[Definition:Sequence|sequence]] $t_1, s_1, t_2, s_2, t_3 \ldots$ as follows.
Consider any... | Cantor-Bernstein-Schröder Theorem/Proof 2 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_2 | https://proofwiki.org/wiki/Cantor-Bernstein-Schröder_Theorem/Proof_2 | [
"Cantor-Bernstein-Schröder Theorem"
] | [
"Definition:Set",
"Definition:Dominate (Set Theory)",
"Definition:Dominate (Set Theory)",
"Definition:Set Equivalence"
] | [
"Definition:Dominate (Set Theory)",
"Definition:Injection",
"Definition:Sequence",
"Law of Excluded Middle",
"Definition:Injection",
"Law of Excluded Middle",
"Definition:Injection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Set Partition",
"Definition:Subset",
"Definition:... |
proofwiki-2810 | Dominance Relation is Ordering | Let $S$ and $T$ be cardinals.
Let $S \preccurlyeq T$ denote that $S$ is dominated by $T$.
Let $\mathbb S$ be any set of cardinals.
Then the relational structure $\struct {\mathbb S, \preccurlyeq}$ is an ordered set.
That is, $\preccurlyeq$ is an ordering (at least partial) on $\mathbb S$. | From the definition, a cardinal is a set, so standard set theoretic results apply.
So, checking in turn each of the criteria for an ordering: | Let $S$ and $T$ be [[Definition:Cardinal|cardinals]].
Let $S \preccurlyeq T$ denote that $S$ [[Definition:Dominate (Set Theory)|is dominated by]] $T$.
Let $\mathbb S$ be any [[Definition:Set|set]] of [[Definition:Cardinal|cardinals]].
Then the [[Definition:Relational Structure|relational structure]] $\struct {\math... | From the definition, a [[Definition:Cardinal|cardinal]] is a [[Definition:Set|set]], so standard set theoretic results apply.
So, checking in turn each of the criteria for an [[Definition:Ordering|ordering]]: | Dominance Relation is Ordering | https://proofwiki.org/wiki/Dominance_Relation_is_Ordering | https://proofwiki.org/wiki/Dominance_Relation_is_Ordering | [
"Set Theory",
"Order Theory"
] | [
"Definition:Cardinal",
"Definition:Dominate (Set Theory)",
"Definition:Set",
"Definition:Cardinal",
"Definition:Relational Structure",
"Definition:Ordered Set",
"Definition:Ordering",
"Definition:Partial Ordering"
] | [
"Definition:Cardinal",
"Definition:Set",
"Definition:Ordering",
"Definition:Cardinal",
"Definition:Cardinal"
] |
proofwiki-2811 | Huygens-Steiner Theorem | Let $B$ be a body of mass $M$.
Let $I_0$ be the moment of inertia of $B$ about some axis $A$ through the centre of mass of $B$.
Let $I$ be the moment of inertia of $B$ about another axis $A'$ parallel to $A$.
Then $I_0$ and $I$ are related by:
:$I = I_0 + M l^2$
where $l$ is the distance between $A$ and $A'$. | {{WLOG}}, suppose $I$ is oriented along the $z$-axis.
By definition of moment of inertia:
:$\ds I = \sum m_j \lambda_j^2$
:$\ds I_0 = \sum m_j \lambda_j'^2$
where:
:$\lambda_j$ is the position vector to the $j$th particle from the $z$-axis
:$\lambda_j'$ is related to $\lambda_j$ by:
::$\lambda_j = \lambda_j' + R_\perp$... | Let $B$ be a [[Definition:Body|body]] of [[Definition:Mass|mass]] $M$.
Let $I_0$ be the [[Definition:Moment of Inertia|moment of inertia]] of $B$ about some [[Definition:Axis of Rotation|axis]] $A$ through the [[Definition:Center of Mass|centre of mass]] of $B$.
Let $I$ be the [[Definition:Moment of Inertia|moment of... | {{WLOG}}, suppose $I$ is oriented along the [[Definition:Z-Axis|$z$-axis]].
By definition of [[Definition:Moment of Inertia|moment of inertia]]:
:$\ds I = \sum m_j \lambda_j^2$
:$\ds I_0 = \sum m_j \lambda_j'^2$
where:
:$\lambda_j$ is the [[Definition:Position Vector|position vector]] to the $j$th [[Definition:Part... | Huygens-Steiner Theorem | https://proofwiki.org/wiki/Huygens-Steiner_Theorem | https://proofwiki.org/wiki/Huygens-Steiner_Theorem | [
"Huygens-Steiner Theorem",
"Moments of Inertia"
] | [
"Definition:Body",
"Definition:Mass",
"Definition:Moment of Inertia",
"Definition:Rotation (Geometry)/Axis",
"Definition:Center of Mass",
"Definition:Moment of Inertia",
"Definition:Rotation (Geometry)/Axis",
"Definition:Parallel (Geometry)/Lines",
"Definition:Distance between Parallel Lines"
] | [
"Definition:Axis/Z-Axis",
"Definition:Moment of Inertia",
"Definition:Position Vector",
"Definition:Particle",
"Definition:Axis/Z-Axis",
"Definition:Perpendicular Distance between Point and Straight Line",
"Definition:Center of Mass"
] |
proofwiki-2812 | Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2 | Let $\struct {S, \preceq}$ be an ordered set such that:
:$\forall T \subseteq S: \exists a \in T: \forall x \in T: a \preceq x$
That is, such that $\preceq$ is a well-ordering by definition 1.
Then $\preceq$ is a total ordering. | Consider $X = \set {a, b}$ where $a, b \in S$.
We have {{hypothesis}} that $X$ has a smallest element.
So either $\min X = a$ or $\min X = b$.
If $\min X = a$, then $a \preceq b$.
If $\min X = b$, then $b \preceq a$.
So either $a \preceq b$ or $b \preceq a$.
That is, $a$ and $b$ are comparable.
As this applies to all $... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]] such that:
:$\forall T \subseteq S: \exists a \in T: \forall x \in T: a \preceq x$
That is, such that $\preceq$ is a [[Definition:Well-Ordering/Definition 1|well-ordering by definition 1]].
Then $\preceq$ is a [[Definition:Total Ordering|total o... | Consider $X = \set {a, b}$ where $a, b \in S$.
We have {{hypothesis}} that $X$ has a [[Definition:Smallest Element|smallest element]].
So either $\min X = a$ or $\min X = b$.
If $\min X = a$, then $a \preceq b$.
If $\min X = b$, then $b \preceq a$.
So either $a \preceq b$ or $b \preceq a$.
That is, $a$ and $b$ ar... | Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Well-Ordering/Definition_1_implies_Definition_2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Well-Ordering/Definition_1_implies_Definition_2 | [
"Equivalence of Definitions of Well-Ordering"
] | [
"Definition:Ordered Set",
"Definition:Well-Ordering/Definition 1",
"Definition:Total Ordering"
] | [
"Definition:Smallest Element",
"Definition:Comparable Elements",
"Definition:Ordering",
"Definition:Total Ordering"
] |
proofwiki-2813 | Choice Function Exists for Well-Orderable Union of Sets | Let $\mathbb S$ be a set of sets such that:
:$\forall S \in \mathbb S: S \ne \O$
that is, none of the sets in $\mathbb S$ may be empty.
Let the union $\bigcup \mathbb S$ be well-orderable.
Then there exists a choice function $f: \mathbb S \to \bigcup \mathbb S$ defined as:
:$\forall S \in \mathbb S: \exists x \in S: \m... | Suppose $T = \bigcup \mathbb S$ is well-orderable.
Then we can create a well-ordering $\preceq$ on $T$ so as to make $\struct {T, \preceq}$ a well-ordered set.
From the definition of well-ordered set, every subset of $T$ is itself well-ordered.
From Set is Subset of Union we have that:
:$\forall S \in \mathbb S: S \sub... | Let $\mathbb S$ be a [[Definition:Set of Sets|set of sets]] such that:
:$\forall S \in \mathbb S: S \ne \O$
that is, none of the sets in $\mathbb S$ may be [[Definition:Empty Set|empty]].
Let the [[Definition:Set Union|union]] $\bigcup \mathbb S$ be [[Definition:Well-Orderable Class|well-orderable]].
Then there exis... | Suppose $T = \bigcup \mathbb S$ is [[Definition:Well-Orderable Class|well-orderable]].
Then we can create a [[Definition:Well-Ordering|well-ordering]] $\preceq$ on $T$ so as to make $\struct {T, \preceq}$ a [[Definition:Well-Ordered Set|well-ordered set]].
From the definition of [[Definition:Well-Ordered Set|well-ord... | Choice Function Exists for Well-Orderable Union of Sets | https://proofwiki.org/wiki/Choice_Function_Exists_for_Well-Orderable_Union_of_Sets | https://proofwiki.org/wiki/Choice_Function_Exists_for_Well-Orderable_Union_of_Sets | [
"Set Union",
"Well-Orderings",
"Choice Functions"
] | [
"Definition:Set of Sets",
"Definition:Empty Set",
"Definition:Set Union",
"Definition:Well-Orderable Set/Class Theory",
"Definition:Choice Function"
] | [
"Definition:Well-Orderable Set/Class Theory",
"Definition:Well-Ordering",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered Set",
"Set is Subset of Union/General Result",
"Choice Function Exists for Set of Well-Ordered Sets",
"Definition:Element",
"Definition:Wel... |
proofwiki-2814 | Principle of Finite Choice | Let $I$ be a non-empty finite indexing set.
Let $\family {S_i}_{i \mathop \in I}$ be an $I$-indexed family of non-empty sets.
Then there exists an $I$-indexed family $\family {x_i}_{i \mathop \in I}$ such that:
:$\forall i \in I: x_i \in S_i$
That is, there exists a mapping:
:$\ds f: I \to \bigcup_{i \mathop \in I} S_i... | We use the Principle of Mathematical Induction on the cardinality of $I$. | Let $I$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Indexing Set|indexing set]].
Let $\family {S_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family of Sets|$I$-indexed family]] of [[Definition:Non-Empty Set|non-empty sets]].
Then there exists an [[Definition:Indexed... | We use the [[Principle of Mathematical Induction]] on the [[Definition:Cardinality|cardinality]] of $I$. | Principle of Finite Choice | https://proofwiki.org/wiki/Principle_of_Finite_Choice | https://proofwiki.org/wiki/Principle_of_Finite_Choice | [
"Choice Functions",
"Named Theorems",
"Proofs by Induction"
] | [
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Non-Empty Set",
"Definition:Indexing Set/Family",
"Definition:Mapping"
] | [
"Principle of Mathematical Induction",
"Definition:Cardinality"
] |
proofwiki-2815 | Choice Function Exists for Set of Well-Ordered Sets | Let $\mathbb S$ be a set of sets such that:
:$\forall S \in \mathbb S: S \ne \O$
that is, none of the sets in $\mathbb S$ may be empty.
Let every element of $\mathbb S$ be well-ordered.
Then there exists a choice function $f: \mathbb S \to \bigcup \mathbb S$ satisfying:
:$\forall S \in \mathbb S: \exists x \in S: \map ... | Every member of $\mathbb S$ is a well-ordered set.
Thus, for $S \in \mathbb S$, there is a minimal element $s$ for $S$ (with respect to the ordering of $S$).
By Well-Ordering Minimal Elements are Unique, $s$ is unique.
Therefore, we can define $f$ by:
:$\forall S \in \mathbb S: \map f S = s$
{{qed}} | Let $\mathbb S$ be a [[Definition:Set of Sets|set of sets]] such that:
:$\forall S \in \mathbb S: S \ne \O$
that is, none of the sets in $\mathbb S$ may be [[Definition:Empty Set|empty]].
Let every [[Definition:Element|element]] of $\mathbb S$ be [[Definition:Well-Ordered Set|well-ordered]].
Then there exists a [[De... | Every member of $\mathbb S$ is a [[Definition:Well-Ordered Set|well-ordered set]].
Thus, for $S \in \mathbb S$, there is a [[Definition:Minimal Element|minimal element]] $s$ for $S$ (with respect to the ordering of $S$).
By [[Well-Ordering Minimal Elements are Unique]], $s$ is unique.
Therefore, we can define $f$ ... | Choice Function Exists for Set of Well-Ordered Sets | https://proofwiki.org/wiki/Choice_Function_Exists_for_Set_of_Well-Ordered_Sets | https://proofwiki.org/wiki/Choice_Function_Exists_for_Set_of_Well-Ordered_Sets | [
"Well-Orderings",
"Choice Functions"
] | [
"Definition:Set of Sets",
"Definition:Empty Set",
"Definition:Element",
"Definition:Well-Ordered Set",
"Definition:Choice Function"
] | [
"Definition:Well-Ordered Set",
"Definition:Minimal/Element",
"Well-Ordering Minimal Elements are Unique"
] |
proofwiki-2816 | Identity of Power Set with Union | Let $S$ be a set and let $\powerset S$ be its power set.
Consider the algebraic structure $\struct {\powerset S, \cup}$, where $\cup$ denotes set union.
Then the empty set $\O$ serves as the identity for $\struct {\powerset S, \cup}$. | From Empty Set is Element of Power Set:
:$\O \in \powerset S$
From Union with Empty Set:
:$\forall A \subseteq S: A \cup \O = A = \O \cup A$
By definition of power set:
:$A \subseteq S \iff A \in \powerset S$
So:
:$\forall A \in \powerset S: A \cup \O = A = \O \cup A$
Thus we see that $\O$ acts as the identity for $\st... | Let $S$ be a [[Definition:Set|set]] and let $\powerset S$ be its [[Definition:Power Set|power set]].
Consider the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\powerset S, \cup}$, where $\cup$ denotes [[Definition:Set Union|set union]].
Then the [[Definition:Empty Set|empty set]] $\O$ serves as t... | From [[Empty Set is Element of Power Set]]:
:$\O \in \powerset S$
From [[Union with Empty Set]]:
:$\forall A \subseteq S: A \cup \O = A = \O \cup A$
By definition of [[Definition:Power Set|power set]]:
:$A \subseteq S \iff A \in \powerset S$
So:
:$\forall A \in \powerset S: A \cup \O = A = \O \cup A$
Thus we see t... | Identity of Power Set with Union | https://proofwiki.org/wiki/Identity_of_Power_Set_with_Union | https://proofwiki.org/wiki/Identity_of_Power_Set_with_Union | [
"Set Union",
"Power Set",
"Empty Set",
"Examples of Identity Elements"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Algebraic Structure",
"Definition:Set Union",
"Definition:Empty Set",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Empty Set is Element of Power Set",
"Union with Empty Set",
"Definition:Power Set",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-2817 | Identity of Power Set with Intersection | Let $S$ be a set and let $\powerset S$ be its power set.
Consider the algebraic structure $\struct {\powerset S, \cap}$, where $\cap$ denotes set intersection.
Then $S$ serves as the identity for $\struct {\powerset S, \cap}$. | We note that by Set is Subset of Itself, $S \subseteq S$ and so $S \in \powerset S$ from the definition of the power set.
From Intersection with Subset is Subset, we have:
:$A \subseteq S \iff A \cap S = A = S \cap A$
By definition of power set:
:$A \subseteq S \iff A \in \powerset S$
So:
:$\forall A \in \powerset S: A... | Let $S$ be a [[Definition:Set|set]] and let $\powerset S$ be its [[Definition:Power Set|power set]].
Consider the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\powerset S, \cap}$, where $\cap$ denotes [[Definition:Set Intersection|set intersection]].
Then $S$ serves as the [[Definition:Identity E... | We note that by [[Set is Subset of Itself]], $S \subseteq S$ and so $S \in \powerset S$ from the definition of the [[Definition:Power Set|power set]].
From [[Intersection with Subset is Subset]], we have:
:$A \subseteq S \iff A \cap S = A = S \cap A$
By definition of [[Definition:Power Set|power set]]:
:$A \subseteq ... | Identity of Power Set with Intersection | https://proofwiki.org/wiki/Identity_of_Power_Set_with_Intersection | https://proofwiki.org/wiki/Identity_of_Power_Set_with_Intersection | [
"Set Intersection",
"Power Set",
"Examples of Identity Elements"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Algebraic Structure",
"Definition:Set Intersection",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Set is Subset of Itself",
"Definition:Power Set",
"Intersection with Subset is Subset",
"Definition:Power Set",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-2818 | Commutation with Inverse in Monoid | Let $x, y \in S$ such that $y$ is invertible.
Then $x$ commutes with $y$ {{iff}} $x$ commutes with $y^{-1}$. | === Necessary Condition ===
Let $x$ commute with $y$.
Then:
{{begin-eqn}}
{{eqn | l = y^{-1} \circ x
| r = \paren {y^{-1} \circ x} \circ e
| c = {{MonoidAxiom|2}}
}}
{{eqn | r = y^{-1} \circ \paren {x \circ e}
| c = {{MonoidAxiom|1}}
}}
{{eqn | r = y^{-1} \circ \paren {x \circ \paren {y \circ y^{-1} }... | Let $x, y \in S$ such that $y$ is [[Definition:Invertible Element|invertible]].
Then $x$ [[Definition:Commuting Elements|commutes]] with $y$ {{iff}} $x$ [[Definition:Commuting Elements|commutes]] with $y^{-1}$. | === Necessary Condition ===
Let $x$ [[Definition:Commuting Elements|commute]] with $y$.
Then:
{{begin-eqn}}
{{eqn | l = y^{-1} \circ x
| r = \paren {y^{-1} \circ x} \circ e
| c = {{MonoidAxiom|2}}
}}
{{eqn | r = y^{-1} \circ \paren {x \circ e}
| c = {{MonoidAxiom|1}}
}}
{{eqn | r = y^{-1} \circ \pa... | Commutation with Inverse in Monoid | https://proofwiki.org/wiki/Commutation_with_Inverse_in_Monoid | https://proofwiki.org/wiki/Commutation_with_Inverse_in_Monoid | [
"Inverse Elements",
"Monoids",
"Commutativity"
] | [
"Definition:Invertible Element",
"Definition:Commutative/Elements",
"Definition:Commutative/Elements"
] | [
"Definition:Commutative/Elements",
"Definition:Invertible Element",
"Definition:Commutative/Elements",
"Definition:Invertible Element",
"Definition:Commutative/Elements",
"Definition:Commutative/Elements"
] |
proofwiki-2819 | Commutation of Inverses in Monoid | Let $x, y \in S$ such that $x$ and $y$ are both invertible.
Then $x$ commutes with $y$ {{iff}} $x^{-1}$ commutes with $y^{-1}$. | === Necessary Condition ===
Let $x$ commute with $y$.
Then:
{{begin-eqn}}
{{eqn | l = x^{-1} \circ y^{-1}
| r = \paren {y \circ x}^{-1}
| c = Inverse of Product
}}
{{eqn | r = \paren {x \circ y}^{-1}
| c = $x$ commutes with $y$
}}
{{eqn | r = y^{-1} \circ x^{-1}
| c = Inverse of Product
}}
{{en... | Let $x, y \in S$ such that $x$ and $y$ are both [[Definition:Invertible Element|invertible]].
Then $x$ [[Definition:Commute|commutes]] with $y$ {{iff}} $x^{-1}$ [[Definition:Commute|commutes]] with $y^{-1}$. | === Necessary Condition ===
Let $x$ [[Definition:Commute|commute]] with $y$.
Then:
{{begin-eqn}}
{{eqn | l = x^{-1} \circ y^{-1}
| r = \paren {y \circ x}^{-1}
| c = [[Inverse of Product]]
}}
{{eqn | r = \paren {x \circ y}^{-1}
| c = $x$ [[Definition:Commute|commutes]] with $y$
}}
{{eqn | r = y^{-1... | Commutation of Inverses in Monoid | https://proofwiki.org/wiki/Commutation_of_Inverses_in_Monoid | https://proofwiki.org/wiki/Commutation_of_Inverses_in_Monoid | [
"Inverse Elements",
"Commutativity",
"Monoids"
] | [
"Definition:Invertible Element",
"Definition:Commutative/Elements",
"Definition:Commutative/Elements"
] | [
"Definition:Commutative/Elements",
"Inverse of Product",
"Definition:Commutative/Elements",
"Inverse of Product",
"Definition:Commutative/Elements",
"Definition:Commutative/Elements",
"Definition:Commutative/Elements",
"Definition:Commutative/Elements"
] |
proofwiki-2820 | Inverse of Commuting Pair | Let $x, y \in S$ such that $x$ and $y$ are both invertible.
Then $x$ commutes with $y$ {{iff}}:
: $\struct {x \circ y}^{-1} = x^{-1} \circ y^{-1}$ | {{begin-eqn}}
{{eqn | l = x \circ y
| r = y \circ x
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \struct {x \circ y}^{-1}
| r = \struct {y \circ x}^{-1}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \struct {x \circ y}^{-1}
| r = x^{-1} \circ y^{-1}
| c = Inverse of Product
... | Let $x, y \in S$ such that $x$ and $y$ are both [[Definition:Invertible Element|invertible]].
Then $x$ [[Definition:Commuting Elements|commutes]] with $y$ {{iff}}:
: $\struct {x \circ y}^{-1} = x^{-1} \circ y^{-1}$ | {{begin-eqn}}
{{eqn | l = x \circ y
| r = y \circ x
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \struct {x \circ y}^{-1}
| r = \struct {y \circ x}^{-1}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \struct {x \circ y}^{-1}
| r = x^{-1} \circ y^{-1}
| c = [[Inverse of Produc... | Inverse of Commuting Pair | https://proofwiki.org/wiki/Inverse_of_Commuting_Pair | https://proofwiki.org/wiki/Inverse_of_Commuting_Pair | [
"Monoids",
"Group Theory",
"Commutativity",
"Inverse of Commuting Pair"
] | [
"Definition:Invertible Element",
"Definition:Commutative/Elements"
] | [
"Inverse of Product"
] |
proofwiki-2821 | Conjugate of Commuting Elements | Let $x, y \in S$ such that $x$ and $y$ are both invertible.
Then $x \circ y \circ x^{-1} = y$ {{iff}} $x$ and $y$ commute. | As $\struct {S, \circ}$ is a monoid, it is by definition a semigroup.
Therefore it is taken for granted that $\circ$ is associative, so we can dispense with parentheses.
We also take for granted the fact that $x$ and $y$ are cancellable from Invertible Element of Monoid is Cancellable.
So:
{{begin-eqn}}
{{eqn | l = x \... | Let $x, y \in S$ such that $x$ and $y$ are both [[Definition:Invertible Element|invertible]].
Then $x \circ y \circ x^{-1} = y$ {{iff}} $x$ and $y$ [[Definition:Commute|commute]]. | As $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], it is by definition a [[Definition:Semigroup|semigroup]].
Therefore it is taken for granted that $\circ$ is [[Definition:Associative Operation|associative]], so we can dispense with [[Definition:Parenthesis|parentheses]].
We also take for granted the fact tha... | Conjugate of Commuting Elements | https://proofwiki.org/wiki/Conjugate_of_Commuting_Elements | https://proofwiki.org/wiki/Conjugate_of_Commuting_Elements | [
"Monoids",
"Commutativity",
"Conjugacy"
] | [
"Definition:Invertible Element",
"Definition:Commutative/Elements"
] | [
"Definition:Monoid",
"Definition:Semigroup",
"Definition:Associative Operation",
"Definition:Parenthesis",
"Definition:Cancellable Element",
"Invertible Element of Associative Structure is Cancellable/Corollary",
"Category:Monoids",
"Category:Commutativity",
"Category:Conjugacy"
] |
proofwiki-2822 | Product of Commuting Elements with Inverses | Let $x, y \in S$ such that $x$ and $y$ are both invertible.
Then:
:$x \circ y \circ x^{-1} \circ y^{-1} = e_S = x^{-1} \circ y^{-1} \circ x \circ y$
{{iff}} $x$ and $y$ commute. | As $\struct {S, \circ}$ is a monoid, it is by definition a semigroup.
Therefore $\circ$ is associative, so we can dispense with parentheses.
From Invertible Element of Monoid is Cancellable, we also have that $x, y, x^{-1}, y^{-1}$ are cancellable.
So let :
{{begin-eqn}}
{{eqn | l = x \circ y
| r = y \circ x
... | Let $x, y \in S$ such that $x$ and $y$ are both [[Definition:Invertible Element|invertible]].
Then:
:$x \circ y \circ x^{-1} \circ y^{-1} = e_S = x^{-1} \circ y^{-1} \circ x \circ y$
{{iff}} $x$ and $y$ [[Definition:Commuting Elements|commute]]. | As $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], it is by definition a [[Definition:Semigroup|semigroup]].
Therefore $\circ$ is [[Definition:Associative Operation|associative]], so we can dispense with [[Definition:Parenthesis|parentheses]].
From [[Invertible Element of Monoid is Cancellable]], we also have... | Product of Commuting Elements with Inverses | https://proofwiki.org/wiki/Product_of_Commuting_Elements_with_Inverses | https://proofwiki.org/wiki/Product_of_Commuting_Elements_with_Inverses | [
"Monoids",
"Commutativity"
] | [
"Definition:Invertible Element",
"Definition:Commutative/Elements"
] | [
"Definition:Monoid",
"Definition:Semigroup",
"Definition:Associative Operation",
"Definition:Parenthesis",
"Invertible Element of Associative Structure is Cancellable/Corollary",
"Definition:Cancellable Element",
"Definition:Commutative/Elements",
"Conjugate of Commuting Elements",
"Definition:Inver... |
proofwiki-2823 | Exists Element Not in Set | Let $S$ be a set.
Then:
:$\exists x: x \notin S$
That is, for any set, there exists some element which is not in that set. | Consider the power set $\powerset S$ of $S$.
{{AimForCont}} $\forall x \in \powerset S: x \in S$.
Then the identity mapping $I_S: S \to \powerset S$ would be a surjection.
But from Cantor's Theorem, there is no surjection $f: S \to \powerset S$.
From this contradiction it follows that:
:$\exists x \in \powerset S: x \n... | Let $S$ be a [[Definition:Set|set]].
Then:
:$\exists x: x \notin S$
That is, for any [[Definition:Set|set]], there exists some [[Definition:Element|element]] which is not in that [[Definition:Set|set]]. | Consider the [[Definition:Power Set|power set]] $\powerset S$ of $S$.
{{AimForCont}} $\forall x \in \powerset S: x \in S$.
Then the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to \powerset S$ would be a [[Definition:Surjection|surjection]].
But from [[Cantor's Theorem]], there is no [[Definition:Surjec... | Exists Element Not in Set/Proof 1 | https://proofwiki.org/wiki/Exists_Element_Not_in_Set | https://proofwiki.org/wiki/Exists_Element_Not_in_Set/Proof_1 | [
"Exists Element Not in Set",
"Set Theory"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Element",
"Definition:Set"
] | [
"Definition:Power Set",
"Definition:Identity Mapping",
"Definition:Surjection",
"Cantor's Theorem",
"Definition:Surjection",
"Definition:Contradiction"
] |
proofwiki-2824 | Exists Element Not in Set | Let $S$ be a set.
Then:
:$\exists x: x \notin S$
That is, for any set, there exists some element which is not in that set. | By the {{Axiom-link|Specification|Sets}}, we can construct the set:
:$T = \set {x \in S: x \notin x}$
Then for all $y$:
:$(*) \quad y \in T$ {{iff}} $\paren {y \in S \land y \notin y}$.
{{AimForCont}} $T \in S$.
By Law of Excluded Middle, either $T \in T$ or $T \notin T$.
Suppose $T \in T$.
By $(*)$, $T \in S \land T \... | Let $S$ be a [[Definition:Set|set]].
Then:
:$\exists x: x \notin S$
That is, for any [[Definition:Set|set]], there exists some [[Definition:Element|element]] which is not in that [[Definition:Set|set]]. | By the {{Axiom-link|Specification|Sets}}, we can construct the [[Definition:Set|set]]:
:$T = \set {x \in S: x \notin x}$
Then for all $y$:
:$(*) \quad y \in T$ {{iff}} $\paren {y \in S \land y \notin y}$.
{{AimForCont}} $T \in S$.
By [[Law of Excluded Middle]], either $T \in T$ or $T \notin T$.
Suppose $T \in T$.... | Exists Element Not in Set/Proof 2 | https://proofwiki.org/wiki/Exists_Element_Not_in_Set | https://proofwiki.org/wiki/Exists_Element_Not_in_Set/Proof_2 | [
"Exists Element Not in Set",
"Set Theory"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Element",
"Definition:Set"
] | [
"Definition:Set",
"Law of Excluded Middle",
"Rule of Simplification",
"Definition:Contradiction",
"Modus Ponendo Tollens",
"Definition:Contradiction"
] |
proofwiki-2825 | Rule of Material Implication | ==== Formulation 1 ====
{{:Rule of Material Implication/Formulation 1}}
==== Formulation 2 ====
{{:Rule of Material Implication/Formulation 2}} | {{BeginTableau|p \implies q \vdash \neg p \lor q}}
{{Premise|1|p \implies q}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p}}
{{Addition|4|3|\neg p \lor q|3|1}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|1|5}}
{{Addition|7|1, 5|\neg p \lor q|6|2}}
{{ProofByCases|8|1|\neg p \lor q|2|3|4|5|7}}
{{EndTableau}}
{{... | ==== [[Rule of Material Implication/Formulation 1|Formulation 1]] ====
{{:Rule of Material Implication/Formulation 1}}
==== [[Rule of Material Implication/Formulation 2|Formulation 2]] ====
{{:Rule of Material Implication/Formulation 2}} | {{BeginTableau|p \implies q \vdash \neg p \lor q}}
{{Premise|1|p \implies q}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p}}
{{Addition|4|3|\neg p \lor q|3|1}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|1|5}}
{{Addition|7|1, 5|\neg p \lor q|6|2}}
{{ProofByCases|8|1|\neg p \lor q|2|3|4|5|7}}
{{EndTableau}}
{{... | Rule of Material Implication/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Material_Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Forward_Implication/Proof | [
"Rule of Material Implication",
"Disjunction",
"Conditional",
"Logical Negation"
] | [
"Rule of Material Implication/Formulation 1",
"Rule of Material Implication/Formulation 2"
] | [] |
proofwiki-2826 | Rule of Material Implication | ==== Formulation 1 ====
{{:Rule of Material Implication/Formulation 1}}
==== Formulation 2 ====
{{:Rule of Material Implication/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.
:<nowiki>$\begin {array} {|ccc||cccc|} \hline
p & \implies & q & \neg & p & \lor & q \\
\hline
\F & \T & \F & \T & \F & \T & \F \\
\F & \T & \T & \T & \F & \T & \T \\
\T ... | ==== [[Rule of Material Implication/Formulation 1|Formulation 1]] ====
{{:Rule of Material Implication/Formulation 1}}
==== [[Rule of Material Implication/Formulation 2|Formulation 2]] ====
{{:Rule of Material Implication/Formulation 2}} | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin {array} {|ccc||cccc|} \hline
... | Rule of Material Implication/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Material_Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Proof_by_Truth_Table | [
"Rule of Material Implication",
"Disjunction",
"Conditional",
"Logical Negation"
] | [
"Rule of Material Implication/Formulation 1",
"Rule of Material Implication/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-2827 | Rule of Material Implication | ==== Formulation 1 ====
{{:Rule of Material Implication/Formulation 1}}
==== Formulation 2 ====
{{:Rule of Material Implication/Formulation 2}} | {{BeginTableau|\neg p \lor q \vdash p \implies q}}
{{Premise|1|\neg p \lor q}}
{{Assumption|2|\neg p|Pick the first of the disjuncts ...}}
{{Assumption|3|p|Assume its negation ...}}
{{NonContradiction|4|2, 3|3|2| ... and demonstrate a contradiction}}
{{Explosion|5|2, 3|q|4| ... from a falsehood, ''any'' statement can b... | ==== [[Rule of Material Implication/Formulation 1|Formulation 1]] ====
{{:Rule of Material Implication/Formulation 1}}
==== [[Rule of Material Implication/Formulation 2|Formulation 2]] ====
{{:Rule of Material Implication/Formulation 2}} | {{BeginTableau|\neg p \lor q \vdash p \implies q}}
{{Premise|1|\neg p \lor q}}
{{Assumption|2|\neg p|Pick the first of the disjuncts ...}}
{{Assumption|3|p|Assume its negation ...}}
{{NonContradiction|4|2, 3|3|2| ... and demonstrate a contradiction}}
{{Explosion|5|2, 3|q|4| ... from a falsehood, ''any'' statement can b... | Rule of Material Implication/Formulation 1/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Rule_of_Material_Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Reverse_Implication/Proof_1 | [
"Rule of Material Implication",
"Disjunction",
"Conditional",
"Logical Negation"
] | [
"Rule of Material Implication/Formulation 1",
"Rule of Material Implication/Formulation 2"
] | [] |
proofwiki-2828 | Rule of Material Implication | ==== Formulation 1 ====
{{:Rule of Material Implication/Formulation 1}}
==== Formulation 2 ====
{{:Rule of Material Implication/Formulation 2}} | {{BeginTableau |\neg p \lor q \vdash p \implies q}}
{{Premise |1|\neg p \lor q}}
{{DeMorgan |2|1|\neg \paren {\neg \neg p \land \neg q}|1|Disjunction}}
{{SequentIntro |3|1
|\neg \neg p \implies q
|2
|Conditional is Equivalent to Negation of Conjunction with Negative: $\neg \paren {p \land \neg q}... | ==== [[Rule of Material Implication/Formulation 1|Formulation 1]] ====
{{:Rule of Material Implication/Formulation 1}}
==== [[Rule of Material Implication/Formulation 2|Formulation 2]] ====
{{:Rule of Material Implication/Formulation 2}} | {{BeginTableau |\neg p \lor q \vdash p \implies q}}
{{Premise |1|\neg p \lor q}}
{{DeMorgan |2|1|\neg \paren {\neg \neg p \land \neg q}|1|Disjunction}}
{{SequentIntro |3|1
|\neg \neg p \implies q
|2
|[[Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implic... | Rule of Material Implication/Formulation 1/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Rule_of_Material_Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Reverse_Implication/Proof_2 | [
"Rule of Material Implication",
"Disjunction",
"Conditional",
"Logical Negation"
] | [
"Rule of Material Implication/Formulation 1",
"Rule of Material Implication/Formulation 2"
] | [
"Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implication"
] |
proofwiki-2829 | Rule of Material Implication | ==== Formulation 1 ====
{{:Rule of Material Implication/Formulation 1}}
==== Formulation 2 ====
{{:Rule of Material Implication/Formulation 2}} | {{BeginTableau|\paren {p \implies q} \iff \paren {\neg p \lor q} }}
{{Assumption|1|p \implies q}}
{{SequentIntro|2|1|\neg p \lor q|1|Rule of Material Implication: Formulation 1}}
{{Implication|3||\paren {p \implies q} \implies \paren {\neg p \lor q}|1|2}}
{{Assumption|4|\neg p \lor q}}
{{SequentIntro|5|4|p \implies q|4... | ==== [[Rule of Material Implication/Formulation 1|Formulation 1]] ====
{{:Rule of Material Implication/Formulation 1}}
==== [[Rule of Material Implication/Formulation 2|Formulation 2]] ====
{{:Rule of Material Implication/Formulation 2}} | {{BeginTableau|\paren {p \implies q} \iff \paren {\neg p \lor q} }}
{{Assumption|1|p \implies q}}
{{SequentIntro|2|1|\neg p \lor q|1|[[Rule of Material Implication/Formulation 1/Forward Implication|Rule of Material Implication: Formulation 1]]}}
{{Implication|3||\paren {p \implies q} \implies \paren {\neg p \lor q}|1|2... | Rule of Material Implication/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Material_Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2/Proof_1 | [
"Rule of Material Implication",
"Disjunction",
"Conditional",
"Logical Negation"
] | [
"Rule of Material Implication/Formulation 1",
"Rule of Material Implication/Formulation 2"
] | [
"Rule of Material Implication/Formulation 1/Forward Implication",
"Rule of Material Implication/Formulation 1/Reverse Implication"
] |
proofwiki-2830 | Rule of Material Implication | ==== Formulation 1 ====
{{:Rule of Material Implication/Formulation 1}}
==== Formulation 2 ====
{{:Rule of Material Implication/Formulation 2}} | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
:<nowiki>$\begin{array}{|ccc|c|cccc|} \hline
(p & \implies & q) & \iff & (\neg & p & \lor & q) \\
\hline
\F & \T & \F & \T & \T & \F & \T & \F \\
\F & \T & \T & \T & \T... | ==== [[Rule of Material Implication/Formulation 1|Formulation 1]] ====
{{:Rule of Material Implication/Formulation 1}}
==== [[Rule of Material Implication/Formulation 2|Formulation 2]] ====
{{:Rule of Material Implication/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 connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{... | Rule of Material Implication/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Material_Implication | https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2/Proof_by_Truth_Table | [
"Rule of Material Implication",
"Disjunction",
"Conditional",
"Logical Negation"
] | [
"Rule of Material Implication/Formulation 1",
"Rule of Material Implication/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-2831 | De Morgan's Laws (Set Theory)/Proof by Induction | Let $\mathbb T = \left\{{T_i: i \mathop \in I}\right\}$, where each $T_i$ is a set and $I$ is some finite indexing set.
Then: | Let the cardinality $\card I$ of the indexing set $I$ be $n$.
Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
as:
:$\ds S \setminus \bigcap_{i \mathop = 1}^n ... | Let $\mathbb T = \left\{{T_i: i \mathop \in I}\right\}$, where each $T_i$ is a [[Definition:Set|set]] and $I$ is some [[Definition:Finite Set|finite]] [[Definition:Indexing Set|indexing set]].
Then: | Let the [[Definition:Cardinality|cardinality]] $\card I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \math... | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Intersection/Proof | [
"De Morgan's Laws"
] | [
"Definition:Set",
"Definition:Finite Set",
"Definition:Indexing Set"
] | [
"Definition:Cardinality",
"Definition:Indexing Set",
"Definition:Cardinality",
"Principle of Mathematical Induction",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"In... |
proofwiki-2832 | De Morgan's Laws (Set Theory)/Proof by Induction | Let $\mathbb T = \left\{{T_i: i \mathop \in I}\right\}$, where each $T_i$ is a set and $I$ is some finite indexing set.
Then: | Let the cardinality $\size I$ of the indexing set $I$ be $n$.
Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
as:
:$\ds S \setminus \bigcup_{i \mathop = 1}^n ... | Let $\mathbb T = \left\{{T_i: i \mathop \in I}\right\}$, where each $T_i$ is a [[Definition:Set|set]] and $I$ is some [[Definition:Finite Set|finite]] [[Definition:Indexing Set|indexing set]].
Then: | Let the [[Definition:Cardinality|cardinality]] $\size I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
:$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \math... | De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union/Proof | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction | https://proofwiki.org/wiki/De_Morgan's_Laws_(Set_Theory)/Proof_by_Induction/Difference_with_Union/Proof | [
"De Morgan's Laws"
] | [
"Definition:Set",
"Definition:Finite Set",
"Definition:Indexing Set"
] | [
"Definition:Cardinality",
"Definition:Indexing Set",
"Definition:Cardinality",
"Principle of Mathematical Induction",
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Union",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Union is ... |
proofwiki-2833 | Union equals Intersection iff Sets are Equal | Let $S$ and $T$ be sets.
Then:
:$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S = T$
where:
:$S \cup T$ denotes set union
:$S \cap T$ denotes set intersection. | From Intersection with Subset is Subset:
:$S \subseteq T \iff S \cap T = S$
From Union with Superset is Superset:
:$S \subseteq T \iff S \cup T = T$
That is:
:$T \subseteq S \iff S \cup T = S$
Thus:
:$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S \subseteq T \subseteq S$
By definition of set equality:
:$S = ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Then:
:$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S = T$
where:
:$S \cup T$ denotes [[Definition:Set Union|set union]]
:$S \cap T$ denotes [[Definition:Set Intersection|set intersection]]. | From [[Intersection with Subset is Subset]]:
:$S \subseteq T \iff S \cap T = S$
From [[Union with Superset is Superset]]:
:$S \subseteq T \iff S \cup T = T$
That is:
:$T \subseteq S \iff S \cup T = S$
Thus:
:$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S \subseteq T \subseteq S$
By definition of [[Defin... | Union equals Intersection iff Sets are Equal | https://proofwiki.org/wiki/Union_equals_Intersection_iff_Sets_are_Equal | https://proofwiki.org/wiki/Union_equals_Intersection_iff_Sets_are_Equal | [
"Set Union",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Set Union",
"Definition:Set Intersection"
] | [
"Intersection with Subset is Subset",
"Union with Superset is Superset",
"Definition:Set Equality/Definition 2"
] |
proofwiki-2834 | Intersection with Set Difference is Set Difference with Intersection | : $\left({R \setminus S}\right) \cap T = \left({R \cap T}\right) \setminus S$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe.
Then:
{{begin-eqn}}
{{eqn | l = \paren {R \setminus S} \cap T
| r = \paren {R \cap \map \complement S} \cap T
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = \paren {R \cap T} \cap \map \complement S... | : $\left({R \setminus S}\right) \cap T = \left({R \cap T}\right) \setminus S$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universal Set|universe]].
Then:
{{begin-eqn}}
{{eqn | l = \paren {R \setminus S} \cap T
| r = \paren {R \cap \map \complement S} \cap T
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | r = \pare... | Intersection with Set Difference is Set Difference with Intersection/Proof 1 | https://proofwiki.org/wiki/Intersection_with_Set_Difference_is_Set_Difference_with_Intersection | https://proofwiki.org/wiki/Intersection_with_Set_Difference_is_Set_Difference_with_Intersection/Proof_1 | [
"Intersection",
"Set Intersection",
"Set Difference",
"Set Intersection",
"Intersection with Set Difference is Set Difference with Intersection"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Intersection is Commutative",
"Intersection is Associative",
"Set Difference as Intersection with Complement"
] |
proofwiki-2835 | Intersection with Set Difference is Set Difference with Intersection | : $\left({R \setminus S}\right) \cap T = \left({R \cap T}\right) \setminus S$ | {{begin-eqn}}
{{eqn | o =
| r = x \in \paren {R \setminus S} \cap T
| c =
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \paren {x \in R \land x \notin S} \land x \in T
| c = {{Defof|Set Intersection}} and {{Defof|Set Difference}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \par... | : $\left({R \setminus S}\right) \cap T = \left({R \cap T}\right) \setminus S$ | {{begin-eqn}}
{{eqn | o =
| r = x \in \paren {R \setminus S} \cap T
| c =
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \paren {x \in R \land x \notin S} \land x \in T
| c = {{Defof|Set Intersection}} and {{Defof|Set Difference}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| r = \par... | Intersection with Set Difference is Set Difference with Intersection/Proof 2 | https://proofwiki.org/wiki/Intersection_with_Set_Difference_is_Set_Difference_with_Intersection | https://proofwiki.org/wiki/Intersection_with_Set_Difference_is_Set_Difference_with_Intersection/Proof_2 | [
"Intersection",
"Set Intersection",
"Set Difference",
"Set Intersection",
"Intersection with Set Difference is Set Difference with Intersection"
] | [] | [
"Rule of Commutation",
"Rule of Association"
] |
proofwiki-2836 | Set Difference is Right Distributive over Set Intersection | :$\paren {A \cap B} \setminus C = \paren {A \setminus C} \cap \paren {B \setminus C}$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{X \mathop \in U} \paren {X \setminus T}
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall X \in U
| l = x
| o = \in
| r = X \setminus T
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadstoandfrom
| q = \fora... | :$\paren {A \cap B} \setminus C = \paren {A \setminus C} \cap \paren {B \setminus C}$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcap_{X \mathop \in U} \paren {X \setminus T}
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall X \in U
| l = x
| o = \in
| r = X \setminus T
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadstoandfrom
| q = \fora... | Set Difference is Right Distributive over Set Intersection/General Case/Proof | https://proofwiki.org/wiki/Set_Difference_is_Right_Distributive_over_Set_Intersection | https://proofwiki.org/wiki/Set_Difference_is_Right_Distributive_over_Set_Intersection/General_Case/Proof | [
"Set Difference",
"Set Intersection",
"Set Difference is Right Distributive over Set Intersection"
] | [] | [] |
proofwiki-2837 | Set Difference is Right Distributive over Set Intersection | :$\paren {A \cap B} \setminus C = \paren {A \setminus C} \cap \paren {B \setminus C}$ | Consider $A, B, C \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
{{begin-eqn}}
{{eqn | l = \paren {A \cap B} \setminus C
| r = \paren {A \cap B} \cap \map \complement C
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = \paren {A \cap B} \cap \paren {\map \compl... | :$\paren {A \cap B} \setminus C = \paren {A \setminus C} \cap \paren {B \setminus C}$ | Consider $A, B, C \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universal Set|universal set]].
{{begin-eqn}}
{{eqn | l = \paren {A \cap B} \setminus C
| r = \paren {A \cap B} \cap \map \complement C
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | r = \paren ... | Set Difference is Right Distributive over Set Intersection/Proof 1 | https://proofwiki.org/wiki/Set_Difference_is_Right_Distributive_over_Set_Intersection | https://proofwiki.org/wiki/Set_Difference_is_Right_Distributive_over_Set_Intersection/Proof_1 | [
"Set Difference",
"Set Intersection",
"Set Difference is Right Distributive over Set Intersection"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Set Intersection is Idempotent",
"Intersection is Associative",
"Intersection is Commutative",
"Set Difference as Intersection with Complement"
] |
proofwiki-2838 | Set Difference is Right Distributive over Set Intersection | :$\paren {A \cap B} \setminus C = \paren {A \setminus C} \cap \paren {B \setminus C}$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \paren {A \cap B} \setminus C
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = A \land x \in B
| c = {{Defof|Set Intersection}}
}}
{{eqn | lo= \land
| l = x
| o = \notin
| r = C
| c = {{Defof|Set... | :$\paren {A \cap B} \setminus C = \paren {A \setminus C} \cap \paren {B \setminus C}$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \paren {A \cap B} \setminus C
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = A \land x \in B
| c = {{Defof|Set Intersection}}
}}
{{eqn | lo= \land
| l = x
| o = \notin
| r = C
| c = {{Defof|Set... | Set Difference is Right Distributive over Set Intersection/Proof 2 | https://proofwiki.org/wiki/Set_Difference_is_Right_Distributive_over_Set_Intersection | https://proofwiki.org/wiki/Set_Difference_is_Right_Distributive_over_Set_Intersection/Proof_2 | [
"Set Difference",
"Set Intersection",
"Set Difference is Right Distributive over Set Intersection"
] | [] | [
"Rule of Idempotence",
"Rule of Idempotence",
"Rule of Association",
"Rule of Commutation",
"Rule of Association"
] |
proofwiki-2839 | Set Intersection Distributes over Set Difference | :$\paren {R \setminus S} \cap T = \paren {R \cap T} \setminus \paren {S \cap T}$
:$R \cap \paren {S \setminus T} = \paren {R \cap S} \setminus \paren {R \cap T}$ | {{begin-eqn}}
{{eqn | l = \paren {R \cap T} \setminus \paren {S \cap T}
| r = \paren {\paren {R \cap T} \setminus S} \cup \paren {\paren {R \cap T} \setminus T}
| c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \paren {\paren {R \cap T} \setminus S} \cup \O
| c = Set Difference of In... | :$\paren {R \setminus S} \cap T = \paren {R \cap T} \setminus \paren {S \cap T}$
:$R \cap \paren {S \setminus T} = \paren {R \cap S} \setminus \paren {R \cap T}$ | {{begin-eqn}}
{{eqn | l = \paren {R \cap T} \setminus \paren {S \cap T}
| r = \paren {\paren {R \cap T} \setminus S} \cup \paren {\paren {R \cap T} \setminus T}
| c = [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
}}
{{eqn | r = \... | Set Intersection Distributes over Set Difference | https://proofwiki.org/wiki/Set_Intersection_Distributes_over_Set_Difference | https://proofwiki.org/wiki/Set_Intersection_Distributes_over_Set_Difference | [
"Set Difference",
"Set Intersection",
"Examples of Distributive Operations"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection",
"Set Difference of Intersection with Set is Empty Set",
"Union with Empty Set",
"Intersection with Set Difference is Set Difference with Intersection",
"Intersection is Commutative",
"Intersection is Commutative"
] |
proofwiki-2840 | Set Difference with Intersection is Difference | :$S \setminus \paren {S \cap T} = S \setminus T$ | {{begin-eqn}}
{{eqn | l = S \setminus \paren {S \cap T}
| r = \paren {S \setminus S} \cup \paren {S \setminus T}
| c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \O \cup \paren {S \setminus T}
| c = Set Difference with Self is Empty Set
}}
{{eqn | r = S \setminus T
| c = Union... | :$S \setminus \paren {S \cap T} = S \setminus T$ | {{begin-eqn}}
{{eqn | l = S \setminus \paren {S \cap T}
| r = \paren {S \setminus S} \cup \paren {S \setminus T}
| c = [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
}}
{{eqn | r = \O \cup \paren {S \setminus T}
| c = [[Set ... | Set Difference with Intersection is Difference | https://proofwiki.org/wiki/Set_Difference_with_Intersection_is_Difference | https://proofwiki.org/wiki/Set_Difference_with_Intersection_is_Difference | [
"Set Difference",
"Set Intersection"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection",
"Set Difference with Self is Empty Set",
"Union with Empty Set"
] |
proofwiki-2841 | Set Difference of Intersection with Set is Empty Set | :$\paren {S \cap T} \setminus S = \O$
:$\paren {S \cap T} \setminus T = \O$ | From Set Difference is Right Distributive over Set Intersection:
: $\paren {R \cap S} \setminus T = \paren {R \setminus T} \cap \paren {S \setminus T}$
Hence:
{{begin-eqn}}
{{eqn | l=\paren {S \cap T} \setminus S
| r=\paren {S \setminus S} \cap \paren {T \setminus S}
| c=Set Difference is Right Distributive... | :$\paren {S \cap T} \setminus S = \O$
:$\paren {S \cap T} \setminus T = \O$ | From [[Set Difference is Right Distributive over Set Intersection]]:
: $\paren {R \cap S} \setminus T = \paren {R \setminus T} \cap \paren {S \setminus T}$
Hence:
{{begin-eqn}}
{{eqn | l=\paren {S \cap T} \setminus S
| r=\paren {S \setminus S} \cap \paren {T \setminus S}
| c=[[Set Difference is Right Dis... | Set Difference of Intersection with Set is Empty Set | https://proofwiki.org/wiki/Set_Difference_of_Intersection_with_Set_is_Empty_Set | https://proofwiki.org/wiki/Set_Difference_of_Intersection_with_Set_is_Empty_Set | [
"Set Difference",
"Set Intersection",
"Empty Set"
] | [] | [
"Set Difference is Right Distributive over Set Intersection",
"Set Difference is Right Distributive over Set Intersection",
"Set Difference with Self is Empty Set",
"Intersection with Empty Set",
"Intersection is Commutative",
"Category:Set Difference",
"Category:Set Intersection",
"Category:Empty Set... |
proofwiki-2842 | Set Difference Intersection with First Set is Set Difference | :$\paren {S \setminus T} \cap S = S \setminus T$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T}
| o = \subseteq
| r = S
| c = Set Difference is Subset
}}
{{eqn | ll= \leadsto
| l = \paren {S \setminus T} \cap S
| r = S \setminus T
| c = Intersection with Subset is Subset
}}
{{end-eqn}}
{{qed}} | :$\paren {S \setminus T} \cap S = S \setminus T$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T}
| o = \subseteq
| r = S
| c = [[Set Difference is Subset]]
}}
{{eqn | ll= \leadsto
| l = \paren {S \setminus T} \cap S
| r = S \setminus T
| c = [[Intersection with Subset is Subset]]
}}
{{end-eqn}}
{{qed}} | Set Difference Intersection with First Set is Set Difference/Proof 1 | https://proofwiki.org/wiki/Set_Difference_Intersection_with_First_Set_is_Set_Difference | https://proofwiki.org/wiki/Set_Difference_Intersection_with_First_Set_is_Set_Difference/Proof_1 | [
"Set Difference",
"Set Intersection",
"Set Difference Intersection with First Set is Set Difference"
] | [] | [
"Set Difference is Subset",
"Intersection with Subset is Subset"
] |
proofwiki-2843 | Set Difference Intersection with First Set is Set Difference | :$\paren {S \setminus T} \cap S = S \setminus T$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cap S
| r = \paren {S \cap S} \setminus T
| c = Intersection with Set Difference is Set Difference with Intersection
}}
{{eqn | r = S \setminus T
| c = Set Intersection is Idempotent
}}
{{end-eqn}}
{{qed}} | :$\paren {S \setminus T} \cap S = S \setminus T$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cap S
| r = \paren {S \cap S} \setminus T
| c = [[Intersection with Set Difference is Set Difference with Intersection]]
}}
{{eqn | r = S \setminus T
| c = [[Set Intersection is Idempotent]]
}}
{{end-eqn}}
{{qed}} | Set Difference Intersection with First Set is Set Difference/Proof 2 | https://proofwiki.org/wiki/Set_Difference_Intersection_with_First_Set_is_Set_Difference | https://proofwiki.org/wiki/Set_Difference_Intersection_with_First_Set_is_Set_Difference/Proof_2 | [
"Set Difference",
"Set Intersection",
"Set Difference Intersection with First Set is Set Difference"
] | [] | [
"Intersection with Set Difference is Set Difference with Intersection",
"Set Intersection is Idempotent"
] |
proofwiki-2844 | Set Difference Intersection with Second Set is Empty Set | :$\paren {S \setminus T} \cap T = \O$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cap T
| r = \paren {S \cap T} \setminus \paren {T \cap T}
| c = Set Intersection Distributes over Set Difference
}}
{{eqn | r = \paren {S \cap T} \setminus T
| c = Set Intersection is Idempotent
}}
{{eqn | r = \O
| c = Set Difference of Intersect... | :$\paren {S \setminus T} \cap T = \O$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cap T
| r = \paren {S \cap T} \setminus \paren {T \cap T}
| c = [[Set Intersection Distributes over Set Difference]]
}}
{{eqn | r = \paren {S \cap T} \setminus T
| c = [[Set Intersection is Idempotent]]
}}
{{eqn | r = \O
| c = [[Set Difference of... | Set Difference Intersection with Second Set is Empty Set/Proof 1 | https://proofwiki.org/wiki/Set_Difference_Intersection_with_Second_Set_is_Empty_Set | https://proofwiki.org/wiki/Set_Difference_Intersection_with_Second_Set_is_Empty_Set/Proof_1 | [
"Set Difference",
"Set Intersection",
"Set Difference Intersection with Second Set is Empty Set"
] | [] | [
"Set Intersection Distributes over Set Difference",
"Set Intersection is Idempotent",
"Set Difference of Intersection with Set is Empty Set"
] |
proofwiki-2845 | Set Difference Intersection with Second Set is Empty Set | :$\paren {S \setminus T} \cap T = \O$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cap T
| r = \paren {S \cap T} \setminus T
| c = Intersection with Set Difference is Set Difference with Intersection
}}
{{eqn | r = \O
| c = Set Difference of Intersection with Set is Empty Set
}}
{{end-eqn}}
{{qed}} | :$\paren {S \setminus T} \cap T = \O$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cap T
| r = \paren {S \cap T} \setminus T
| c = [[Intersection with Set Difference is Set Difference with Intersection]]
}}
{{eqn | r = \O
| c = [[Set Difference of Intersection with Set is Empty Set]]
}}
{{end-eqn}}
{{qed}} | Set Difference Intersection with Second Set is Empty Set/Proof 2 | https://proofwiki.org/wiki/Set_Difference_Intersection_with_Second_Set_is_Empty_Set | https://proofwiki.org/wiki/Set_Difference_Intersection_with_Second_Set_is_Empty_Set/Proof_2 | [
"Set Difference",
"Set Intersection",
"Set Difference Intersection with Second Set is Empty Set"
] | [] | [
"Intersection with Set Difference is Set Difference with Intersection",
"Set Difference of Intersection with Set is Empty Set"
] |
proofwiki-2846 | Set Difference with Union is Set Difference | :$\paren {S \cup T} \setminus T = S \setminus T$ | Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
{{begin-eqn}}
{{eqn | l = \paren {S \cup T} \setminus T
| r = \paren {S \cup T} \cap \overline T
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = \paren {S \cap \overline T} \cup \paren {T \cap \ove... | :$\paren {S \cup T} \setminus T = S \setminus T$ | Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universal Set|universal set]].
{{begin-eqn}}
{{eqn | l = \paren {S \cup T} \setminus T
| r = \paren {S \cup T} \cap \overline T
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | r = \paren {S \cap \o... | Set Difference with Union is Set Difference | https://proofwiki.org/wiki/Set_Difference_with_Union_is_Set_Difference | https://proofwiki.org/wiki/Set_Difference_with_Union_is_Set_Difference | [
"Set Difference",
"Set Union"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Intersection Distributes over Union",
"Set Difference as Intersection with Complement",
"Set Difference with Self is Empty Set",
"Union with Empty Set",
"Category:Set Difference",
"Category:Set Union"
] |
proofwiki-2847 | Set Difference Union First Set is First Set | :$\paren {S \setminus T} \cup S = S$ | Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
{{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cup S
| r = \paren {S \cap \map \complement T} \cup S
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = S
| c = Union Absorbs Intersection
}}
{... | :$\paren {S \setminus T} \cup S = S$ | Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universal Set|universal set]].
{{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cup S
| r = \paren {S \cap \map \complement T} \cup S
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | r = S
| c... | Set Difference Union First Set is First Set | https://proofwiki.org/wiki/Set_Difference_Union_First_Set_is_First_Set | https://proofwiki.org/wiki/Set_Difference_Union_First_Set_is_First_Set | [
"Set Difference",
"Set Union"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Absorption Laws (Set Theory)/Union Absorbs Intersection",
"Category:Set Difference",
"Category:Set Union"
] |
proofwiki-2848 | Set Difference Union Second Set is Union | :$\paren {S \setminus T} \cup T = S \cup T$ | Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
{{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cup T
| r = \paren {S \cap \map \complement T} \cup T
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = \paren {S \cup T} \cap \paren {\map \compleme... | :$\paren {S \setminus T} \cup T = S \cup T$ | Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universal Set|universal set]].
{{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cup T
| r = \paren {S \cap \map \complement T} \cup T
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | r = \paren {S ... | Set Difference Union Second Set is Union | https://proofwiki.org/wiki/Set_Difference_Union_Second_Set_is_Union | https://proofwiki.org/wiki/Set_Difference_Union_Second_Set_is_Union | [
"Set Difference",
"Set Union"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Union Distributes over Intersection",
"Union with Complement",
"Intersection with Universal Set"
] |
proofwiki-2849 | Set Difference with Union | :$R \setminus \paren {S \cup T} = \paren {R \cup T} \setminus \paren {S \cup T} = \paren {R \setminus S} \setminus T = \paren {R \setminus T} \setminus S$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
{{begin-eqn}}
{{eqn | l = \paren {R \cup T} \setminus \paren {S \cup T}
| r = \paren {R \cup T} \cap \overline {\paren {S \cup T} }
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = \paren {R \cup... | :$R \setminus \paren {S \cup T} = \paren {R \cup T} \setminus \paren {S \cup T} = \paren {R \setminus S} \setminus T = \paren {R \setminus T} \setminus S$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universal Set|universal set]].
{{begin-eqn}}
{{eqn | l = \paren {R \cup T} \setminus \paren {S \cup T}
| r = \paren {R \cup T} \cap \overline {\paren {S \cup T} }
| c = [[Set Difference as Intersection with Complem... | Set Difference with Union | https://proofwiki.org/wiki/Set_Difference_with_Union | https://proofwiki.org/wiki/Set_Difference_with_Union | [
"Set Difference",
"Set Union",
"Set Difference with Union"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"De Morgan's Laws (Set Theory)/Set Complement/Complement of Union",
"Intersection is Associative",
"Intersection is Commutative",
"Set Difference as Intersection with Complement",
"Set Difference with Union is Set Difference",... |
proofwiki-2850 | Set Difference is Right Distributive over Union | :$\paren {R \cup S} \setminus T = \paren {R \setminus T} \cup \paren {S \setminus T}$ | {{begin-eqn}}
{{eqn | l = \paren {R \cup S} \setminus T
| r = \paren {R \cup S} \cap \overline T
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = \paren {R \cap \overline T} \cup \paren {S \cap \overline T}
| c = Intersection Distributes over Union
}}
{{eqn | r = \paren {R \setminus... | :$\paren {R \cup S} \setminus T = \paren {R \setminus T} \cup \paren {S \setminus T}$ | {{begin-eqn}}
{{eqn | l = \paren {R \cup S} \setminus T
| r = \paren {R \cup S} \cap \overline T
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | r = \paren {R \cap \overline T} \cup \paren {S \cap \overline T}
| c = [[Intersection Distributes over Union]]
}}
{{eqn | r = \paren {R \... | Set Difference is Right Distributive over Union | https://proofwiki.org/wiki/Set_Difference_is_Right_Distributive_over_Union | https://proofwiki.org/wiki/Set_Difference_is_Right_Distributive_over_Union | [
"Set Difference",
"Set Union",
"Examples of Distributive Operations"
] | [] | [
"Set Difference as Intersection with Complement",
"Intersection Distributes over Union",
"Set Difference as Intersection with Complement",
"Category:Set Difference",
"Category:Set Union",
"Category:Examples of Distributive Operations"
] |
proofwiki-2851 | Set Difference with Set Difference is Union of Set Difference with Intersection | :$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \cup \paren {R \cap T}$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
{{begin-eqn}}
{{eqn | l = R \setminus \paren {S \setminus T}
| r = R \cap \overline {\paren {S \cap \overline T} }
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = R \cap \paren {\overline S \cup... | :$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \cup \paren {R \cap T}$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universal Set|universal set]].
{{begin-eqn}}
{{eqn | l = R \setminus \paren {S \setminus T}
| r = R \cap \overline {\paren {S \cap \overline T} }
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | ... | Set Difference with Set Difference is Union of Set Difference with Intersection | https://proofwiki.org/wiki/Set_Difference_with_Set_Difference_is_Union_of_Set_Difference_with_Intersection | https://proofwiki.org/wiki/Set_Difference_with_Set_Difference_is_Union_of_Set_Difference_with_Intersection | [
"Set Difference",
"Set Union",
"Set Intersection",
"Set Difference with Set Difference is Union of Set Difference with Intersection"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"De Morgan's Laws (Set Theory)",
"Complement of Complement",
"Intersection Distributes over Union",
"Set Difference as Intersection with Complement"
] |
proofwiki-2852 | Set Difference is Subset of Union of Differences | :$R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
{{begin-eqn}}
{{eqn | l = R \setminus S
| r = R \cap \overline S
| c = Set Difference as Intersection with Complement
}}
{{eqn | r = \paren {R \cap \overline S} \cap \mathbb U
| c = Intersection with Universa... | :$R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$ | Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the [[Definition:Universal Set|universal set]].
{{begin-eqn}}
{{eqn | l = R \setminus S
| r = R \cap \overline S
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | r = \paren {R \cap \overline S} \cap \mathbb U
... | Set Difference is Subset of Union of Differences | https://proofwiki.org/wiki/Set_Difference_is_Subset_of_Union_of_Differences | https://proofwiki.org/wiki/Set_Difference_is_Subset_of_Union_of_Differences | [
"Set Difference",
"Set Union"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Intersection with Universal Set",
"Union with Complement",
"Intersection Distributes over Union",
"Union of Intersections",
"Set Difference as Intersection with Complement",
"Category:Set Difference",
"Category:Set Union"... |
proofwiki-2853 | Ordered Sum of Tosets is Totally Ordered Set | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be tosets.
Let $S + T = \struct {S \cup T, \preceq}$ be the ordered sum of $S$ and $T$.
Then $\struct {S \cup T, \preceq}$ is itself a toset. | By definition of ordered sum, we have that:
* If $a, b \in S$, then $a \preceq b \iff a \preceq_1 b$.
* Otherwise, if $a, b \in T$, then $a \preceq b \iff a \preceq_2 b$.
* If neither of these is the case, then $a \in S, b \in T \iff a \preceq b$.
First we show that $\preceq$ is connected.
We note that as $\struct {S, ... | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Toset|tosets]].
Let $S + T = \struct {S \cup T, \preceq}$ be the [[Definition:Ordered Sum|ordered sum]] of $S$ and $T$.
Then $\struct {S \cup T, \preceq}$ is itself a [[Definition:Toset|toset]]. | By definition of [[Definition:Ordered Sum|ordered sum]], we have that:
* If $a, b \in S$, then $a \preceq b \iff a \preceq_1 b$.
* Otherwise, if $a, b \in T$, then $a \preceq b \iff a \preceq_2 b$.
* If neither of these is the case, then $a \in S, b \in T \iff a \preceq b$.
First we show that $\preceq$ is [[Definiti... | Ordered Sum of Tosets is Totally Ordered Set | https://proofwiki.org/wiki/Ordered_Sum_of_Tosets_is_Totally_Ordered_Set | https://proofwiki.org/wiki/Ordered_Sum_of_Tosets_is_Totally_Ordered_Set | [
"Total Orderings",
"Ordered Sums"
] | [
"Symbols:Abbreviations/T/Toset",
"Definition:Ordered Sum",
"Symbols:Abbreviations/T/Toset"
] | [
"Definition:Ordered Sum",
"Definition:Connected Relation",
"Symbols:Abbreviations/T/Toset",
"Definition:Connected Relation",
"Definition:Connected Relation",
"Definition:Ordering",
"Definition:Connected Relation",
"Symbols:Abbreviations/T/Toset"
] |
proofwiki-2854 | Antilexicographic Product of Totally Ordered Sets is Totally Ordered | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be totally ordered sets.
Let $S_1 \otimes^a S_2 = \struct {S_1 \times S_2, \preccurlyeq_a}$ be the antilexicographic product of $S_1$ and $S_2$.
Then $\struct {S_1 \times S_2, \preccurlyeq_a}$ is itself a totally ordered set. | From Antilexicographic Order is Ordering, we have that $\preccurlyeq_a$ is an ordering.
It remains to be shown that $\preccurlyeq_a$ is connected.
By definition of antilexicographic product, we have that:
::$\tuple {a, b} \preccurlyeq_a \tuple {c, d} \iff \tuple {c \prec_2 d} \lor \paren {c = d \land a \preccurlyeq_1 b... | Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be [[Definition:Totally Ordered Set|totally ordered sets]].
Let $S_1 \otimes^a S_2 = \struct {S_1 \times S_2, \preccurlyeq_a}$ be the [[Definition:Antilexicographic Order|antilexicographic product]] of $S_1$ and $S_2$.
Then $\struct {S_1 \times ... | From [[Antilexicographic Order is Ordering]], we have that $\preccurlyeq_a$ is an [[Definition:Ordering|ordering]].
It remains to be shown that $\preccurlyeq_a$ is [[Definition:Connected Relation|connected]].
By definition of [[Definition:Antilexicographic Order|antilexicographic product]], we have that:
::$\tuple {... | Antilexicographic Product of Totally Ordered Sets is Totally Ordered | https://proofwiki.org/wiki/Antilexicographic_Product_of_Totally_Ordered_Sets_is_Totally_Ordered | https://proofwiki.org/wiki/Antilexicographic_Product_of_Totally_Ordered_Sets_is_Totally_Ordered | [
"Total Orderings",
"Antilexicographic Order",
"Antilexicographic Product of Totally Ordered Sets is Totally Ordered"
] | [
"Definition:Totally Ordered Set",
"Definition:Antilexicographic Order",
"Definition:Totally Ordered Set"
] | [
"Antilexicographic Order is Ordering",
"Definition:Ordering",
"Definition:Connected Relation",
"Definition:Antilexicographic Order",
"Definition:Totally Ordered Set",
"Definition:Connected Relation",
"Definition:Connected Relation",
"Definition:Connected Relation",
"Definition:Reflexive Relation",
... |
proofwiki-2855 | Restriction of Mapping is Mapping | Let $f: S \to T$ be a mapping.
Let $X \subseteq S$.
Let $f \restriction_X$ be the restriction of $f$ to $X$.
Then $f \restriction_X: X \to T$ is a mapping:
: whose domain is $X$
: whose preimage is $X$. | As $f: S \to T$ is a mapping, we have that:
:$\forall x \in S: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$
From the definition of a subset, $x \in X \implies x \in S$, and so:
:$\forall x \in X: \tuple {x, y_1} \in f \restriction_X \land \tuple {x, y_2} \in f \restriction_X \implies y_1 = y_2$... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $X \subseteq S$.
Let $f \restriction_X$ be the [[Definition:Restriction of Mapping|restriction of $f$ to $X$]].
Then $f \restriction_X: X \to T$ is a [[Definition:Mapping|mapping]]:
: whose [[Definition:Domain of Mapping|domain]] is $X$
: whose [[Definition:... | As $f: S \to T$ is a [[Definition:Mapping|mapping]], we have that:
:$\forall x \in S: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$
From the [[Definition:Subset|definition of a subset]], $x \in X \implies x \in S$, and so:
:$\forall x \in X: \tuple {x, y_1} \in f \restriction_X \land \tuple {x,... | Restriction of Mapping is Mapping | https://proofwiki.org/wiki/Restriction_of_Mapping_is_Mapping | https://proofwiki.org/wiki/Restriction_of_Mapping_is_Mapping | [
"Restrictions"
] | [
"Definition:Mapping",
"Definition:Restriction/Mapping",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Preimage/Mapping/Mapping"
] | [
"Definition:Mapping",
"Definition:Subset",
"Definition:Mapping",
"Definition:Subset",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Preimage/Mapping/Mapping",
"Preimage of Mapping equals Domain"
] |
proofwiki-2856 | Image of Subset is Image of Restriction | Let $f: S \to T$ be a mapping.
Let $X \subseteq S$.
Let $f {\restriction_X}$ be the restriction of $f$ to $X$.
Then:
:$f \sqbrk X = \Img {f {\restriction_X} }$
where $\Img f$ denotes the image of $f$, defined as:
:$\Img f = \set {t \in T: \exists s \in S: t = \map f s}$ | Let $y \in f \sqbrk X$.
Then by definition of the image of a subset:
:$\exists x \in X: \map f x = y$
or equivalently:
:$\exists x \in X: \tuple {x, y} \in f$
But then by definition of restriction of $f$ to $X$:
:$f {\restriction_X} = \set {\tuple {x, y} \in f: x \in X}$
Thus by definition of the image set of $f {\rest... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $X \subseteq S$.
Let $f {\restriction_X}$ be the [[Definition:Restriction of Mapping|restriction of $f$ to $X$]].
Then:
:$f \sqbrk X = \Img {f {\restriction_X} }$
where $\Img f$ denotes the [[Definition:Image of Mapping|image]] of $f$, defined as:
:$\Img f =... | Let $y \in f \sqbrk X$.
Then by definition of the [[Definition:Image of Subset under Mapping|image of a subset]]:
:$\exists x \in X: \map f x = y$
or equivalently:
:$\exists x \in X: \tuple {x, y} \in f$
But then by definition of [[Definition:Restriction of Mapping|restriction of $f$ to $X$]]:
:$f {\restriction_X} =... | Image of Subset is Image of Restriction | https://proofwiki.org/wiki/Image_of_Subset_is_Image_of_Restriction | https://proofwiki.org/wiki/Image_of_Subset_is_Image_of_Restriction | [
"Restrictions"
] | [
"Definition:Mapping",
"Definition:Restriction/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping"
] | [
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Restriction/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Restriction/Mapping",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Set Equality/Definition 2"
] |
proofwiki-2857 | Probability Mass Function of Binomial Distribution | The probability mass function (pmf) of a binomially distributed random variable $X$ is equal to:
:$\map \Pr {X = x} = \dbinom n x p^x \paren {1 - p}^{n - x}$
where $n$ is the number of trials and $p$ is the probability of success. | Let $B_i: i = 1, 2, \ldots, \dbinom n x$ be events such that:
:$(1): \quad B_i$ is the $i$th possible way to see $x$ successes in $n$ Bernoulli trials
:$(2): \quad \forall i \ne j: B_i \cap B_j = \O$
We can see that:
:$\forall i: \map \Pr {B_i} = p^x \paren {1 - p}^{n - x}$
This is true since there will be $x$ successe... | The [[Definition:Probability Mass Function|probability mass function (pmf)]] of a [[Definition:Binomial Distribution|binomially distributed random variable]] $X$ is equal to:
:$\map \Pr {X = x} = \dbinom n x p^x \paren {1 - p}^{n - x}$
where $n$ is the number of trials and $p$ is the [[Definition:Probability|probabil... | Let $B_i: i = 1, 2, \ldots, \dbinom n x$ be [[Definition:Event|events]] such that:
:$(1): \quad B_i$ is the $i$th possible way to see $x$ successes in $n$ [[Definition:Bernoulli Trial|Bernoulli trials]]
:$(2): \quad \forall i \ne j: B_i \cap B_j = \O$
We can see that:
:$\forall i: \map \Pr {B_i} = p^x \paren {1 - ... | Probability Mass Function of Binomial Distribution | https://proofwiki.org/wiki/Probability_Mass_Function_of_Binomial_Distribution | https://proofwiki.org/wiki/Probability_Mass_Function_of_Binomial_Distribution | [
"Probability Mass Function of Binomial Distribution",
"Binomial Distribution",
"Examples of Probability Mass Functions"
] | [
"Definition:Probability Mass Function",
"Definition:Binomial Distribution",
"Definition:Probability",
"Definition:Bernoulli Distribution"
] | [
"Definition:Event",
"Definition:Bernoulli Trial",
"Definition:Independent Events",
"Binomial Experiment has Binomial Distribution",
"Inclusion-Exclusion Principle",
"Addition Law of Probability",
"Definition:Set Union/Countable Union",
"Definition:Event",
"Definition:Pairwise Disjoint"
] |
proofwiki-2858 | Semantic Consequence Union Negation | Let $U$ be a set of propositional formulas.
Let $P$ be a propositional formula.
Let $U \models P$ denote that $U$ is a semantic consequence $P$.
Then:
:$U \models P$
iff:
:$U \cup \set {\neg P}$ has no models. | === Sufficient Condition ===
{{AimForCont}} $U \models P$.
Let $\MM$ be a model such that:
:$\MM \models U \cup \set {\neg P}$
Then by definition of $\MM$ being a model, we have that:
:$\MM \models U$
and by definition of semantic consequence:
:$\MM \models P$
So we have that $\MM \models P$ and $\MM \models \neg P$
By... | Let $U$ be a [[Definition:Set|set]] of [[Definition:Propositional Formula|propositional formulas]].
Let $P$ be a [[Definition:Propositional Formula|propositional formula]].
Let $U \models P$ denote that $U$ is a [[Definition:Semantic Consequence|semantic consequence]] $P$.
Then:
:$U \models P$
iff:
:$U \cup \set {\... | === Sufficient Condition ===
{{AimForCont}} $U \models P$.
Let $\MM$ be a [[Definition:Model (Boolean Interpretations)|model]] such that:
:$\MM \models U \cup \set {\neg P}$
Then by definition of $\MM$ being a [[Definition:Model (Boolean Interpretations)|model]], we have that:
:$\MM \models U$
and by definition of ... | Semantic Consequence Union Negation | https://proofwiki.org/wiki/Semantic_Consequence_Union_Negation | https://proofwiki.org/wiki/Semantic_Consequence_Union_Negation | [
"Propositional Logic"
] | [
"Definition:Set",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Semantic Consequence",
"Definition:Model (Boolean Interpretations)"
] | [
"Definition:Model (Boolean Interpretations)",
"Definition:Model (Boolean Interpretations)",
"Definition:Semantic Consequence",
"Definition:Contradiction",
"Definition:Model (Boolean Interpretations)",
"Definition:Conditional/Sufficient Condition",
"Definition:Model (Boolean Interpretations)",
"Definit... |
proofwiki-2859 | Union of Singleton | Consider the set of sets $A$ such that $A$ consists of just one set $x$:
:$A = \set x$
Then the union of $A$ is $x$:
:$\bigcup A = x$ | Let $A = \set x$.
From the definition of set union:
:$\bigcup \set x = \set {y: \exists z \in \set x: y \in z}$
from which it follows directly that:
:$\bigcup \set x = \set {y: y \in x}$
as $x$ is the only set in $\set x$.
That is:
:$\bigcup A = x$
{{qed}} | Consider the [[Definition:Set of Sets|set of sets]] $A$ such that $A$ consists of just one [[Definition:Set|set]] $x$:
:$A = \set x$
Then the [[Definition:Union of Set of Sets|union]] of $A$ is $x$:
:$\bigcup A = x$ | Let $A = \set x$.
From the definition of [[Definition:Union of Set of Sets|set union]]:
:$\bigcup \set x = \set {y: \exists z \in \set x: y \in z}$
from which it follows directly that:
:$\bigcup \set x = \set {y: y \in x}$
as $x$ is the only set in $\set x$.
That is:
:$\bigcup A = x$
{{qed}} | Union of Singleton | https://proofwiki.org/wiki/Union_of_Singleton | https://proofwiki.org/wiki/Union_of_Singleton | [
"Set Union",
"Singletons"
] | [
"Definition:Set of Sets",
"Definition:Set",
"Definition:Set Union/Set of Sets"
] | [
"Definition:Set Union/Set of Sets"
] |
proofwiki-2860 | Intersection of Singleton | Consider the set of sets $A$ such that $A$ consists of just one set $x$:
:$A = \set x$
Then the intersection of $A$ is $x$:
:$\bigcap A = x$ | Let $A = \set x$.
Then from the definition:
:$\bigcap \set x = \set {y: \forall z \in \set x: y \in z}$
from which it follows directly:
:$\bigcap \set x = \set {y: y \in x}$
as $x$ is the only set in $\set x$.
That is:
:$\bigcap A = x$
{{qed}} | Consider the [[Definition:Set of Sets|set of sets]] $A$ such that $A$ consists of just one [[Definition:Set|set]] $x$:
:$A = \set x$
Then the [[Definition:Intersection of Set of Sets|intersection]] of $A$ is $x$:
:$\bigcap A = x$ | Let $A = \set x$.
Then from the definition:
:$\bigcap \set x = \set {y: \forall z \in \set x: y \in z}$
from which it follows directly:
:$\bigcap \set x = \set {y: y \in x}$
as $x$ is the only set in $\set x$.
That is:
:$\bigcap A = x$
{{qed}} | Intersection of Singleton | https://proofwiki.org/wiki/Intersection_of_Singleton | https://proofwiki.org/wiki/Intersection_of_Singleton | [
"Set Intersection",
"Singletons"
] | [
"Definition:Set of Sets",
"Definition:Set",
"Definition:Set Intersection/Set of Sets"
] | [] |
proofwiki-2861 | Union of Empty Set | Consider the set of sets $\mathbb S$ such that $\mathbb S$ is the empty set $\O$.
Then the union of $\mathbb S$ is $\O$:
:$\mathbb S = \O \implies \bigcup \mathbb S = \O$ | Let $\mathbb S = \O$.
Then from the definition:
:$\bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$
from which it follows directly:
:$\bigcup \mathbb S = \O$
as there are no sets in $\mathbb S$.
{{qed}} | Consider the [[Definition:Set of Sets|set of sets]] $\mathbb S$ such that $\mathbb S$ is the [[Definition:Empty Set|empty set]] $\O$.
Then the [[Definition:Union of Set of Sets|union]] of $\mathbb S$ is $\O$:
:$\mathbb S = \O \implies \bigcup \mathbb S = \O$ | Let $\mathbb S = \O$.
Then from the definition:
:$\bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$
from which it follows directly:
:$\bigcup \mathbb S = \O$
as there are no [[Definition:Set|sets]] in $\mathbb S$.
{{qed}} | Union of Empty Set | https://proofwiki.org/wiki/Union_of_Empty_Set | https://proofwiki.org/wiki/Union_of_Empty_Set | [
"Set Union",
"Empty Set"
] | [
"Definition:Set of Sets",
"Definition:Empty Set",
"Definition:Set Union/Set of Sets"
] | [
"Definition:Set"
] |
proofwiki-2862 | Intersection of Empty Set | Consider the set of sets $\mathbb S$ such that $\mathbb S$ is the empty set $\O$.
Then the intersection of $\mathbb S$ is $\mathbb U$:
:$\mathbb S = \O \implies \bigcap \mathbb S = \mathbb U$
where $\mathbb U$ is the universal set.
A paradoxical result. | Let $\mathbb S = \O$.
Then from the definition:
:$ \bigcap \mathbb S = \set {x: \forall X \in \mathbb S: x \in X}$
Consider any $x \in \mathbb U$.
Then as $\mathbb S = \O$, it follows that:
:$\forall X \in \mathbb S: x \in X$
from the definition of vacuous truth.
It follows directly that:
:$\bigcap \mathbb S = \set {x:... | Consider the [[Definition:Set of Sets|set of sets]] $\mathbb S$ such that $\mathbb S$ is the [[Definition:Empty Set|empty set]] $\O$.
Then the [[Definition:Intersection of Set of Sets|intersection]] of $\mathbb S$ is $\mathbb U$:
:$\mathbb S = \O \implies \bigcap \mathbb S = \mathbb U$
where $\mathbb U$ is the [[Def... | Let $\mathbb S = \O$.
Then from the definition:
:$ \bigcap \mathbb S = \set {x: \forall X \in \mathbb S: x \in X}$
Consider any $x \in \mathbb U$.
Then as $\mathbb S = \O$, it follows that:
:$\forall X \in \mathbb S: x \in X$
from the definition of [[Definition:Vacuous Truth|vacuous truth]].
It follows directly tha... | Intersection of Empty Set | https://proofwiki.org/wiki/Intersection_of_Empty_Set | https://proofwiki.org/wiki/Intersection_of_Empty_Set | [
"Set Intersection",
"Veridical Paradoxes",
"Intersection of Empty Set"
] | [
"Definition:Set of Sets",
"Definition:Empty Set",
"Definition:Set Intersection/Set of Sets",
"Definition:Universal Set",
"Definition:Veridical Paradox"
] | [
"Definition:Vacuous Truth"
] |
proofwiki-2863 | Inverse of Identity Mapping | Let $S$ be a set.
Let $I_S: S \to S$ be the identity mapping on $S$.
Then the inverse of $I_S$ is itself:
:$\paren {I_S}^{-1} = I_S$ | From the nature of the identity mapping, we have:
:$I_S \circ I_S = I_S$
from which it follows by definition that $I_S$ is the inverse of itself.
{{Qed}} | Let $S$ be a [[Definition:Set|set]].
Let $I_S: S \to S$ be the [[Definition:Identity Mapping|identity mapping on $S$]].
Then the [[Definition:Inverse Mapping|inverse]] of $I_S$ is itself:
:$\paren {I_S}^{-1} = I_S$ | From the nature of the [[Definition:Identity Mapping|identity mapping]], we have:
:$I_S \circ I_S = I_S$
from which it follows by definition that $I_S$ is the [[Definition:Inverse Mapping/Definition 2|inverse]] of itself.
{{Qed}} | Inverse of Identity Mapping | https://proofwiki.org/wiki/Inverse_of_Identity_Mapping | https://proofwiki.org/wiki/Inverse_of_Identity_Mapping | [
"Inverse Mappings",
"Identity Mappings"
] | [
"Definition:Set",
"Definition:Identity Mapping",
"Definition:Inverse Mapping"
] | [
"Definition:Identity Mapping",
"Definition:Inverse Mapping/Definition 2"
] |
proofwiki-2864 | Factoring Mapping into Quotient and Injection | Let $f: S \to T$ be a mapping.
Then $f$ can be uniquely '''factored into''' a quotient mapping, followed by an injection.
Thus:
:$f = h \circ q_{\RR_f}$
where:
:$q_{\RR_f}: S \to S / \RR_f: \map {q_{\RR_f} } s = \eqclass s {\RR_f}$
:$h: S / \RR_f \to T: \map h {\eqclass s {\RR_f} } = \map f s$
:$\eqclass s {\RR_f}$ den... | The mapping $q_{\RR_f}: S \to S / \RR_f$ follows from the definition of quotient mapping.
The mapping $h$ is justified by Condition for Mapping from Quotient Set to be Well-Defined.
{{qed}} | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then $f$ can be uniquely '''factored into''' a [[Definition:Quotient Mapping|quotient mapping]], followed by an [[Definition:Injection|injection]].
Thus:
:$f = h \circ q_{\RR_f}$
where:
:$q_{\RR_f}: S \to S / \RR_f: \map {q_{\RR_f} } s = \eqclass s {\RR_f}$
:... | The mapping $q_{\RR_f}: S \to S / \RR_f$ follows from the definition of [[Definition:Quotient Mapping|quotient mapping]].
The mapping $h$ is justified by [[Condition for Mapping from Quotient Set to be Well-Defined]].
{{qed}} | Factoring Mapping into Quotient and Injection | https://proofwiki.org/wiki/Factoring_Mapping_into_Quotient_and_Injection | https://proofwiki.org/wiki/Factoring_Mapping_into_Quotient_and_Injection | [
"Injections",
"Quotient Mappings"
] | [
"Definition:Mapping",
"Definition:Quotient Mapping",
"Definition:Injection",
"Definition:Equivalence Class",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Commutative Diagram"
] | [
"Definition:Quotient Mapping",
"Condition for Mapping from Quotient Set to be Well-Defined"
] |
proofwiki-2865 | Subsets in Increasing Union | Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be a nested sequence of sets, that is:
:$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$
Let $S$ be the increasing union of $S_0, S_1, S_2, \ldots, S_i, \ldots$:
:$\ds S = \bigcup_{i \mathop \in \N} S_i$
Then:
:$\forall s \in S: \exists k \in \N:... | Let $k \in \N$.
Let $j \ge k$.
Then by as many applications as necessary of Subset Relation is Transitive, we have:
:$S_k \subseteq S_j$
Now $s \in S$ means, by definition of set union, that:
:$\exists S_k \subseteq S: s \in S_k$
Then from above:
:$j \ge k \implies S_k \subseteq S_j$
it follows directly that:
:$\forall... | Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be a [[Definition:Nested Sequence|nested sequence]] of [[Definition:Set|sets]], that is:
:$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$
Let $S$ be the [[Definition:Increasing Union|increasing union]] of $S_0, S_1, S_2, \ldots, S_i, \ldots$:
:... | Let $k \in \N$.
Let $j \ge k$.
Then by as many applications as necessary of [[Subset Relation is Transitive]], we have:
:$S_k \subseteq S_j$
Now $s \in S$ means, by definition of [[Definition:Set Union|set union]], that:
:$\exists S_k \subseteq S: s \in S_k$
Then from above:
:$j \ge k \implies S_k \subseteq S_j$
... | Subsets in Increasing Union | https://proofwiki.org/wiki/Subsets_in_Increasing_Union | https://proofwiki.org/wiki/Subsets_in_Increasing_Union | [
"Subsets",
"Set Union"
] | [
"Definition:Nested Sequence",
"Definition:Set",
"Definition:Increasing Union"
] | [
"Subset Relation is Transitive",
"Definition:Set Union",
"Definition:Subset"
] |
proofwiki-2866 | Mapping on Increasing Union | Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be sets such that:
:$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$
that is, each set is contained in the next as a subset.
Let $S$ be the increasing union of $S_0, S_1, S_2, \ldots, S_i, \ldots$:
:$\ds S = \bigcup_{i \mathop \in \N} S_i$
For ea... | Suppose $f: S \to T$ and $g: S \to T$ are both extensions of $f_i$ to $S$ for all $i \in \N$ such that $f \ne g$.
We have been given the domain and codomain of $f$.
Suppose there exists any $g: S \to T$ which is different from $f$.
This must be because $f$ and $g$ do not agree throughout their entire domain.
Thus:
:$\e... | Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be [[Definition:Set|sets]] such that:
:$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$
that is, each set is contained in the next as a [[Definition:Subset|subset]].
Let $S$ be the [[Definition:Increasing Union|increasing union]] of $S_0, S_1, ... | Suppose $f: S \to T$ and $g: S \to T$ are both [[Definition:Extension of Mapping|extensions]] of $f_i$ to $S$ for all $i \in \N$ such that $f \ne g$.
We have been given the [[Definition:Domain of Mapping|domain]] and [[Definition:Codomain of Mapping|codomain]] of $f$.
Suppose there exists any $g: S \to T$ which is di... | Mapping on Increasing Union | https://proofwiki.org/wiki/Mapping_on_Increasing_Union | https://proofwiki.org/wiki/Mapping_on_Increasing_Union | [
"Mapping Theory",
"Set Union"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Increasing Union",
"Definition:Mapping",
"Definition:Restriction/Mapping",
"Definition:Mapping",
"Definition:Extension of Mapping"
] | [
"Definition:Extension of Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Agreement/Mappings",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Set Union",
"Definition:Extension of Mapping",
"Definition:Agreement/Mappings"
] |
proofwiki-2867 | Futurama Theorem | Let $A_{n - 2} \subset A_n$ be a subgroup of the alternating group on $n$ letters $A_n$.
For any element $x \in A_{n - 2}$, let $x = x_1 x_2 \dots x_k$, where $x_i \in H$ is a transposition.
{{explain|What is $H$?}}
Then there exists $y$ which can be represented as a series of transpositions $y_1 y_2 \dots y_j \in A_n$... | Let $w = (n [n - 1])$, that is, the transposition of the $n^{th}$ and $n - 1^{th}$ letters that we consider $A_n$ acting on.
Then the permutation $x^{-1} w$ is the $y$ of the theorem.
{{qed}} | Let $A_{n - 2} \subset A_n$ be a [[Definition:Subgroup|subgroup]] of the [[Definition:Alternating Group|alternating group on $n$ letters]] $A_n$.
For any [[Definition:Element|element]] $x \in A_{n - 2}$, let $x = x_1 x_2 \dots x_k$, where $x_i \in H$ is a [[Definition:Transposition|transposition]].
{{explain|What is ... | Let $w = (n [n - 1])$, that is, the [[Definition:Transposition|transposition]] of the $n^{th}$ and $n - 1^{th}$ letters that we consider $A_n$ acting on.
Then the permutation $x^{-1} w$ is the $y$ of the theorem.
{{qed}} | Futurama Theorem | https://proofwiki.org/wiki/Futurama_Theorem | https://proofwiki.org/wiki/Futurama_Theorem | [
"Permutations",
"Named Theorems"
] | [
"Definition:Subgroup",
"Definition:Alternating Group",
"Definition:Element",
"Definition:Transposition",
"Definition:Transposition",
"Definition:Transposition"
] | [
"Definition:Transposition"
] |
proofwiki-2868 | Pointwise Addition on Continuous Real Functions forms Group | Let $C$ be the set of all continuous real functions on the set of real numbers $\R$.
Let $f, g \in C$.
Let $f + g$ be the pointwise sum of $f$ and $g$:
:$\forall x \in R: \map {\paren {f + g} } x = \map f x + \map g x$
Then $\struct {C, +}$, the algebraic structure on $C$ induced by $+$, forms a group. | Taking the group axioms in turn: | Let $C$ be the [[Definition:Set|set]] of all [[Definition:Continuous Real Function|continuous real functions]] on the set of [[Definition:Real Number|real numbers]] $\R$.
Let $f, g \in C$.
Let $f + g$ be the [[Definition:Pointwise Addition|pointwise sum]] of $f$ and $g$:
:$\forall x \in R: \map {\paren {f + g} } x = ... | Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Pointwise Addition on Continuous Real Functions forms Group | https://proofwiki.org/wiki/Pointwise_Addition_on_Continuous_Real_Functions_forms_Group | https://proofwiki.org/wiki/Pointwise_Addition_on_Continuous_Real_Functions_forms_Group | [
"Real Analysis",
"Examples of Groups",
"Pointwise Addition"
] | [
"Definition:Set",
"Definition:Continuous Real Function",
"Definition:Real Number",
"Definition:Pointwise Addition",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Group"
] | [
"Axiom:Group Axioms",
"Axiom:Group Axioms"
] |
proofwiki-2869 | General Intersection Property of Topological Space | Let $\struct {S, \tau}$ be a topological space.
Let $S_1, S_2, \ldots, S_n$ be open sets of $\struct {S, \tau}$.
Then:
:$\ds \bigcap_{i \mathop = 1}^n S_i$
is also an open set of $\struct {S, \tau}$.
That is, the intersection of any finite number of open sets of a topology is also in $\tau$.
Conversely, if the intersec... | Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:For any sets $S_1, S_2, \ldots, S_n \in \tau$, it follows that $\ds \bigcap_{i \mathop = 1}^n S_i \in \tau$.
Let $\mathbb S$ be any finite subset of $\tau$.
From Intersection of Empty Set, we have that:
:$\mathbb S = \O \implies \bigcap \m... | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $S_1, S_2, \ldots, S_n$ be [[Definition:Open Set (Topology)|open sets]] of $\struct {S, \tau}$.
Then:
:$\ds \bigcap_{i \mathop = 1}^n S_i$
is also an [[Definition:Open Set (Topology)|open set]] of $\struct {S, \tau}$.
That is, the ... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:For any sets $S_1, S_2, \ldots, S_n \in \tau$, it follows that $\ds \bigcap_{i \mathop = 1}^n S_i \in \tau$.
Let $\mathbb S$ be any [[Definition:Finite Set|finite]] [[De... | General Intersection Property of Topological Space | https://proofwiki.org/wiki/General_Intersection_Property_of_Topological_Space | https://proofwiki.org/wiki/General_Intersection_Property_of_Topological_Space | [
"Topology"
] | [
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Set Intersection",
"Definition:Finite Set",
"Definition:Open Set/Topology",
"Definition:Topology",
"Definition:Set Intersection",
"Definition:Finite Set",
"Definition:Open Set/Topology",
"... | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Finite Set",
"Definition:Subset",
"Intersection of Empty Set",
"Definition:Topology",
"Intersection of Singleton",
"Principle of Mathematical Induction",
"Intersection of Empty Set"
] |
proofwiki-2870 | Empty Set is Element of Topology | Let $T = \struct {X, \tau}$ be a topological space.
Then the empty set $\O$ is an open set of $T$. | By Empty Set is Subset of All Sets:
:$\O \subseteq \tau$
By Union of Empty Set:
:$\O = \bigcup \O$
By {{Open-set-axiom|1}}:
:$\O \in \tau$
{{qed}}
Category:Topology
Category:Empty Set
6b0ibvhzp9mr16tg9angho7kn8cv958 | Let $T = \struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Then the [[Definition:Empty Set|empty set]] $\O$ is an [[Definition:Open Set (Topology)|open set]] of $T$. | By [[Empty Set is Subset of All Sets]]:
:$\O \subseteq \tau$
By [[Union of Empty Set]]:
:$\O = \bigcup \O$
By {{Open-set-axiom|1}}:
:$\O \in \tau$
{{qed}}
[[Category:Topology]]
[[Category:Empty Set]]
6b0ibvhzp9mr16tg9angho7kn8cv958 | Empty Set is Element of Topology | https://proofwiki.org/wiki/Empty_Set_is_Element_of_Topology | https://proofwiki.org/wiki/Empty_Set_is_Element_of_Topology | [
"Topology",
"Empty Set"
] | [
"Definition:Topological Space",
"Definition:Empty Set",
"Definition:Open Set/Topology"
] | [
"Empty Set is Subset of All Sets",
"Union of Empty Set",
"Category:Topology",
"Category:Empty Set"
] |
proofwiki-2871 | Coarseness Relation on Topologies is Partial Ordering | Let $S$ be a set.
Let $\mathbb T$ be the set of all topologies on $S$.
Let $\le$ be the relation on $\mathbb T$ defined as:
:$\forall \tau_1, \tau_2 \in \mathbb T: \tau_1 \le \tau_2$ {{iff}} $\tau_1$ is coarser than $\tau_2$.
Then $\le$ is a partial ordering on $\mathbb T$. | It follows directly from the definition of being coarser that:
:$\tau_1 \le \tau_2 \iff \tau_1 \subseteq \tau_2$
By Subset Relation is Ordering it thus follows that $\le$ is an ordering.
From Topologies are not necessarily Comparable by Coarseness, it follows that such an ordering is not always total.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $\mathbb T$ be the set of all [[Definition:Topology|topologies]] on $S$.
Let $\le$ be the [[Definition:Relation|relation]] on $\mathbb T$ defined as:
:$\forall \tau_1, \tau_2 \in \mathbb T: \tau_1 \le \tau_2$ {{iff}} $\tau_1$ is [[Definition:Coarser Topology|coarser]] than $\... | It follows directly from the definition of being [[Definition:Coarser Topology|coarser]] that:
:$\tau_1 \le \tau_2 \iff \tau_1 \subseteq \tau_2$
By [[Subset Relation is Ordering]] it thus follows that $\le$ is an [[Definition:Ordering|ordering]].
From [[Topologies are not necessarily Comparable by Coarseness]], it ... | Coarseness Relation on Topologies is Partial Ordering | https://proofwiki.org/wiki/Coarseness_Relation_on_Topologies_is_Partial_Ordering | https://proofwiki.org/wiki/Coarseness_Relation_on_Topologies_is_Partial_Ordering | [
"Coarser Topology",
"Partial Orderings"
] | [
"Definition:Set",
"Definition:Topology",
"Definition:Relation",
"Definition:Coarser Topology",
"Definition:Partial Ordering"
] | [
"Definition:Coarser Topology",
"Subset Relation is Ordering",
"Definition:Ordering",
"Topologies are not necessarily Comparable by Coarseness",
"Definition:Total Ordering"
] |
proofwiki-2872 | Topology Defined by Closed Sets | Let $S$ be a set.
Let $\tau$ be a set of subsets of $S$.
Then $\tau$ is a topology on $S$ {{iff}}:
:$(1): \quad$ Any intersection of arbitrarily many closed sets of $S$ under $\tau$ is a closed set of $S$ under $\tau$
:$(2): \quad$ The union of any finite number of closed sets of $S$ under $\tau$ is a closed set of $S$... | From the definition, if $V$ is a closed set of $S$, then $S \setminus V$ is an open set of $S$.
Let $\mathbb V$ be any arbitrary set of closed sets of $S$.
Then by De Morgan's Laws: Difference with Intersection, we have:
:$\ds S \setminus \bigcap \mathbb V = \bigcup_{V \mathop \in \mathbb V} \paren {S \setminus V}$
Fir... | Let $S$ be a [[Definition:Set|set]].
Let $\tau$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
Then $\tau$ is a [[Definition:Topology|topology]] on $S$ {{iff}}:
:$(1): \quad$ Any [[Definition:Set Intersection|intersection]] of arbitrarily many [[Definition:Closed Set (Topology)|closed sets]] o... | From the definition, if $V$ is a [[Definition:Closed Set (Topology)|closed set]] of $S$, then $S \setminus V$ is an [[Definition:Open Set (Topology)|open set]] of $S$.
Let $\mathbb V$ be any arbitrary [[Definition:Set|set]] of [[Definition:Closed Set (Topology)|closed sets]] of $S$.
Then by [[De Morgan's Laws (Set Th... | Topology Defined by Closed Sets | https://proofwiki.org/wiki/Topology_Defined_by_Closed_Sets | https://proofwiki.org/wiki/Topology_Defined_by_Closed_Sets | [
"Closed Sets",
"Topologies"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Subset",
"Definition:Topology",
"Definition:Set Intersection",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Set Union",
"Definition:Finite Set",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
... | [
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Set",
"Definition:Closed Set/Topology",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection",
"Definition:Topology",
"Intersection of Closed Sets is Closed/Topology",
"Finite Union of Clos... |
proofwiki-2873 | Complement of F-Sigma Set is G-Delta Set | Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an $F_\sigma$ set of $T$.
Then its complement $S \setminus X$ is a $G_\delta$ set of $S$. | Let $X$ be an $F_\sigma$ set of $T$.
Then $X = \ds \bigcup \VV$ where $\VV$ is a countable set of closed sets in $T$.
Then from De Morgan's Laws: Difference with Union we have:
:$\ds S \setminus X = S \setminus \bigcup \VV = \bigcap_{V \mathop \in \VV} \paren {S \setminus V}$
By definition of closed set, each of the $S... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $X$ be an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
Then its [[Definition:Relative Complement|complement]] $S \setminus X$ is a [[Definition:G-Delta Set|$G_\delta$ set]] of $S$. | Let $X$ be an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
Then $X = \ds \bigcup \VV$ where $\VV$ is a [[Definition:Countable Set|countable]] [[Definition:Set|set]] of [[Definition:Closed Set (Topology)|closed sets]] in $T$.
Then from [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Un... | Complement of F-Sigma Set is G-Delta Set | https://proofwiki.org/wiki/Complement_of_F-Sigma_Set_is_G-Delta_Set | https://proofwiki.org/wiki/Complement_of_F-Sigma_Set_is_G-Delta_Set | [
"F-Sigma Sets",
"G-Delta Sets"
] | [
"Definition:Topological Space",
"Definition:F-Sigma Set",
"Definition:Relative Complement",
"Definition:G-Delta Set"
] | [
"Definition:F-Sigma Set",
"Definition:Countable Set",
"Definition:Set",
"Definition:Closed Set/Topology",
"De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Countable Set",
"Definition:Set I... |
proofwiki-2874 | Closed Set is F-Sigma Set | Let $T = \struct {S, \tau}$ be a topological space.
Let $V$ be a closed set of $T$.
Then $V$ is an $F_\sigma$ set of $T$. | $V$ is the union of a singleton.
So $V$ is trivially the union of a countable number of closed sets of $T$.
The result follows by definition of $F_\sigma$ set.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $V$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
Then $V$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$. | $V$ is the [[Union of Singleton|union of a singleton]].
So $V$ is trivially the [[Definition:Set Union|union]] of a [[Definition:Countable|countable]] number of [[Definition:Closed Set (Topology)|closed sets]] of $T$.
The result follows by [[Definition:F-Sigma Set|definition of $F_\sigma$ set]].
{{qed}} | Closed Set is F-Sigma Set | https://proofwiki.org/wiki/Closed_Set_is_F-Sigma_Set | https://proofwiki.org/wiki/Closed_Set_is_F-Sigma_Set | [
"F-Sigma Sets",
"Closed Sets"
] | [
"Definition:Topological Space",
"Definition:Closed Set/Topology",
"Definition:F-Sigma Set"
] | [
"Union of Singleton",
"Definition:Set Union",
"Definition:Countable Set",
"Definition:Closed Set/Topology",
"Definition:F-Sigma Set"
] |
proofwiki-2875 | Open Set is G-Delta Set | Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ be an open set of $T$.
Then $U$ is a $G_\delta$ set of $T$. | $U$ is the intersection of a singleton.
So $U$ is trivially the intersection of a countable number of open sets of $T$.
The result follows by definition of $G_\delta$ set.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $U$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
Then $U$ is a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$. | $U$ is the [[Intersection of Singleton|intersection of a singleton]].
So $U$ is trivially the [[Definition:Set Intersection|intersection]] of a [[Definition:Countable Set|countable number]] of [[Definition:Open Set (Topology)|open sets]] of $T$.
The result follows by [[Definition:G-Delta Set|definition of $G_\delta$ ... | Open Set is G-Delta Set | https://proofwiki.org/wiki/Open_Set_is_G-Delta_Set | https://proofwiki.org/wiki/Open_Set_is_G-Delta_Set | [
"G-Delta Sets",
"Open Sets"
] | [
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Definition:G-Delta Set"
] | [
"Intersection of Singleton",
"Definition:Set Intersection",
"Definition:Countable Set",
"Definition:Open Set/Topology",
"Definition:G-Delta Set"
] |
proofwiki-2876 | Set is Open iff Neighborhood of all its Points | Let $T = \struct {S, \tau}$ be a topological space.
Let $V \subseteq S$ be a subset of $T$.
Then:
:$V$ is an open set of $T$
{{iff}}:
:$V$ is a neighborhood of all the points in $V$. | === Necessary Condition ===
Let $V$ be open in $T$.
Let $z \in V$.
By definition, a neighborhood of $z$ is any subset of $S$ containing an open set which itself contains $z$.
But $V$ is itself an open set which itself contains $z$.
Hence by Set is Subset of Itself, $V$ is a subset of $S$ which contains an open set whic... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $V \subseteq S$ be a [[Definition:Subset|subset]] of $T$.
Then:
:$V$ is an [[Definition:Open Set (Topology)|open set]] of $T$
{{iff}}:
:$V$ is a [[Definition:Neighborhood of Point|neighborhood]] of all the [[Definition:Element|p... | === Necessary Condition ===
Let $V$ be [[Definition:Open Set (Topology)|open]] in $T$.
Let $z \in V$.
By definition, a [[Definition:Neighborhood of Set|neighborhood]] of $z$ is any [[Definition:Subset|subset]] of $S$ containing an [[Definition:Open Set (Topology)|open set]] which itself contains $z$.
But $V$ is its... | Set is Open iff Neighborhood of all its Points | https://proofwiki.org/wiki/Set_is_Open_iff_Neighborhood_of_all_its_Points | https://proofwiki.org/wiki/Set_is_Open_iff_Neighborhood_of_all_its_Points | [
"Open Sets",
"Neighborhoods"
] | [
"Definition:Topological Space",
"Definition:Subset",
"Definition:Open Set/Topology",
"Definition:Neighborhood (Topology)/Point",
"Definition:Element"
] | [
"Definition:Open Set/Topology",
"Definition:Neighborhood (Topology)/Set",
"Definition:Subset",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Set is Subset of Itself",
"Definition:Subset",
"Definition:Open Set/Topology",
"Definition:Neighborhood (Topology)/Set",
"Definition:Neigh... |
proofwiki-2877 | Relationship between Limit Point Types | Let $T = \struct {X, \tau}$ be a topological space.
Let $A \subseteq X$.
Let:
:$C$ be the set of condensation points of $A$
:$W$ be the set of $\omega$-accumulation points of $A$
:$L$ be the set of limit points of $A$
:$D$ be the set of adherent points of $A$.
Then:
:$C \subseteq W \subseteq L \subseteq D$
That is:
:Ev... | Let $x \in C$.
By definition of condensation point, every open set of $T$ containing $x$ also contains an uncountable number of points of $A$.
As an uncountable number is also an infinite number, we could also say that every open set of $T$ containing $x$ also contains an infinite number of points of $A$.
That is, $x$ ... | Let $T = \struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $A \subseteq X$.
Let:
:$C$ be the set of [[Definition:Condensation Point|condensation points]] of $A$
:$W$ be the set of [[Definition:Omega-Accumulation Point|$\omega$-accumulation points]] of $A$
:$L$ be the set of [[Definiti... | Let $x \in C$.
By [[Definition:Condensation Point|definition of condensation point]], every [[Definition:Open Set (Topology)|open set]] of $T$ containing $x$ also contains an [[Definition:Uncountable Set|uncountable number]] of points of $A$.
As an [[Definition:Uncountable Set|uncountable number]] is also an [[Defini... | Relationship between Limit Point Types | https://proofwiki.org/wiki/Relationship_between_Limit_Point_Types | https://proofwiki.org/wiki/Relationship_between_Limit_Point_Types | [
"Limit Points of Sets",
"Omega-Accumulation Points",
"Adherent Points of Sets",
"Condensation Points"
] | [
"Definition:Topological Space",
"Definition:Condensation Point",
"Definition:Omega-Accumulation Point",
"Definition:Limit Point/Topology/Set",
"Definition:Adherent Point of Set",
"Definition:Condensation Point",
"Definition:Omega-Accumulation Point",
"Definition:Omega-Accumulation Point",
"Definitio... | [
"Definition:Condensation Point",
"Definition:Open Set/Topology",
"Definition:Uncountable/Set",
"Definition:Uncountable/Set",
"Definition:Infinite",
"Definition:Open Set/Topology",
"Definition:Infinite",
"Definition:Omega-Accumulation Point",
"Definition:Subset",
"Definition:Open Set/Topology",
"... |
proofwiki-2878 | Limit of Sequence is Accumulation Point | Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
Let $\sequence {x_n}$ be a sequence in $A$.
Let $\alpha$ be a limit of $\sequence {x_n}$.
Then $\alpha$ is also an accumulation point of $\sequence {x_n}$. | Let $\alpha$ be a limit of $\sequence {x_n}$.
Then by definition of limit, $\sequence {x_n}$ converges to $\alpha$.
By definition of convergence that means:
:for any open set $U \subseteq T$ such that $\alpha \in U$: $\exists N \in \R: n > N \implies x_n \in U$.
As there is an infinite number of values of $n > N$, ther... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $A \subseteq S$.
Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] in $A$.
Let $\alpha$ be a [[Definition:Limit of Sequence (Topology)|limit of $\sequence {x_n}$]].
Then $\alpha$ is also an [[Definition:Accumulation... | Let $\alpha$ be a [[Definition:Limit of Sequence (Topology)|limit of $\sequence {x_n}$]].
Then by definition of [[Definition:Limit of Sequence (Topology)|limit]], $\sequence {x_n}$ [[Definition:Convergent Sequence (Topology)|converges]] to $\alpha$.
By definition of [[Definition:Convergent Sequence (Topology)|converg... | Limit of Sequence is Accumulation Point | https://proofwiki.org/wiki/Limit_of_Sequence_is_Accumulation_Point | https://proofwiki.org/wiki/Limit_of_Sequence_is_Accumulation_Point | [
"Limit Points",
"Accumulation Points"
] | [
"Definition:Topological Space",
"Definition:Sequence",
"Definition:Limit of Sequence/Topological Space",
"Definition:Accumulation Point/Sequence"
] | [
"Definition:Limit of Sequence/Topological Space",
"Definition:Limit of Sequence/Topological Space",
"Definition:Convergent Sequence/Topology",
"Definition:Convergent Sequence/Topology",
"Definition:Open Set/Topology",
"Definition:Infinite Set",
"Definition:Infinite Set",
"Definition:Accumulation Point... |
proofwiki-2879 | Sum of Independent Poisson Random Variables is Poisson | Let $X$ and $Y$ be independent discrete random variables with:
:$X \sim \Poisson {\lambda_1}$
and
:$Y \sim \Poisson {\lambda_2}$
for some $\lambda_1, \lambda_2 \in \R_{> 0}$.
Then their sum $X + Y$ is distributed:
:$X + Y \sim \Poisson {\lambda_1 + \lambda_2}$ | From Moment Generating Function of Poisson Distribution, the moment generating functions $X$ and $Y$, $M_X$ and $M_Y$, are given by:
:$\map {M_X} t = e^{\lambda_1 \paren {e^t - 1} }$
and
:$\map {M_Y} t = e^{\lambda_2 \paren {e^t - 1} }$
As $X$ and $Y$ are independent, we may apply Moment Generating Function of Linear ... | Let $X$ and $Y$ be [[Definition:Independent Random Variables|independent]] [[Definition:Discrete Random Variable|discrete random variables]] with:
:$X \sim \Poisson {\lambda_1}$
and
:$Y \sim \Poisson {\lambda_2}$
for some $\lambda_1, \lambda_2 \in \R_{> 0}$.
Then their sum $X + Y$ is distributed:
:$X + Y \sim \Pois... | From [[Moment Generating Function of Poisson Distribution]], the [[Definition:Moment Generating Function|moment generating functions]] $X$ and $Y$, $M_X$ and $M_Y$, are given by:
:$\map {M_X} t = e^{\lambda_1 \paren {e^t - 1} }$
and
:$\map {M_Y} t = e^{\lambda_2 \paren {e^t - 1} }$
As $X$ and $Y$ are independent, ... | Sum of Independent Poisson Random Variables is Poisson/Proof 2 | https://proofwiki.org/wiki/Sum_of_Independent_Poisson_Random_Variables_is_Poisson | https://proofwiki.org/wiki/Sum_of_Independent_Poisson_Random_Variables_is_Poisson/Proof_2 | [
"Poisson Distribution",
"Probability Generating Functions",
"Sum of Independent Poisson Random Variables is Poisson"
] | [
"Definition:Independent Random Variables",
"Definition:Random Variable/Discrete"
] | [
"Moment Generating Function of Poisson Distribution",
"Definition:Moment Generating Function",
"Moment Generating Function of Linear Combination of Independent Random Variables",
"Moment Generating Function of Linear Combination of Independent Random Variables",
"Exponential of Sum"
] |
proofwiki-2880 | Filter on Set is Proper Filter | Let $S$ be a set.
Let $\powerset S$ denote the power set of $S$.
Let $\struct {\powerset S, \subseteq}$ be the poset defined on $\powerset S$ by the subset relation.
Let $\FF$ be a filter on $S$.
Then $\FF$ is a proper filter on $\struct {\powerset S, \subseteq}$. | From the general definition of a filter, we have:
A '''filter on $\struct {S, \preccurlyeq}$''' is a subset $\FF \subseteq S$ which satisfies the following conditions:
:$\FF \ne \O$
:$x, y \in \FF \implies \exists z \in \FF: z \preccurlyeq x, z \preccurlyeq y$
:$\forall x \in \FF: \forall y \in S: x \preccurlyeq y \imp... | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$.
Let $\struct {\powerset S, \subseteq}$ be the [[Subset Relation on Power Set is Partial Ordering|poset defined on $\powerset S$]] by the [[Definition:Subset|subset relation]].
Let $\FF$ be a [[Definition:Fil... | From the general definition of a [[Definition:Filter|filter]], we have:
A '''filter on $\struct {S, \preccurlyeq}$''' is a [[Definition:Subset|subset]] $\FF \subseteq S$ which satisfies the following conditions:
:$\FF \ne \O$
:$x, y \in \FF \implies \exists z \in \FF: z \preccurlyeq x, z \preccurlyeq y$
:$\forall x... | Filter on Set is Proper Filter | https://proofwiki.org/wiki/Filter_on_Set_is_Proper_Filter | https://proofwiki.org/wiki/Filter_on_Set_is_Proper_Filter | [
"Filter Theory",
"Filter on Set is Proper Filter"
] | [
"Definition:Set",
"Definition:Power Set",
"Subset Relation on Power Set is Partial Ordering",
"Definition:Subset",
"Definition:Filter on Set",
"Definition:Filter/Proper Filter"
] | [
"Definition:Filter",
"Definition:Subset",
"Definition:Filter/Proper Filter",
"Definition:Filter on Set",
"Definition:Filter/Proper Filter"
] |
proofwiki-2881 | Peirce's Law is Equivalent to Law of Excluded Middle | Peirce's Law:
:$\paren {p \implies q} \implies p \vdash p$
is logically equivalent to the Law of Excluded Middle:
:$\vdash p \lor \neg p$
That is, Peirce's Law holds {{iff}} the Law of Excluded Middle holds. | === Law of Excluded Middle implies Peirce's Law ===
{{:Law of Excluded Middle implies Peirce's Law}}
{{qed|lemma}} | [[Peirce's Law]]:
:$\paren {p \implies q} \implies p \vdash p$
is [[Definition:Logical Equivalence|logically equivalent]] to the [[Law of Excluded Middle]]:
:$\vdash p \lor \neg p$
That is, [[Peirce's Law]] holds {{iff}} the [[Law of Excluded Middle]] holds. | === [[Law of Excluded Middle implies Peirce's Law]] ===
{{:Law of Excluded Middle implies Peirce's Law}}
{{qed|lemma}} | Peirce's Law is Equivalent to Law of Excluded Middle | https://proofwiki.org/wiki/Peirce's_Law_is_Equivalent_to_Law_of_Excluded_Middle | https://proofwiki.org/wiki/Peirce's_Law_is_Equivalent_to_Law_of_Excluded_Middle | [
"Peirce's Law",
"Law of Excluded Middle"
] | [
"Peirce's Law",
"Definition:Logical Equivalence",
"Law of Excluded Middle",
"Peirce's Law",
"Law of Excluded Middle"
] | [
"Law of Excluded Middle implies Peirce's Law"
] |
proofwiki-2882 | Conditional is Left Distributive over Conjunction | The implication operator is left distributive over the conjunction operator:
=== Formulation 1 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2}} | {{BeginTableau|p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \land r} }}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|q \land r|1|2}}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|r|3|2}}
{{Implication|6|1|p \implies q|2|4}}
{{Implication... | The [[Definition:Conditional|implication operator]] is [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Conjunction|conjunction operator]]:
=== [[Conditional is Left Distributive over Conjunction/Formulation 1|Formulation 1]] ===
{{:Conditional is Left Distributive over Conjunction/Fo... | {{BeginTableau|p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r} }}
{{Premise|1|p \implies \paren {q \land r} }}
{{Assumption|2|p}}
{{ModusPonens|3|1, 2|q \land r|1|2}}
{{Simplification|4|1, 2|q|3|1}}
{{Simplification|5|1, 2|r|3|2}}
{{Implication|6|1|p \implies q|2|4}}
{{Implication... | Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Forward_Implication/Proof | [
"Conditional is Left Distributive over Conjunction",
"Conditional",
"Conjunction"
] | [
"Definition:Conditional",
"Definition:Distributive Operation/Left",
"Definition:Conjunction",
"Conditional is Left Distributive over Conjunction/Formulation 1",
"Conditional is Left Distributive over Conjunction/Formulation 2"
] | [] |
proofwiki-2883 | Conditional is Left Distributive over Conjunction | The implication operator is left distributive over the conjunction operator:
=== Formulation 1 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2}} | === Proof of Forward Implication ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof}}
=== Proof of Reverse Implication ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication/Proof}} | The [[Definition:Conditional|implication operator]] is [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Conjunction|conjunction operator]]:
=== [[Conditional is Left Distributive over Conjunction/Formulation 1|Formulation 1]] ===
{{:Conditional is Left Distributive over Conjunction/Fo... | === [[Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof|Proof of Forward Implication]] ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof}}
=== [[Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication/Pro... | Conditional is Left Distributive over Conjunction/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Proof_1 | [
"Conditional is Left Distributive over Conjunction",
"Conditional",
"Conjunction"
] | [
"Definition:Conditional",
"Definition:Distributive Operation/Left",
"Definition:Conjunction",
"Conditional is Left Distributive over Conjunction/Formulation 1",
"Conditional is Left Distributive over Conjunction/Formulation 2"
] | [
"Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication/Proof",
"Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication/Proof"
] |
proofwiki-2884 | Conditional is Left Distributive over Conjunction | The implication operator is left distributive over the conjunction operator:
=== Formulation 1 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Conditional is Left Distributive over Conjunction/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}{|ccccc||ccccccc|} \hline
p & \implies & (q & \land & r) & (p & \implies & q) & \land & (p & \implies & r) \\
\hline
\F & \T & \F & \F & \F & \F & \T & \F &... | The [[Definition:Conditional|implication operator]] is [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Conjunction|conjunction operator]]:
=== [[Conditional is Left Distributive over Conjunction/Formulation 1|Formulation 1]] ===
{{:Conditional is Left Distributive over Conjunction/Fo... | 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}{|ccccc||ccccccc|} \hline
p & \i... | Conditional is Left Distributive over Conjunction/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Proof_by_Truth_Table | [
"Conditional is Left Distributive over Conjunction",
"Conditional",
"Conjunction"
] | [
"Definition:Conditional",
"Definition:Distributive Operation/Left",
"Definition:Conjunction",
"Conditional is Left Distributive over Conjunction/Formulation 1",
"Conditional is Left Distributive over Conjunction/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-2885 | Conditional is Left Distributive over Conjunction | The implication operator is left distributive over the conjunction operator:
=== Formulation 1 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2}} | {{BeginTableau|\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r} }}
{{Premise|1|\paren {p \implies q} \land \paren {p \implies r} }}
{{SequentIntro|2|1|\paren {p \land p} \implies \paren {q \land r}|1|Praeclarum Theorema}}
{{Assumption|3|p}}
{{Idempotence|4|3|p \land p|3|Conjunction... | The [[Definition:Conditional|implication operator]] is [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Conjunction|conjunction operator]]:
=== [[Conditional is Left Distributive over Conjunction/Formulation 1|Formulation 1]] ===
{{:Conditional is Left Distributive over Conjunction/Fo... | {{BeginTableau|\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r} }}
{{Premise|1|\paren {p \implies q} \land \paren {p \implies r} }}
{{SequentIntro|2|1|\paren {p \land p} \implies \paren {q \land r}|1|[[Praeclarum Theorema]]}}
{{Assumption|3|p}}
{{Idempotence|4|3|p \land p|3|Conjunc... | Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_1/Reverse_Implication/Proof | [
"Conditional is Left Distributive over Conjunction",
"Conditional",
"Conjunction"
] | [
"Definition:Conditional",
"Definition:Distributive Operation/Left",
"Definition:Conjunction",
"Conditional is Left Distributive over Conjunction/Formulation 1",
"Conditional is Left Distributive over Conjunction/Formulation 2"
] | [
"Praeclarum Theorema"
] |
proofwiki-2886 | Conditional is Left Distributive over Conjunction | The implication operator is left distributive over the conjunction operator:
=== Formulation 1 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2}} | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implies \... | The [[Definition:Conditional|implication operator]] is [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Conjunction|conjunction operator]]:
=== [[Conditional is Left Distributive over Conjunction/Formulation 1|Formulation 1]] ===
{{:Conditional is Left Distributive over Conjunction/Fo... | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implie... | Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Forward_Implication/Proof_2 | [
"Conditional is Left Distributive over Conjunction",
"Conditional",
"Conjunction"
] | [
"Definition:Conditional",
"Definition:Distributive Operation/Left",
"Definition:Conjunction",
"Conditional is Left Distributive over Conjunction/Formulation 1",
"Conditional is Left Distributive over Conjunction/Formulation 2"
] | [
"Conditional is Left Distributive over Conjunction/Formulation 1/Forward Implication"
] |
proofwiki-2887 | Conditional is Left Distributive over Conjunction | The implication operator is left distributive over the conjunction operator:
=== Formulation 1 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2}} | === Proof of Forward Implication ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 1}}
=== Proof of Reverse Implication ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication/Proof 1}}
{{BeginTableau|\vdash \paren {p \implies \paren {q \... | The [[Definition:Conditional|implication operator]] is [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Conjunction|conjunction operator]]:
=== [[Conditional is Left Distributive over Conjunction/Formulation 1|Formulation 1]] ===
{{:Conditional is Left Distributive over Conjunction/Fo... | === [[Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 1|Proof of Forward Implication]] ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 1}}
=== [[Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication... | Conditional is Left Distributive over Conjunction/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Proof_1 | [
"Conditional is Left Distributive over Conjunction",
"Conditional",
"Conjunction"
] | [
"Definition:Conditional",
"Definition:Distributive Operation/Left",
"Definition:Conjunction",
"Conditional is Left Distributive over Conjunction/Formulation 1",
"Conditional is Left Distributive over Conjunction/Formulation 2"
] | [
"Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication/Proof 1",
"Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication/Proof 1",
"Conditional is Left Distributive over Conjunction/Formulation 2/Forward Implication",
"Conditional is Left Distributive... |
proofwiki-2888 | Conditional is Left Distributive over Conjunction | The implication operator is left distributive over the conjunction operator:
=== Formulation 1 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2}} | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc|c|ccccccc|} \hline
(p & \implies & (q & \land & r)) & \iff & ((p & \implies & q) & \land & (p & \implies & r)) \\
\hline
\F & \T & \F & \F... | The [[Definition:Conditional|implication operator]] is [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Conjunction|conjunction operator]]:
=== [[Conditional is Left Distributive over Conjunction/Formulation 1|Formulation 1]] ===
{{:Conditional is Left Distributive over Conjunction/Fo... | 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]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|... | Conditional is Left Distributive over Conjunction/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Proof_by_Truth_Table | [
"Conditional is Left Distributive over Conjunction",
"Conditional",
"Conjunction"
] | [
"Definition:Conditional",
"Definition:Distributive Operation/Left",
"Definition:Conjunction",
"Conditional is Left Distributive over Conjunction/Formulation 1",
"Conditional is Left Distributive over Conjunction/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-2889 | Conditional is Left Distributive over Conjunction | The implication operator is left distributive over the conjunction operator:
=== Formulation 1 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 1}}
=== Formulation 2 ===
{{:Conditional is Left Distributive over Conjunction/Formulation 2}} | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implies \... | The [[Definition:Conditional|implication operator]] is [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Conjunction|conjunction operator]]:
=== [[Conditional is Left Distributive over Conjunction/Formulation 1|Formulation 1]] ===
{{:Conditional is Left Distributive over Conjunction/Fo... | Let us use the following abbreviations
{{begin-eqn}}
{{eqn | l = \phi
| o = \text {for}
| r = \paren {p \implies q} \land \paren {p \implies r}
| c =
}}
{{eqn | l = \psi
| o = \text {for}
| r = p \implies \paren {q \land r}
| c =
}}
{{end-eqn}}
{{BeginTableau|\vdash \phi \implie... | Conditional is Left Distributive over Conjunction/Formulation 2/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction | https://proofwiki.org/wiki/Conditional_is_Left_Distributive_over_Conjunction/Formulation_2/Reverse_Implication/Proof_2 | [
"Conditional is Left Distributive over Conjunction",
"Conditional",
"Conjunction"
] | [
"Definition:Conditional",
"Definition:Distributive Operation/Left",
"Definition:Conjunction",
"Conditional is Left Distributive over Conjunction/Formulation 1",
"Conditional is Left Distributive over Conjunction/Formulation 2"
] | [
"Conditional is Left Distributive over Conjunction/Formulation 1/Reverse Implication"
] |
proofwiki-2890 | Existence of Negation Normal Form of Statement | Any propositional formula can be expressed in negation normal form (NNF). | A propositional variable is already trivially in negation normal form (NNF).
So we consider the general propositional formula $S$.
In the following, $P$ and $Q$ are to stand for general propositional formulas.
First, from Functionally Complete Logical Connectives: Conjunction, Negation and Disjunction we have that:
:$\... | Any [[Definition:Propositional Formula|propositional formula]] can be expressed in [[Definition:Negation Normal Form|negation normal form (NNF)]]. | A [[Definition:Propositional Variable|propositional variable]] is already trivially in [[Definition:Negation Normal Form|negation normal form (NNF)]].
So we consider the general [[Definition:Propositional Formula|propositional formula]] $S$.
In the following, $P$ and $Q$ are to stand for general [[Definition:Proposi... | Existence of Negation Normal Form of Statement | https://proofwiki.org/wiki/Existence_of_Negation_Normal_Form_of_Statement | https://proofwiki.org/wiki/Existence_of_Negation_Normal_Form_of_Statement | [
"Negation Normal Form"
] | [
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Negation Normal Form"
] | [
"Definition:Variable/Propositional Logic",
"Definition:Negation Normal Form",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Functionally Complete Logical Connectives/Conjunction, Negation and Disjunction",
"Definition:Fun... |
proofwiki-2891 | Existence of Conjunctive Normal Form of Statement | Any propositional formula can be expressed in conjunctive normal form (CNF). | A propositional variable is already trivially in conjunctive normal form (CNF).
So we consider the general propositional formula $S$.
First we convert to negation normal form (NNF).
This is always possible, by Existence of Negation Normal Form of Statement.
Now $S$ will be of the form:
:$P_1 \land P_2 \land \cdots \lan... | Any [[Definition:Propositional Formula|propositional formula]] can be expressed in [[Definition:Conjunctive Normal Form|conjunctive normal form (CNF)]]. | A [[Definition:Propositional Variable|propositional variable]] is already trivially in [[Definition:Conjunctive Normal Form|conjunctive normal form (CNF)]].
So we consider the general [[Definition:Propositional Formula|propositional formula]] $S$.
First we convert to [[Definition:Negation Normal Form|negation normal... | Existence of Conjunctive Normal Form of Statement | https://proofwiki.org/wiki/Existence_of_Conjunctive_Normal_Form_of_Statement | https://proofwiki.org/wiki/Existence_of_Conjunctive_Normal_Form_of_Statement | [
"Conjunctive Normal Form",
"Propositional Logic"
] | [
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Conjunctive Normal Form"
] | [
"Definition:Variable/Propositional Logic",
"Definition:Conjunctive Normal Form",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Negation Normal Form",
"Existence of Negation Normal Form of Statement",
"Definition:Literal",
"Definition:Literal",
"Rule of Distribution/Disju... |
proofwiki-2892 | Existence of Disjunctive Normal Form of Statement | Any propositional formula can be expressed in disjunctive normal form (DNF). | A propositional variable is already trivially in disjunctive normal form (DNF).
So we consider the general propositional formula $S$.
First we convert to negation normal form (NNF).
This is always possible, by Existence of Negation Normal Form of Statement.
Now $S$ will be of the form:
:$P_1 \lor P_2 \lor \cdots \lor P... | Any [[Definition:Propositional Formula|propositional formula]] can be expressed in [[Definition:Disjunctive Normal Form|disjunctive normal form (DNF)]]. | A [[Definition:Propositional Variable|propositional variable]] is already trivially in [[Definition:Disjunctive Normal Form|disjunctive normal form (DNF)]].
So we consider the general [[Definition:Propositional Formula|propositional formula]] $S$.
First we convert to [[Definition:Negation Normal Form|negation normal... | Existence of Disjunctive Normal Form of Statement | https://proofwiki.org/wiki/Existence_of_Disjunctive_Normal_Form_of_Statement | https://proofwiki.org/wiki/Existence_of_Disjunctive_Normal_Form_of_Statement | [
"Disjunctive Normal Form"
] | [
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Disjunctive Normal Form"
] | [
"Definition:Variable/Propositional Logic",
"Definition:Disjunctive Normal Form",
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Negation Normal Form",
"Existence of Negation Normal Form of Statement",
"Definition:Literal",
"Definition:Literal",
"Rule of Distribution/Conju... |
proofwiki-2893 | Existence of Base-N Representation | {{refactor|lemmata and sub proofs|level = medium}}
{{tidy|Could be improved in presentational style}}
Given a number $x \in \hointr 0 1$, there exists a representation of that number in a base-$p$ positional system.
Specifically, there exists a sequence $\sequence {a_n}$ such that:
:$0 \le a_n < p$, and
:$\ds \sum_{n \... | === Existence of Representation ===
Define:
:$\ds a_j = \floor {\paren {x - \sum_{i \mathop = 1}^{j - 1} \frac {a_i} {p^i} } p^j}$
where we accept the vacuous summation $\ds \sum_{i \mathop = 1}^0 a_i p^{-i} = 0$.
This recursive definition allows for all $a_n$ to be computed. | {{refactor|lemmata and sub proofs|level = medium}}
{{tidy|Could be improved in presentational style}}
Given a number $x \in \hointr 0 1$, there exists a representation of that number in a [[Definition:Number Base|base-$p$ positional system]].
Specifically, there exists a [[Definition:Sequence|sequence]] $\sequence {a... | === Existence of Representation ===
Define:
:$\ds a_j = \floor {\paren {x - \sum_{i \mathop = 1}^{j - 1} \frac {a_i} {p^i} } p^j}$
where we accept the [[Definition:Vacuous Summation|vacuous summation]] $\ds \sum_{i \mathop = 1}^0 a_i p^{-i} = 0$.
This recursive definition allows for all $a_n$ to be computed. | Existence of Base-N Representation | https://proofwiki.org/wiki/Existence_of_Base-N_Representation | https://proofwiki.org/wiki/Existence_of_Base-N_Representation | [
"Number Bases"
] | [
"Definition:Number Base",
"Definition:Sequence",
"Definition:Convergent Series",
"Definition:Unique",
"Definition:Sequence"
] | [
"Definition:Summation/Vacuous Summation"
] |
proofwiki-2894 | Symmetry Group of Equilateral Triangle is Group | The symmetry group of the equilateral triangle is a group. | Let us refer to this group as $D_3$.
Taking the group axioms in turn: | The [[Definition:Symmetry Group of Equilateral Triangle|symmetry group of the equilateral triangle]] is a [[Definition:Group|group]]. | Let us refer to this group as $D_3$.
Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Symmetry Group of Equilateral Triangle is Group | https://proofwiki.org/wiki/Symmetry_Group_of_Equilateral_Triangle_is_Group | https://proofwiki.org/wiki/Symmetry_Group_of_Equilateral_Triangle_is_Group | [
"Symmetry Group of Equilateral Triangle"
] | [
"Definition:Symmetry Group of Equilateral Triangle",
"Definition:Group",
"Definition:Symmetry Group of Equilateral Triangle",
"Definition:Symmetry Group of Equilateral Triangle"
] | [
"Axiom:Group Axioms"
] |
proofwiki-2895 | Order of General Linear Group over Galois Field | Let $\GF$ be a Galois field with $p$ elements.
Then the order of the general linear group $\GL {n, \GF}$ is:
:$\ds \prod_{j \mathop = 1}^n \paren {p^n - p^{j - 1} }$ | Let $\GF$ be a Galois field with $p$ elements: $\card \GF = p$.
Let $A = \sqbrk {a_{i j} }_{n, n}$ be a matrix such that $\size A \ne 0$ and $a_{i j} \in \GF$.
How many such matrices can be constructed?
In order to avoid a zero determinant, the top row of the matrix, $\sequence {a_{1 j} }_{j \mathop = 1, \dotsc, n}$ mu... | Let $\GF$ be a [[Definition:Galois Field|Galois field]] with $p$ elements.
Then the [[Definition:Order of Structure|order]] of the [[Definition:General Linear Group|general linear group]] $\GL {n, \GF}$ is:
:$\ds \prod_{j \mathop = 1}^n \paren {p^n - p^{j - 1} }$ | Let $\GF$ be a [[Definition:Galois Field|Galois field]] with $p$ [[Definition:Element|elements]]: $\card \GF = p$.
Let $A = \sqbrk {a_{i j} }_{n, n}$ be a [[Definition:Square Matrix|matrix]] such that $\size A \ne 0$ and $a_{i j} \in \GF$.
How many such matrices can be constructed?
In order to avoid a zero determin... | Order of General Linear Group over Galois Field | https://proofwiki.org/wiki/Order_of_General_Linear_Group_over_Galois_Field | https://proofwiki.org/wiki/Order_of_General_Linear_Group_over_Galois_Field | [
"Combinatorics",
"General Linear Group",
"Galois Fields"
] | [
"Definition:Galois Field",
"Definition:Order of Structure",
"Definition:General Linear Group"
] | [
"Definition:Galois Field",
"Definition:Element",
"Definition:Matrix/Square Matrix",
"Definition:Sequence",
"Definition:Sequence",
"Definition:Sequence",
"Definition:Sequence",
"Definition:Sequence",
"Category:Combinatorics",
"Category:General Linear Group",
"Category:Galois Fields"
] |
proofwiki-2896 | Existence of Latin Squares | For each $n \in \Z_{>0}$ there exists at least one Latin square of order $n$. | The Cayley table of a finite group of order $n$ is a Latin square, from Group has Latin Square Property.
For every $n \in \Z_{>0}$ there exists a cyclic group of order $n$.
It follows that for every $n \in \Z_{>0}$ the Cayley table of the cyclic group of order $n$ is a Latin square of order $n$.
{{qed}}
Category:Latin ... | For each $n \in \Z_{>0}$ there exists at least one [[Definition:Latin Square|Latin square]] of [[Definition:Order of Latin Square|order $n$]]. | The [[Definition:Cayley Table|Cayley table]] of a [[Definition:Finite Group|finite group]] of [[Definition:Order of Structure|order $n$]] is a [[Definition:Latin Square|Latin square]], from [[Group has Latin Square Property]].
For every $n \in \Z_{>0}$ there exists a [[Definition:Cyclic Group|cyclic group]] of [[Defin... | Existence of Latin Squares | https://proofwiki.org/wiki/Existence_of_Latin_Squares | https://proofwiki.org/wiki/Existence_of_Latin_Squares | [
"Latin Squares"
] | [
"Definition:Latin Square",
"Definition:Latin Square/Order"
] | [
"Definition:Cayley Table",
"Definition:Finite Group",
"Definition:Order of Structure",
"Definition:Latin Square",
"Group has Latin Square Property",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Cayley Table",
"Definition:Cyclic Group",
"Definition:Order of Structure",
... |
proofwiki-2897 | Hilbert's Basis Theorem | Let $A$ be a Noetherian ring.
Let $A \sqbrk x$ be the ring of polynomial forms over $A$ in the single indeterminate $x$.
Then $A \sqbrk x$ is also a Noetherian ring. | From the definition, a Noetherian ring is also a commutative ring with unity.
Let $f = a_n x^n + \cdots + a_1 x + a_0 \in A \sqbrk x$ be a polynomial over $x$.
Let $I \subseteq A \sqbrk x$ be an ideal of $A \sqbrk x$.
We will show that $I$ is finitely generated.
Let $f_1$ be an element of least degree in $I$, and let ... | Let $A$ be a [[Definition:Noetherian Ring|Noetherian ring]].
Let $A \sqbrk x$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms over $A$ in the single indeterminate $x$]].
Then $A \sqbrk x$ is also a [[Definition:Noetherian Ring|Noetherian ring]]. | From the definition, a [[Definition:Noetherian Ring|Noetherian ring]] is also a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $f = a_n x^n + \cdots + a_1 x + a_0 \in A \sqbrk x$ be a [[Definition:Polynomial (Abstract Algebra)|polynomial]] over $x$.
Let $I \subseteq A \sqbrk x$ be an [[D... | Hilbert's Basis Theorem | https://proofwiki.org/wiki/Hilbert's_Basis_Theorem | https://proofwiki.org/wiki/Hilbert's_Basis_Theorem | [
"Noetherian Rings",
"Commutative Algebra"
] | [
"Definition:Noetherian Ring",
"Definition:Ring of Polynomial Forms",
"Definition:Noetherian Ring"
] | [
"Definition:Noetherian Ring",
"Definition:Commutative and Unitary Ring",
"Definition:Polynomial over Ring",
"Definition:Ideal of Ring",
"Definition:Finitely Generated Ideal of Ring",
"Definition:Degree of Polynomial",
"Definition:Ideal of Ring",
"Definition:Generated Ideal of Ring",
"Definition:Lead... |
proofwiki-2898 | Lipschitz Condition implies Uniform Continuity | Let $\struct {M_1, d_1}$ and $\struct {M_2, d_2}$ be metric spaces.
Let $g: M_1 \to M_2$ satisfy the Lipschitz condition.
Then $g$ is uniformly continuous on $M_1$. | Let $\epsilon > 0$, $x,y \in M_1$.
Let $K$ be a Lipschitz constant for $g$.
First, suppose that $K \le 0$.
Then:
:$\map {d_1} {x, y} \le 0 \map {d_2} {\map g x, \map g y}$ by the Lipschitz condition on $g$.
So $\map {d_1} {x, y} \le 0 \implies \map {d_1} {x, y} = 0 \implies x = y$ for all $x$ and $y$ in $M_1$.
Thus $g... | Let $\struct {M_1, d_1}$ and $\struct {M_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
Let $g: M_1 \to M_2$ satisfy the [[Definition:Lipschitz Condition|Lipschitz condition]].
Then $g$ is [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous]] on $M_1$. | Let $\epsilon > 0$, $x,y \in M_1$.
Let $K$ be a [[Definition:Lipschitz Constant|Lipschitz constant]] for $g$.
First, suppose that $K \le 0$.
Then:
:$\map {d_1} {x, y} \le 0 \map {d_2} {\map g x, \map g y}$ by the [[Definition:Lipschitz Condition|Lipschitz condition]] on $g$.
So $\map {d_1} {x, y} \le 0 \implies \... | Lipschitz Condition implies Uniform Continuity | https://proofwiki.org/wiki/Lipschitz_Condition_implies_Uniform_Continuity | https://proofwiki.org/wiki/Lipschitz_Condition_implies_Uniform_Continuity | [
"Metric Spaces",
"Uniformly Continuous Mappings"
] | [
"Definition:Metric Space",
"Definition:Lipschitz Continuity",
"Definition:Uniform Continuity/Metric Space"
] | [
"Definition:Lipschitz Continuity/Lipschitz Constant",
"Definition:Lipschitz Continuity",
"Constant Function is Uniformly Continuous",
"Definition:Lipschitz Continuity",
"Definition:Uniform Continuity/Metric Space",
"Category:Metric Spaces",
"Category:Uniformly Continuous Mappings"
] |
proofwiki-2899 | Existence of Cyclic Group of Order n | Let $n \in \Z_{>0}$.
Then there exists a cyclic group of order $n$ which is unique up to isomorphism. | Existence follows from Integers Modulo m under Addition form Cyclic Group.
Uniqueness follows from Cyclic Groups of Same Order are Isomorphic.
{{Qed}} | Let $n \in \Z_{>0}$.
Then there exists a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order $n$]] which is [[Definition:Unique|unique]] up to [[Definition:Group Isomorphism|isomorphism]]. | Existence follows from [[Integers Modulo m under Addition form Cyclic Group]].
Uniqueness follows from [[Cyclic Groups of Same Order are Isomorphic]].
{{Qed}} | Existence of Cyclic Group of Order n | https://proofwiki.org/wiki/Existence_of_Cyclic_Group_of_Order_n | https://proofwiki.org/wiki/Existence_of_Cyclic_Group_of_Order_n | [
"Finite Cyclic Groups"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Unique",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Integers Modulo m under Addition form Cyclic Group",
"Cyclic Groups of Same Order are Isomorphic"
] |
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