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-200 | Cancellation Laws | Let $G$ be a group.
Let $a, b, c \in G$.
Then the following hold:
;Right cancellation law
:$b a = c a \implies b = c$
;Left cancellation law
:$a b = a c \implies b = c$ | {{begin-eqn}}
{{eqn | l = h g
| r = g
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {h g} g^{-1}
| r = g g^{-1}
| c = {{Group-axiom|2}}
}}
{{eqn | ll= \leadsto
| l = h \paren {g g^{-1} }
| r = g g^{-1}
| c = {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
| l = h e
| ... | Let $G$ be a [[Definition:Group|group]].
Let $a, b, c \in G$.
Then the following hold:
;Right cancellation law
:$b a = c a \implies b = c$
;Left cancellation law
:$a b = a c \implies b = c$ | {{begin-eqn}}
{{eqn | l = h g
| r = g
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {h g} g^{-1}
| r = g g^{-1}
| c = {{Group-axiom|2}}
}}
{{eqn | ll= \leadsto
| l = h \paren {g g^{-1} }
| r = g g^{-1}
| c = {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
| l = h e
| ... | Cancellation Laws/Corollary 2/Proof 2 | https://proofwiki.org/wiki/Cancellation_Laws | https://proofwiki.org/wiki/Cancellation_Laws/Corollary_2/Proof_2 | [
"Cancellation Laws",
"Group Theory",
"Cancellability",
"Named Theorems"
] | [
"Definition:Group"
] | [] |
proofwiki-201 | Cancellation Laws | Let $G$ be a group.
Let $a, b, c \in G$.
Then the following hold:
;Right cancellation law
:$b a = c a \implies b = c$
;Left cancellation law
:$a b = a c \implies b = c$ | Let $a, b, c \in G$ and let $a^{-1}$ be the inverse of $a$.
Suppose $b a = c a$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {b a} a^{-1}
| r = \paren {c a} a^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = b \paren {a a^{-1} }
| r = c \paren {a a^{-1} }
| c = {{Defof|Associative Operation}}
}}
{{eqn... | Let $G$ be a [[Definition:Group|group]].
Let $a, b, c \in G$.
Then the following hold:
;Right cancellation law
:$b a = c a \implies b = c$
;Left cancellation law
:$a b = a c \implies b = c$ | Let $a, b, c \in G$ and let $a^{-1}$ be the [[Definition:Inverse Element|inverse]] of $a$.
Suppose $b a = c a$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {b a} a^{-1}
| r = \paren {c a} a^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = b \paren {a a^{-1} }
| r = c \paren {a a^{-1} }
| c = {{Defo... | Cancellation Laws/Proof 1 | https://proofwiki.org/wiki/Cancellation_Laws | https://proofwiki.org/wiki/Cancellation_Laws/Proof_1 | [
"Cancellation Laws",
"Group Theory",
"Cancellability",
"Named Theorems"
] | [
"Definition:Group"
] | [
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-202 | Cancellation Laws | Let $G$ be a group.
Let $a, b, c \in G$.
Then the following hold:
;Right cancellation law
:$b a = c a \implies b = c$
;Left cancellation law
:$a b = a c \implies b = c$ | From its definition, a group is a monoid, all of whose elements have inverses and thus are invertible.
From Invertible Element of Monoid is Cancellable, it follows that all its elements are therefore cancellable.
{{qed}} | Let $G$ be a [[Definition:Group|group]].
Let $a, b, c \in G$.
Then the following hold:
;Right cancellation law
:$b a = c a \implies b = c$
;Left cancellation law
:$a b = a c \implies b = c$ | From its [[Definition:Group|definition]], a group is a [[Definition:Monoid|monoid]], all of whose elements have [[Definition:Inverse Element|inverses]] and thus are [[Definition:Invertible Element|invertible]].
From [[Invertible Element of Monoid is Cancellable]], it follows that all its elements are therefore [[Defin... | Cancellation Laws/Proof 2 | https://proofwiki.org/wiki/Cancellation_Laws | https://proofwiki.org/wiki/Cancellation_Laws/Proof_2 | [
"Cancellation Laws",
"Group Theory",
"Cancellability",
"Named Theorems"
] | [
"Definition:Group"
] | [
"Definition:Group",
"Definition:Monoid",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Invertible Element",
"Invertible Element of Associative Structure is Cancellable/Corollary",
"Definition:Cancellable Element"
] |
proofwiki-203 | Cancellation Laws | Let $G$ be a group.
Let $a, b, c \in G$.
Then the following hold:
;Right cancellation law
:$b a = c a \implies b = c$
;Left cancellation law
:$a b = a c \implies b = c$ | Suppose $x = b a = c a$.
By Group has Latin Square Property, there exists exactly one $y \in G$ such that $x = y a$.
That is, $x = b a = c a \implies b = c$.
Similarly, suppose $x = a b = a c$.
Again by Group has Latin Square Property, there exists exactly one $y \in G$ such that $x = a y$.
That is, $a b = a c \implies... | Let $G$ be a [[Definition:Group|group]].
Let $a, b, c \in G$.
Then the following hold:
;Right cancellation law
:$b a = c a \implies b = c$
;Left cancellation law
:$a b = a c \implies b = c$ | Suppose $x = b a = c a$.
By [[Group has Latin Square Property]], there exists [[Definition:Exactly One|exactly one]] $y \in G$ such that $x = y a$.
That is, $x = b a = c a \implies b = c$.
Similarly, suppose $x = a b = a c$.
Again by [[Group has Latin Square Property]], there exists [[Definition:Exactly One|exactl... | Cancellation Laws/Proof 3 | https://proofwiki.org/wiki/Cancellation_Laws | https://proofwiki.org/wiki/Cancellation_Laws/Proof_3 | [
"Cancellation Laws",
"Group Theory",
"Cancellability",
"Named Theorems"
] | [
"Definition:Group"
] | [
"Group has Latin Square Property",
"Definition:Unique",
"Group has Latin Square Property",
"Definition:Unique"
] |
proofwiki-204 | Center of Group is Normal Subgroup | Let $G$ be a group
The center $\map Z G$ of $G$ is a normal subgroup of $G$. | Recall that Center of Group is Abelian Subgroup.
Since $g x = x g$ for each $g \in G$ and $x \in \map Z G$:
:$g \map Z G = \map Z G g$
Thus:
:$\map Z G \lhd G$
{{qed}} | Let $G$ be a [[Definition:Group|group]]
The [[Definition:Center of Group|center]] $\map Z G$ of $G$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$. | Recall that [[Center of Group is Abelian Subgroup]].
Since $g x = x g$ for each $g \in G$ and $x \in \map Z G$:
:$g \map Z G = \map Z G g$
Thus:
:$\map Z G \lhd G$
{{qed}} | Center of Group is Normal Subgroup/Proof 1 | https://proofwiki.org/wiki/Center_of_Group_is_Normal_Subgroup | https://proofwiki.org/wiki/Center_of_Group_is_Normal_Subgroup/Proof_1 | [
"Normal Subgroups",
"Centers of Groups",
"Center of Group is Normal Subgroup"
] | [
"Definition:Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Normal Subgroup"
] | [
"Center of Group is Abelian Subgroup"
] |
proofwiki-205 | Center of Group is Normal Subgroup | Let $G$ be a group
The center $\map Z G$ of $G$ is a normal subgroup of $G$. | We have:
:$\forall a \in G: x \in \map Z G^a \iff a x a^{-1} = x a a^{-1} = x \in \map Z G$
Therefore:
:$\forall a \in G: \map Z G^a = \map Z G$
and $\map Z G$ is a normal subgroup of $G$.
{{qed}} | Let $G$ be a [[Definition:Group|group]]
The [[Definition:Center of Group|center]] $\map Z G$ of $G$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$. | We have:
:$\forall a \in G: x \in \map Z G^a \iff a x a^{-1} = x a a^{-1} = x \in \map Z G$
Therefore:
:$\forall a \in G: \map Z G^a = \map Z G$
and $\map Z G$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
{{qed}} | Center of Group is Normal Subgroup/Proof 2 | https://proofwiki.org/wiki/Center_of_Group_is_Normal_Subgroup | https://proofwiki.org/wiki/Center_of_Group_is_Normal_Subgroup/Proof_2 | [
"Normal Subgroups",
"Centers of Groups",
"Center of Group is Normal Subgroup"
] | [
"Definition:Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Normal Subgroup"
] | [
"Definition:Normal Subgroup"
] |
proofwiki-206 | Composition of Relations is Associative | The composition of relations is an associative binary operation:
:$\paren {\RR_3 \circ \RR_2} \circ \RR_1 = \RR_3 \circ \paren {\RR_2 \circ \RR_1}$ | {{Rewrite|Needs to be rewritten in light of the actual definition of the composition of relations and domain of relations. These equalities may not hold for all relations}}
First, note that from the definition of composition of relations, the following must be the case before the above expression is even to be defined:... | The [[Definition:Composition of Relations|composition of relations]] is an [[Definition:Associative Operation|associative]] [[Definition:Binary Operation|binary operation]]:
:$\paren {\RR_3 \circ \RR_2} \circ \RR_1 = \RR_3 \circ \paren {\RR_2 \circ \RR_1}$ | {{Rewrite|Needs to be rewritten in light of the actual definition of the composition of relations and domain of relations. These equalities may not hold for all relations}}
First, note that from the definition of [[Definition:Composition of Relations|composition of relations]], the following must be the case before th... | Composition of Relations is Associative | https://proofwiki.org/wiki/Composition_of_Relations_is_Associative | https://proofwiki.org/wiki/Composition_of_Relations_is_Associative | [
"Composite Relations"
] | [
"Definition:Composition of Relations",
"Definition:Associative Operation",
"Definition:Operation/Binary Operation"
] | [
"Definition:Composition of Relations",
"Definition:Composition of Relations",
"Definition:Domain (Set Theory)/Relation",
"Domain of Composite Relation",
"Domain of Composite Relation",
"Domain of Composite Relation",
"Definition:Codomain (Set Theory)/Relation",
"Codomain of Composite Relation",
"Cod... |
proofwiki-207 | Intersection of Subsemigroups | Let $\struct {S, \circ}$ be a semigroup.
Let $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ be subsemigroups of $\struct {S, \circ}$.
Then the intersection of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a subsemigroup of that $\struct {S, \circ}$. | {{Proofread| check if proof is correct if $\struct {S, \circ}$ is the empty semigroup}}
Suppose $\struct {S, \circ}$ is a semigroup where $S$ is the empty set.
Suppose $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ are subsemigroups of $\struct {S, \circ}$.
Then it follows that $T_1$ and $T_2$ are both empty.
Sinc... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ be [[Definition:Subsemigroup|subsemigroups]] of $\struct {S, \circ}$.
Then the [[Definition:Set Intersection|intersection]] of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a [[Definit... | {{Proofread| check if proof is correct if $\struct {S, \circ}$ is the empty semigroup}}
Suppose $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]] where $S$ is the [[Definition:Empty Set|empty set]].
Suppose $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ are [[Definition:Subsemigroup|subsemigroups]] of $... | Intersection of Subsemigroups | https://proofwiki.org/wiki/Intersection_of_Subsemigroups | https://proofwiki.org/wiki/Intersection_of_Subsemigroups | [
"Set Intersection",
"Subsemigroups",
"Intersection of Subsemigroups"
] | [
"Definition:Semigroup",
"Definition:Subsemigroup",
"Definition:Set Intersection",
"Definition:Subsemigroup"
] | [
"Definition:Semigroup",
"Definition:Empty Set",
"Definition:Subsemigroup",
"Definition:Empty Set",
"Definition:Empty Set",
"Intersection with Empty Set",
"Definition:Set Intersection",
"Semigroup is Subsemigroup of Itself",
"Definition:Set Intersection",
"Definition:Subsemigroup",
"Definition:Se... |
proofwiki-208 | Intersection of Subsemigroups | Let $\struct {S, \circ}$ be a semigroup.
Let $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ be subsemigroups of $\struct {S, \circ}$.
Then the intersection of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a subsemigroup of that $\struct {S, \circ}$. | Let $T = \bigcap \mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
| l = a, b
| o = \in
| r = K
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ be [[Definition:Subsemigroup|subsemigroups]] of $\struct {S, \circ}$.
Then the [[Definition:Set Intersection|intersection]] of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a [[Definit... | Let $T = \bigcap \mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
| l = a, b
| o = \in
| r = K
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
... | Intersection of Subsemigroups/General Result/Proof 1 | https://proofwiki.org/wiki/Intersection_of_Subsemigroups | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result/Proof_1 | [
"Set Intersection",
"Subsemigroups",
"Intersection of Subsemigroups"
] | [
"Definition:Semigroup",
"Definition:Subsemigroup",
"Definition:Set Intersection",
"Definition:Subsemigroup"
] | [
"Definition:Subsemigroup",
"Subsemigroup Closure Test",
"Definition:Subsemigroup",
"Definition:Subsemigroup",
"Definition:Subsemigroup"
] |
proofwiki-209 | Intersection of Subsemigroups | Let $\struct {S, \circ}$ be a semigroup.
Let $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ be subsemigroups of $\struct {S, \circ}$.
Then the intersection of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a subsemigroup of that $\struct {S, \circ}$. | From Set of Subsemigroups forms Complete Lattice:
:$\struct {\mathbb S, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subsemigroups of $S$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.
Hence the result, by definition of infimum.
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ be [[Definition:Subsemigroup|subsemigroups]] of $\struct {S, \circ}$.
Then the [[Definition:Set Intersection|intersection]] of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a [[Definit... | From [[Set of Subsemigroups forms Complete Lattice]]:
:$\struct {\mathbb S, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
where for every [[Definition:Set|set]] $\mathbb H$ of [[Definition:Subsemigroup|subsemigroups]] of $S$:
:the [[Definition:Infimum of Set|infimum]] of $\mathbb H$ necessarily ad... | Intersection of Subsemigroups/General Result/Proof 2 | https://proofwiki.org/wiki/Intersection_of_Subsemigroups | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result/Proof_2 | [
"Set Intersection",
"Subsemigroups",
"Intersection of Subsemigroups"
] | [
"Definition:Semigroup",
"Definition:Subsemigroup",
"Definition:Set Intersection",
"Definition:Subsemigroup"
] | [
"Set of Subsemigroups forms Complete Lattice",
"Definition:Complete Lattice",
"Definition:Set",
"Definition:Subsemigroup",
"Definition:Infimum of Set",
"Definition:Infimum of Set"
] |
proofwiki-210 | Cancellable Elements of Semigroup form Subsemigroup | Let $\struct {S, \circ}$ be a semigroup.
Let $C$ be the set of cancellable elements of $\struct {S, \circ}$.
Then $\struct {C, \circ}$ is a subsemigroup of $\struct {S, \circ}$. | Now let $C$ be the set of cancellable elements of $\struct {S, \circ}$.
Let $x, y \in C$.
Then $x$ and $y$ are both left cancellable and right cancellable.
Thus by Left Cancellable Elements of Semigroup form Subsemigroup:
:$x \circ y$ is left cancellable
and by Right Cancellable Elements of Semigroup form Subsemigroup:... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $C$ be the set of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$.
Then $\struct {C, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$. | Now let $C$ be the set of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$.
Let $x, y \in C$.
Then $x$ and $y$ are both [[Definition:Left Cancellable Element|left cancellable]] and [[Definition:Right Cancellable Element|right cancellable]].
Thus by [[Left Cancellable Elements of Semig... | Cancellable Elements of Semigroup form Subsemigroup | https://proofwiki.org/wiki/Cancellable_Elements_of_Semigroup_form_Subsemigroup | https://proofwiki.org/wiki/Cancellable_Elements_of_Semigroup_form_Subsemigroup | [
"Subsemigroups",
"Cancellability"
] | [
"Definition:Semigroup",
"Definition:Cancellable Element",
"Definition:Subsemigroup"
] | [
"Definition:Cancellable Element",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Right Cancellable",
"Left Cancellable Elements of Semigroup form Subsemigroup",
"Definition:Cancellable Element/Left Cancellable",
"Right Cancellable Elements of Semigroup form Subsemigroup... |
proofwiki-211 | Subgroup of Cyclic Group is Cyclic | Let $G$ be a cyclic group.
Let $H$ be a subgroup of $G$.
Then $H$ is cyclic. | Let $G$ be a cyclic group generated by $a$.
Let $H$ be a subgroup of $G$.
If $H = \set e$, then $H$ is a cyclic group subgroup generated by $e$.
Let $H \ne \set e$.
By definition of cyclic group, every element of $G$ has the form $a^n$.
Then as $H$ is a subgroup of $G$, $a^n \in H$ for some $n \in \Z$.
Let $m$ be the s... | Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $H$ is [[Definition:Cyclic Group|cyclic]]. | Let $G$ be a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generator of Cyclic Group|generated]] by $a$.
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
If $H = \set e$, then $H$ is a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generated Subgroup|subgroup generated]] by $e$.
Let $H \ne \set ... | Subgroup of Cyclic Group is Cyclic/Proof 1 | https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic | https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic/Proof_1 | [
"Subgroup of Cyclic Group is Cyclic",
"Cyclic Groups",
"Subgroups"
] | [
"Definition:Cyclic Group",
"Definition:Subgroup",
"Definition:Cyclic Group"
] | [
"Definition:Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Subgroup",
"Definition:Cyclic Group",
"Definition:Generated Subgroup",
"Definition:Cyclic Group",
"Definition:Element",
"Definition:Subgroup",
"Definition:Strictly Positive/Integer",
"Definition:Element",
"Definition:Sub... |
proofwiki-212 | Subgroup of Cyclic Group is Cyclic | Let $G$ be a cyclic group.
Let $H$ be a subgroup of $G$.
Then $H$ is cyclic. | Let $G$ be a cyclic group generated by $a$.
=== Finite Group ===
Let $G$ be finite.
By Bijection from Divisors to Subgroups of Cyclic Group there are exactly as many subgroups of $G$ as divisors of the order of $G$.
As each one of these is cyclic by Subgroup of Finite Cyclic Group is Determined by Order, the result fol... | Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $H$ is [[Definition:Cyclic Group|cyclic]]. | Let $G$ be a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generator of Cyclic Group|generated]] by $a$.
=== Finite Group ===
Let $G$ be [[Definition:Finite Group|finite]].
By [[Bijection from Divisors to Subgroups of Cyclic Group]] there are exactly as many [[Definition:Subgroup|subgroups]] of $G$ as [[Def... | Subgroup of Cyclic Group is Cyclic/Proof 2 | https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic | https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic/Proof_2 | [
"Subgroup of Cyclic Group is Cyclic",
"Cyclic Groups",
"Subgroups"
] | [
"Definition:Cyclic Group",
"Definition:Subgroup",
"Definition:Cyclic Group"
] | [
"Definition:Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Finite Group",
"Bijection from Divisors to Subgroups of Cyclic Group",
"Definition:Subgroup",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Cyclic Group",
"Subgroup of Finite Cyclic Gro... |
proofwiki-213 | Subgroup of Cyclic Group is Cyclic | Let $G$ be a cyclic group.
Let $H$ be a subgroup of $G$.
Then $H$ is cyclic. | Let $G$ be a cyclic group generated by $a$.
Let $H$ be a subgroup of $G$.
By Cyclic Group is Abelian, $G$ is abelian.
By Subgroup of Abelian Group is Normal, $H$ is normal in $G$.
Let $G / H$ be the quotient group of $G$ by $H$.
Let $q_H: G \to G / H$ be the quotient epimorphism from $G$ to $G / H$:
:$\forall x \in G: ... | Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $H$ is [[Definition:Cyclic Group|cyclic]]. | Let $G$ be a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generator of Cyclic Group|generated]] by $a$.
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
By [[Cyclic Group is Abelian]], $G$ is [[Definition:Abelian Group|abelian]].
By [[Subgroup of Abelian Group is Normal]], $H$ is [[Definition:Normal Su... | Subgroup of Cyclic Group is Cyclic/Proof 3 | https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic | https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic/Proof_3 | [
"Subgroup of Cyclic Group is Cyclic",
"Cyclic Groups",
"Subgroups"
] | [
"Definition:Cyclic Group",
"Definition:Subgroup",
"Definition:Cyclic Group"
] | [
"Definition:Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Subgroup",
"Cyclic Group is Abelian",
"Definition:Abelian Group",
"Subgroup of Abelian Group is Normal",
"Definition:Normal Subgroup",
"Definition:Quotient Group",
"Definition:Quotient Epimorphism/Group",
"Quotient Epimorp... |
proofwiki-214 | Empty Set is Subset of All Sets | The empty set $\O$ is a subset of every set (including itself).
That is:
:$\forall S: \O \subseteq S$ | By the definition of subset, $\O \subseteq S$ means:
:$\forall x: \paren {x \in \O \implies x \in S}$
By the definition of the empty set:
:$\forall x: \neg \paren {x \in \O}$
Thus $\O \subseteq S$ is vacuously true.
{{qed}} | The [[Definition:Empty Set|empty set]] $\O$ is a [[Definition:Subset|subset]] of every [[Definition:Set|set]] (including itself).
That is:
:$\forall S: \O \subseteq S$ | By the definition of [[Definition:Subset|subset]], $\O \subseteq S$ means:
:$\forall x: \paren {x \in \O \implies x \in S}$
By the definition of the [[Definition:Empty Set|empty set]]:
:$\forall x: \neg \paren {x \in \O}$
Thus $\O \subseteq S$ is [[Definition:Vacuous Truth|vacuously true]].
{{qed}} | Empty Set is Subset of All Sets/Proof 1 | https://proofwiki.org/wiki/Empty_Set_is_Subset_of_All_Sets | https://proofwiki.org/wiki/Empty_Set_is_Subset_of_All_Sets/Proof_1 | [
"Subsets",
"Empty Set",
"Empty Set is Subset of All Sets"
] | [
"Definition:Empty Set",
"Definition:Subset",
"Definition:Set"
] | [
"Definition:Subset",
"Definition:Empty Set",
"Definition:Vacuous Truth"
] |
proofwiki-215 | Empty Set is Subset of All Sets | The empty set $\O$ is a subset of every set (including itself).
That is:
:$\forall S: \O \subseteq S$ | $S \subseteq T$ means:
:'''every element of $S$ is also in $T$'''
or, equivalently:
:'''every element that is ''not'' in $T$ is not in $S$ either.'''
Thus:
{{begin-eqn}}
{{eqn | o =
| r = S \subseteq T
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \forall x \in S: x \in T
| c = {{Defof|Subset}}
}... | The [[Definition:Empty Set|empty set]] $\O$ is a [[Definition:Subset|subset]] of every [[Definition:Set|set]] (including itself).
That is:
:$\forall S: \O \subseteq S$ | $S \subseteq T$ means:
:'''every [[Definition:Element|element]] of $S$ is also in $T$'''
or, equivalently:
:'''every [[Definition:Element|element]] that is ''not'' in $T$ is not in $S$ either.'''
Thus:
{{begin-eqn}}
{{eqn | o =
| r = S \subseteq T
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \foral... | Empty Set is Subset of All Sets/Proof 2 | https://proofwiki.org/wiki/Empty_Set_is_Subset_of_All_Sets | https://proofwiki.org/wiki/Empty_Set_is_Subset_of_All_Sets/Proof_2 | [
"Subsets",
"Empty Set",
"Empty Set is Subset of All Sets"
] | [
"Definition:Empty Set",
"Definition:Subset",
"Definition:Set"
] | [
"Definition:Element",
"Definition:Element",
"De Morgan's Laws (Predicate Logic)",
"Definition:Element",
"Definition:Element",
"Definition:Empty Set",
"Definition:Element",
"Definition:Set",
"Definition:Element",
"Definition:Vacuous Truth",
"Definition:Set"
] |
proofwiki-216 | Relation Reflexivity | Every relation has exactly one of these properties: it is either:
:reflexive,
:antireflexive or
:non-reflexive. | A reflexive relation can not be antireflexive, and vice versa:
:$\tuple {x, x} \in \RR \iff \neg \paren {\tuple {x, x} \notin \RR}$
By the definition of non-reflexive, a reflexive relation can not also be non-reflexive.
So a reflexive relation is neither antireflexive nor non-reflexive.
An antireflexive relation can be... | Every [[Definition:Relation|relation]] has exactly one of these properties: it is either:
:[[Definition:Reflexive Relation|reflexive]],
:[[Definition:Antireflexive Relation|antireflexive]] or
:[[Definition:Non-Reflexive Relation|non-reflexive]]. | A [[Definition:Reflexive Relation|reflexive relation]] can not be [[Definition:Antireflexive Relation|antireflexive]], and vice versa:
:$\tuple {x, x} \in \RR \iff \neg \paren {\tuple {x, x} \notin \RR}$
By the definition of [[Definition:Non-Reflexive Relation|non-reflexive]], a [[Definition:Reflexive Relation|refle... | Relation Reflexivity | https://proofwiki.org/wiki/Relation_Reflexivity | https://proofwiki.org/wiki/Relation_Reflexivity | [
"Reflexive Relations"
] | [
"Definition:Relation",
"Definition:Reflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Non-Reflexive Relation"
] | [
"Definition:Reflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Non-Reflexive Relation",
"Definition:Reflexive Relation",
"Definition:Non-Reflexive Relation",
"Definition:Reflexive Relation",
"Definition:Antireflexive Relation",
"Definition:Non-Reflexive Relation",
"Definition:Anti... |
proofwiki-217 | Set Union is Idempotent | Set union is idempotent:
:$S \cup S = S$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \cup S
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \in S
| o = \lor
| r = x \in S
| c = {{Defof|Set Union}}
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = S
| c = Rule of Idempotence: Disjunction
}}... | [[Definition:Set Union|Set union]] is [[Definition:Idempotent Operation|idempotent]]:
:$S \cup S = S$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \cup S
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \in S
| o = \lor
| r = x \in S
| c = {{Defof|Set Union}}
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = S
| c = [[Rule of Idempotence/Disjunction|R... | Set Union is Idempotent | https://proofwiki.org/wiki/Set_Union_is_Idempotent | https://proofwiki.org/wiki/Set_Union_is_Idempotent | [
"Set Union",
"Examples of Idempotence"
] | [
"Definition:Set Union",
"Definition:Idempotence/Operation"
] | [
"Rule of Idempotence/Disjunction"
] |
proofwiki-218 | Set Intersection is Idempotent | Set intersection is idempotent:
:$S \cap S = S$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \cap S
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \in S
| o = \land
| r = x \in S
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = S
| c = Rule of Idempotence: Conjunc... | [[Definition:Set Intersection|Set intersection]] is [[Definition:Idempotent Operation|idempotent]]:
:$S \cap S = S$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \cap S
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \in S
| o = \land
| r = x \in S
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = S
| c = [[Rule of Idempotence/Conjun... | Set Intersection is Idempotent | https://proofwiki.org/wiki/Set_Intersection_is_Idempotent | https://proofwiki.org/wiki/Set_Intersection_is_Idempotent | [
"Set Intersection",
"Examples of Idempotence"
] | [
"Definition:Set Intersection",
"Definition:Idempotence/Operation"
] | [
"Rule of Idempotence/Conjunction"
] |
proofwiki-219 | Set is Subset of Union | The union of two sets is a superset of each:
:$S \subseteq S \cup T$
:$T \subseteq S \cup T$ | {{begin-eqn}}
{{eqn | l = x \in S
| o = \leadsto
| r = x \in S \lor x \in T
| c = Rule of Addition
}}
{{eqn | o = \leadsto
| r = x \in S \cup T
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadsto
| r = S \subseteq S \cup T
| c = {{Defof|Subset}}
}}
{{end-eqn}}
Similarly for $T$.
... | The [[Definition:Set Union|union]] of two [[Definition:Set|sets]] is a [[Definition:Superset|superset]] of each:
:$S \subseteq S \cup T$
:$T \subseteq S \cup T$ | {{begin-eqn}}
{{eqn | l = x \in S
| o = \leadsto
| r = x \in S \lor x \in T
| c = [[Rule of Addition]]
}}
{{eqn | o = \leadsto
| r = x \in S \cup T
| c = {{Defof|Set Union}}
}}
{{eqn | o = \leadsto
| r = S \subseteq S \cup T
| c = {{Defof|Subset}}
}}
{{end-eqn}}
Similarly for ... | Set is Subset of Union | https://proofwiki.org/wiki/Set_is_Subset_of_Union | https://proofwiki.org/wiki/Set_is_Subset_of_Union | [
"Set is Subset of Union",
"Set Union",
"Subsets"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Subset/Superset"
] | [
"Rule of Addition"
] |
proofwiki-220 | Set is Subset of Union | The union of two sets is a superset of each:
:$S \subseteq S \cup T$
:$T \subseteq S \cup T$ | Let $x \in S_\beta$ for some $\beta \in I$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S_\beta
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \set {x: \exists \alpha \in I: x \in S_\alpha}
| c = {{Defof|Indexed Family of Sets}}
}}
{{eqn | ll= \leadsto
| l =... | The [[Definition:Set Union|union]] of two [[Definition:Set|sets]] is a [[Definition:Superset|superset]] of each:
:$S \subseteq S \cup T$
:$T \subseteq S \cup T$ | Let $x \in S_\beta$ for some $\beta \in I$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S_\beta
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \set {x: \exists \alpha \in I: x \in S_\alpha}
| c = {{Defof|Indexed Family of Sets}}
}}
{{eqn | ll= \leadsto
| l ... | Set is Subset of Union/Family of Sets/Proof 1 | https://proofwiki.org/wiki/Set_is_Subset_of_Union | https://proofwiki.org/wiki/Set_is_Subset_of_Union/Family_of_Sets/Proof_1 | [
"Set is Subset of Union",
"Set Union",
"Subsets"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Subset/Superset"
] | [] |
proofwiki-221 | Set is Subset of Union | The union of two sets is a superset of each:
:$S \subseteq S \cup T$
:$T \subseteq S \cup T$ | Let $\beta \in I$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \beta
| o = \in
| r = I
| c =
}}
{{eqn | ll= \leadsto
| l = \set \beta
| o = \subseteq
| r = I
| c = Singleton of Element is Subset
}}
{{eqn | ll= \leadsto
| l = \bigcup \set {S_\beta}
| o = \subseteq
... | The [[Definition:Set Union|union]] of two [[Definition:Set|sets]] is a [[Definition:Superset|superset]] of each:
:$S \subseteq S \cup T$
:$T \subseteq S \cup T$ | Let $\beta \in I$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \beta
| o = \in
| r = I
| c =
}}
{{eqn | ll= \leadsto
| l = \set \beta
| o = \subseteq
| r = I
| c = [[Singleton of Element is Subset]]
}}
{{eqn | ll= \leadsto
| l = \bigcup \set {S_\beta}
| o = \subs... | Set is Subset of Union/Family of Sets/Proof 2 | https://proofwiki.org/wiki/Set_is_Subset_of_Union | https://proofwiki.org/wiki/Set_is_Subset_of_Union/Family_of_Sets/Proof_2 | [
"Set is Subset of Union",
"Set Union",
"Subsets"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Subset/Superset"
] | [
"Singleton of Element is Subset",
"Union of Subset of Family is Subset of Union of Family"
] |
proofwiki-222 | Set is Subset of Itself | Every set is a subset of itself:
:$\forall S: S \subseteq S$
Thus, by definition, the relation '''is a subset of''' is reflexive. | {{begin-eqn}}
{{eqn | q = \forall x
| l = \leftparen {x \in S}
| o = \implies
| r = \rightparen {x \in S}
| c = Law of Identity:
| cc= a statement implies itself
}}
{{eqn | ll= \leadsto
| l = S
| o = \subseteq
| r = S
| c = {{Defof|Subset}}
}}
{{end-eqn}}
{{qed}} | Every [[Definition:Set|set]] is a [[Definition:Subset|subset]] of itself:
:$\forall S: S \subseteq S$
Thus, by definition, the [[Definition:Relation|relation]] '''is a [[Definition:Subset|subset]] of''' is [[Definition:Reflexive Relation|reflexive]]. | {{begin-eqn}}
{{eqn | q = \forall x
| l = \leftparen {x \in S}
| o = \implies
| r = \rightparen {x \in S}
| c = [[Law of Identity]]:
| cc= a [[Definition:Statement|statement]] implies itself
}}
{{eqn | ll= \leadsto
| l = S
| o = \subseteq
| r = S
| c = {{Defof|Subse... | Set is Subset of Itself | https://proofwiki.org/wiki/Set_is_Subset_of_Itself | https://proofwiki.org/wiki/Set_is_Subset_of_Itself | [
"Subsets"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Relation",
"Definition:Subset",
"Definition:Reflexive Relation"
] | [
"Law of Identity",
"Definition:Statement"
] |
proofwiki-223 | Subset of Set with Propositional Function | Let $S$ be a set.
Let $P: S \to \set {\T, \F}$ be a propositional function on $S$.
Then:
:$\set {x \in S: \map P x} \subseteq S$ | {{begin-eqn}}
{{eqn | l = s
| o = \in
| r = \set {x \in S: \map P x}
| c =
}}
{{eqn | ll= \leadsto
| l = s
| o = \in
| r = \set {x \in S \land \map P x}
| c =
}}
{{eqn | ll= \leadsto
| l = s
| o = \in
| r = \set {x \in S} \land \map P s
| c = {{Defof|S... | Let $S$ be a [[Definition:Set|set]].
Let $P: S \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on $S$.
Then:
:$\set {x \in S: \map P x} \subseteq S$ | {{begin-eqn}}
{{eqn | l = s
| o = \in
| r = \set {x \in S: \map P x}
| c =
}}
{{eqn | ll= \leadsto
| l = s
| o = \in
| r = \set {x \in S \land \map P x}
| c =
}}
{{eqn | ll= \leadsto
| l = s
| o = \in
| r = \set {x \in S} \land \map P s
| c = {{Defof|S... | Subset of Set with Propositional Function | https://proofwiki.org/wiki/Subset_of_Set_with_Propositional_Function | https://proofwiki.org/wiki/Subset_of_Set_with_Propositional_Function | [
"Subsets",
"Mapping Theory"
] | [
"Definition:Set",
"Definition:Propositional Function"
] | [
"Rule of Simplification"
] |
proofwiki-224 | Singleton of Element is Subset | Let $S$ be a set.
Let $\set x$ be the singleton of $x$.
Then:
:$x \in S \iff \set x \subseteq S$ | {{begin-eqn}}
{{eqn | o =
| r = \set x \subseteq A
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \forall y: \paren {y \in \set x \implies y \in A}
| c = {{Defof|Subset}}
}}
{{eqn | o = \leadstoandfrom
| r = \forall y: \paren {y = x \implies y \in A}
| c = {{Defof|Singleton}}
}}
{{eqn ... | Let $S$ be a [[Definition:Set|set]].
Let $\set x$ be the [[Definition:Singleton|singleton of $x$]].
Then:
:$x \in S \iff \set x \subseteq S$ | {{begin-eqn}}
{{eqn | o =
| r = \set x \subseteq A
| c =
}}
{{eqn | o = \leadstoandfrom
| r = \forall y: \paren {y \in \set x \implies y \in A}
| c = {{Defof|Subset}}
}}
{{eqn | o = \leadstoandfrom
| r = \forall y: \paren {y = x \implies y \in A}
| c = {{Defof|Singleton}}
}}
{{eqn ... | Singleton of Element is Subset/Proof 1 | https://proofwiki.org/wiki/Singleton_of_Element_is_Subset | https://proofwiki.org/wiki/Singleton_of_Element_is_Subset/Proof_1 | [
"Subsets",
"Singletons",
"Singleton of Element is Subset"
] | [
"Definition:Set",
"Definition:Singleton"
] | [
"Equality implies Substitution"
] |
proofwiki-225 | Singleton of Element is Subset | Let $S$ be a set.
Let $\set x$ be the singleton of $x$.
Then:
:$x \in S \iff \set x \subseteq S$ | === Necessary Condition ===
Let $x \in S$.
We have:
:$\set x = \set {y \in S: y = x}$
From Subset of Set with Propositional Function:
:$\set {x \in S: \map P x} \subseteq S$
Hence:
:$\set x \subseteq S$
{{qed|lemma}}
=== Sufficient Condition ===
Let $\set x \subseteq S$.
From the definition of a subset:
:$x \in \set x ... | Let $S$ be a [[Definition:Set|set]].
Let $\set x$ be the [[Definition:Singleton|singleton of $x$]].
Then:
:$x \in S \iff \set x \subseteq S$ | === Necessary Condition ===
Let $x \in S$.
We have:
:$\set x = \set {y \in S: y = x}$
From [[Subset of Set with Propositional Function]]:
:$\set {x \in S: \map P x} \subseteq S$
Hence:
:$\set x \subseteq S$
{{qed|lemma}}
=== Sufficient Condition ===
Let $\set x \subseteq S$.
From the definition of a [[Definitio... | Singleton of Element is Subset/Proof 2 | https://proofwiki.org/wiki/Singleton_of_Element_is_Subset | https://proofwiki.org/wiki/Singleton_of_Element_is_Subset/Proof_2 | [
"Subsets",
"Singletons",
"Singleton of Element is Subset"
] | [
"Definition:Set",
"Definition:Singleton"
] | [
"Subset of Set with Propositional Function",
"Definition:Subset"
] |
proofwiki-226 | Subset Relation is Transitive | The subset relation is transitive:
:$\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$ | {{begin-eqn}}
{{eqn | o =
| r = \paren {R \subseteq S} \land \paren {S \subseteq T}
| c =
}}
{{eqn | o = \leadsto
| r = \paren {x \in R \implies x \in S} \land \paren {x \in S \implies x \in T}
| c = {{Defof|Subset}}
}}
{{eqn | o = \leadsto
| r = \paren {x \in R \implies x \in T}
| c... | The [[Definition:Subset Relation|subset relation]] is [[Definition:Transitive Relation|transitive]]:
:$\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$ | {{begin-eqn}}
{{eqn | o =
| r = \paren {R \subseteq S} \land \paren {S \subseteq T}
| c =
}}
{{eqn | o = \leadsto
| r = \paren {x \in R \implies x \in S} \land \paren {x \in S \implies x \in T}
| c = {{Defof|Subset}}
}}
{{eqn | o = \leadsto
| r = \paren {x \in R \implies x \in T}
| c... | Subset Relation is Transitive/Proof 1 | https://proofwiki.org/wiki/Subset_Relation_is_Transitive | https://proofwiki.org/wiki/Subset_Relation_is_Transitive/Proof_1 | [
"Subset Relation",
"Examples of Transitive Relations",
"Subset Relation is Transitive"
] | [
"Definition:Subset Relation",
"Definition:Transitive Relation"
] | [
"Hypothetical Syllogism"
] |
proofwiki-227 | Subset Relation is Transitive | The subset relation is transitive:
:$\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$ | Let $V$ be a basic universe.
By definition of basic universe, $R$, $S$ and $T$ are all elements of $V$.
By the {{axiom-link|Transitivity}}, $R$, $S$ and $T$ are all classes.
We are given that $R \subseteq S$ and $S \subseteq T$.
Hence by Subclass of Subclass is Subclass, $R$ is a subclass of $T$.
By Subclass of Set is ... | The [[Definition:Subset Relation|subset relation]] is [[Definition:Transitive Relation|transitive]]:
:$\paren {R \subseteq S} \land \paren {S \subseteq T} \implies R \subseteq T$ | Let $V$ be a [[Definition:Basic Universe|basic universe]].
By definition of [[Definition:Basic Universe|basic universe]], $R$, $S$ and $T$ are all [[Definition:Element|elements]] of $V$.
By the {{axiom-link|Transitivity}}, $R$, $S$ and $T$ are all [[Definition:Class (Class Theory)|classes]].
We are given that $R \su... | Subset Relation is Transitive/Proof 2 | https://proofwiki.org/wiki/Subset_Relation_is_Transitive | https://proofwiki.org/wiki/Subset_Relation_is_Transitive/Proof_2 | [
"Subset Relation",
"Examples of Transitive Relations",
"Subset Relation is Transitive"
] | [
"Definition:Subset Relation",
"Definition:Transitive Relation"
] | [
"Definition:Basic Universe",
"Definition:Basic Universe",
"Definition:Element",
"Definition:Class (Class Theory)",
"Subclass of Subclass is Subclass",
"Definition:Subclass",
"Subclass of Set is Set",
"Definition:Subset"
] |
proofwiki-228 | Set Inequality | :$S \ne T \iff \paren {S \nsubseteq T} \lor \paren {T \nsubseteq S}$ | {{begin-eqn}}
{{eqn | l=S \ne T
| o=\iff
| r=\neg \paren {S = T}
| c=
}}
{{eqn | o=\iff
| r=\neg \paren {\paren {S \subseteq T} \land \paren {T \subseteq S} }
| c={{Defof|Set Equality|index = 2}}
}}
{{eqn | o=\iff
| r=\neg \paren {S \subseteq T} \lor \neg \paren {T \subseteq S}
... | :$S \ne T \iff \paren {S \nsubseteq T} \lor \paren {T \nsubseteq S}$ | {{begin-eqn}}
{{eqn | l=S \ne T
| o=\iff
| r=\neg \paren {S = T}
| c=
}}
{{eqn | o=\iff
| r=\neg \paren {\paren {S \subseteq T} \land \paren {T \subseteq S} }
| c={{Defof|Set Equality|index = 2}}
}}
{{eqn | o=\iff
| r=\neg \paren {S \subseteq T} \lor \neg \paren {T \subseteq S}
... | Set Inequality | https://proofwiki.org/wiki/Set_Inequality | https://proofwiki.org/wiki/Set_Inequality | [
"Set Theory",
"Subsets"
] | [] | [
"De Morgan's Laws (Logic)/Disjunction of Negations",
"Category:Set Theory",
"Category:Subsets"
] |
proofwiki-229 | Set Equals Itself | All sets are equal to themselves:
:$\forall S: S = S$ | {{begin-eqn}}
{{eqn | l = S \subseteq S
| o = \land
| r = S \supseteq S
| c = Set is Subset of Itself
}}
{{eqn | ll= \leadsto
| l = S
| r = S
| c = {{Defof|Set Equality}}
}}
{{end-eqn}}
{{qed}}
Category:Set Equality
0sgjaz6fxne0e6dc2vi4eh1wrgw4gsw | All [[Definition:Set|sets]] are [[Definition:Set Equality|equal]] to themselves:
:$\forall S: S = S$ | {{begin-eqn}}
{{eqn | l = S \subseteq S
| o = \land
| r = S \supseteq S
| c = [[Set is Subset of Itself]]
}}
{{eqn | ll= \leadsto
| l = S
| r = S
| c = {{Defof|Set Equality}}
}}
{{end-eqn}}
{{qed}}
[[Category:Set Equality]]
0sgjaz6fxne0e6dc2vi4eh1wrgw4gsw | Set Equals Itself | https://proofwiki.org/wiki/Set_Equals_Itself | https://proofwiki.org/wiki/Set_Equals_Itself | [
"Set Equality"
] | [
"Definition:Set",
"Definition:Set Equality"
] | [
"Set is Subset of Itself",
"Category:Set Equality"
] |
proofwiki-230 | Union is Smallest Superset | Let $S_1$ and $S_2$ be sets.
Then $S_1 \cup S_2$ is the smallest set containing both $S_1$ and $S_2$.
That is:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$ | === Necessary Condition ===
From Union of Subsets is Subset:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$
{{qed|lemma}} | Let $S_1$ and $S_2$ be [[Definition:Set|sets]].
Then $S_1 \cup S_2$ is the [[Definition:Smallest Set by Set Inclusion|smallest set]] containing both $S_1$ and $S_2$.
That is:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$ | === Necessary Condition ===
From [[Union of Subsets is Subset]]:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$
{{qed|lemma}} | Union is Smallest Superset | https://proofwiki.org/wiki/Union_is_Smallest_Superset | https://proofwiki.org/wiki/Union_is_Smallest_Superset | [
"Set Union",
"Subsets"
] | [
"Definition:Set",
"Definition:Smallest Set by Set Inclusion"
] | [
"Union of Subsets is Subset"
] |
proofwiki-231 | Union with Empty Set | The union of any set with the empty set is the set itself:
:$S \cup \O = S$ | {{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = S
| c = Set is Subset of Itself
}}
{{eqn | l = \O
| o = \subseteq
| r = S
| c = Empty Set is Subset of All Sets
}}
{{eqn | ll= \leadsto
| l = S \cup \O
| o = \subseteq
| r = S
| c = Union is Smallest Superset
}... | The [[Definition:Set Union|union]] of any [[Definition:Set|set]] with the [[Definition:Empty Set|empty set]] is the [[Definition:Set|set]] itself:
:$S \cup \O = S$ | {{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = S
| c = [[Set is Subset of Itself]]
}}
{{eqn | l = \O
| o = \subseteq
| r = S
| c = [[Empty Set is Subset of All Sets]]
}}
{{eqn | ll= \leadsto
| l = S \cup \O
| o = \subseteq
| r = S
| c = [[Union is Smallest ... | Union with Empty Set/Proof 1 | https://proofwiki.org/wiki/Union_with_Empty_Set | https://proofwiki.org/wiki/Union_with_Empty_Set/Proof_1 | [
"Union with Empty Set",
"Set Union",
"Empty Set"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Empty Set",
"Definition:Set"
] | [
"Set is Subset of Itself",
"Empty Set is Subset of All Sets",
"Union is Smallest Superset",
"Set is Subset of Union"
] |
proofwiki-232 | Union with Empty Set | The union of any set with the empty set is the set itself:
:$S \cup \O = S$ | From Empty Set is Subset of All Sets:
:$\O \subseteq S$
From Union with Superset is Superset:
:$S \cup \O = S$
{{qed}} | The [[Definition:Set Union|union]] of any [[Definition:Set|set]] with the [[Definition:Empty Set|empty set]] is the [[Definition:Set|set]] itself:
:$S \cup \O = S$ | From [[Empty Set is Subset of All Sets]]:
:$\O \subseteq S$
From [[Union with Superset is Superset]]:
:$S \cup \O = S$
{{qed}} | Union with Empty Set/Proof 2 | https://proofwiki.org/wiki/Union_with_Empty_Set | https://proofwiki.org/wiki/Union_with_Empty_Set/Proof_2 | [
"Union with Empty Set",
"Set Union",
"Empty Set"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Empty Set",
"Definition:Set"
] | [
"Empty Set is Subset of All Sets",
"Union with Superset is Superset"
] |
proofwiki-233 | Intersection is Subset | The intersection of two sets is a subset of each:
:$S \cap T \subseteq S$
:$S \cap T \subseteq T$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \cap T
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S \land x \in T
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S
| c = Rule of Simplification
}}
{{eqn | ll= \leadsto
|... | The [[Definition:Set Intersection|intersection]] of two [[Definition:Set|sets]] is a [[Definition:Subset|subset]] of each:
:$S \cap T \subseteq S$
:$S \cap T \subseteq T$ | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S \cap T
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S \land x \in T
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S
| c = [[Rule of Simplification]]
}}
{{eqn | ll= \leadsto
... | Intersection is Subset | https://proofwiki.org/wiki/Intersection_is_Subset | https://proofwiki.org/wiki/Intersection_is_Subset | [
"Set Intersection",
"Subsets"
] | [
"Definition:Set Intersection",
"Definition:Set",
"Definition:Subset"
] | [
"Rule of Simplification"
] |
proofwiki-234 | Intersection with Empty Set | The intersection of any set with the empty set is itself the empty set:
:$S \cap \O = \O$ | {{begin-eqn}}
{{eqn | l = S \cap \O
| o = \subseteq
| r = \O
| c = Intersection is Subset
}}
{{eqn | l = \O
| o = \subseteq
| r = S \cap \O
| c = Empty Set is Subset of All Sets
}}
{{eqn | ll= \leadsto
| l = S \cap \O
| r = \O
| c = {{Defof|Set Equality|index = 2}}
... | The [[Definition:Set Intersection|intersection]] of any [[Definition:Set|set]] with the [[Definition:Empty Set|empty set]] is itself the [[Definition:Empty Set|empty set]]:
:$S \cap \O = \O$ | {{begin-eqn}}
{{eqn | l = S \cap \O
| o = \subseteq
| r = \O
| c = [[Intersection is Subset]]
}}
{{eqn | l = \O
| o = \subseteq
| r = S \cap \O
| c = [[Empty Set is Subset of All Sets]]
}}
{{eqn | ll= \leadsto
| l = S \cap \O
| r = \O
| c = {{Defof|Set Equality|inde... | Intersection with Empty Set | https://proofwiki.org/wiki/Intersection_with_Empty_Set | https://proofwiki.org/wiki/Intersection_with_Empty_Set | [
"Set Intersection",
"Empty Set"
] | [
"Definition:Set Intersection",
"Definition:Set",
"Definition:Empty Set",
"Definition:Empty Set"
] | [
"Intersection is Subset",
"Empty Set is Subset of All Sets"
] |
proofwiki-235 | Intersection is Largest Subset | Let $T_1$ and $T_2$ be sets.
Then $T_1 \cap T_2$ is the largest set contained in both $T_1$ and $T_2$.
That is:
:$S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ | === Sufficient Condition ===
From Set is Subset of Intersection of Supersets we have that:
:$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$
{{qed|lemma}} | Let $T_1$ and $T_2$ be [[Definition:Set|sets]].
Then $T_1 \cap T_2$ is the [[Definition:Greatest Set by Set Inclusion|largest set]] contained in both $T_1$ and $T_2$.
That is:
:$S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ | === Sufficient Condition ===
From [[Set is Subset of Intersection of Supersets]] we have that:
:$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$
{{qed|lemma}} | Intersection is Largest Subset | https://proofwiki.org/wiki/Intersection_is_Largest_Subset | https://proofwiki.org/wiki/Intersection_is_Largest_Subset | [
"Set Intersection",
"Subsets"
] | [
"Definition:Set",
"Definition:Greatest Set by Set Inclusion"
] | [
"Set is Subset of Intersection of Supersets"
] |
proofwiki-236 | Intersection is Subset of Union | The intersection of two sets is a subset of their union:
:$S \cap T \subseteq S \cup T$ | {{begin-eqn}}
{{eqn | l = S \cap T
| o = \subseteq
| r = S
| c = Intersection is Subset
}}
{{eqn | l = S
| o = \subseteq
| r = S \cup T
| c = Set is Subset of Union
}}
{{eqn | ll= \leadsto
| l = S \cap T
| o = \subseteq
| r = S \cup T
| c = Subset Relation is... | The [[Definition:Set Intersection|intersection]] of two [[Definition:Set|sets]] is a [[Definition:Subset|subset]] of their [[Definition:Set Union|union]]:
:$S \cap T \subseteq S \cup T$ | {{begin-eqn}}
{{eqn | l = S \cap T
| o = \subseteq
| r = S
| c = [[Intersection is Subset]]
}}
{{eqn | l = S
| o = \subseteq
| r = S \cup T
| c = [[Set is Subset of Union]]
}}
{{eqn | ll= \leadsto
| l = S \cap T
| o = \subseteq
| r = S \cup T
| c = [[Subset R... | Intersection is Subset of Union | https://proofwiki.org/wiki/Intersection_is_Subset_of_Union | https://proofwiki.org/wiki/Intersection_is_Subset_of_Union | [
"Set Intersection",
"Set Union"
] | [
"Definition:Set Intersection",
"Definition:Set",
"Definition:Subset",
"Definition:Set Union"
] | [
"Intersection is Subset",
"Set is Subset of Union",
"Subset Relation is Transitive"
] |
proofwiki-237 | Union Distributes over Intersection | Set union is distributive over set intersection:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$ | === Union Subset of Intersection ===
Let $\ds x \in S \cup \bigcap \mathbb T$.
We need to show that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and then by definition of subset we will have shown that:
:$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.
S... | [[Definition:Set Union|Set union]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Intersection|set intersection]]:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$ | === Union Subset of Intersection ===
Let $\ds x \in S \cup \bigcap \mathbb T$.
We need to show that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and then by definition of [[Definition:Subset|subset]] we will have shown that:
:$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb... | Union Distributes over Intersection/General Result/Proof | https://proofwiki.org/wiki/Union_Distributes_over_Intersection | https://proofwiki.org/wiki/Union_Distributes_over_Intersection/General_Result/Proof | [
"Union Distributes over Intersection",
"Distributive Laws of Set Theory",
"Set Union",
"Set Intersection",
"Examples of Distributive Operations"
] | [
"Definition:Set Union",
"Definition:Distributive Operation",
"Definition:Set Intersection"
] | [
"Definition:Subset",
"Definition:Set Union",
"Definition:Set Union",
"Definition:Set Intersection",
"Definition:Set Union",
"Proof by Cases",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Set Intersection",
"Definition:Set Union",
"Definition:Set Union",
"Proof by Cases",
"... |
proofwiki-238 | Union Distributes over Intersection | Set union is distributive over set intersection:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in R \cup \paren {S \cap T}
}}
{{eqn | o = \leadstoandfrom
| r = x \in R \lor \paren {x \in S \land x \in T}
| c = {{Defof|Set Union}} and {{Defof|Set Intersection}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in R \lor x \in S} \land \paren {x \in R \lor ... | [[Definition:Set Union|Set union]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Intersection|set intersection]]:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in R \cup \paren {S \cap T}
}}
{{eqn | o = \leadstoandfrom
| r = x \in R \lor \paren {x \in S \land x \in T}
| c = {{Defof|Set Union}} and {{Defof|Set Intersection}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in R \lor x \in S} \land \paren {x \in R \lor ... | Union Distributes over Intersection/Proof 1 | https://proofwiki.org/wiki/Union_Distributes_over_Intersection | https://proofwiki.org/wiki/Union_Distributes_over_Intersection/Proof_1 | [
"Union Distributes over Intersection",
"Distributive Laws of Set Theory",
"Set Union",
"Set Intersection",
"Examples of Distributive Operations"
] | [
"Definition:Set Union",
"Definition:Distributive Operation",
"Definition:Set Intersection"
] | [
"Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive"
] |
proofwiki-239 | Union Distributes over Intersection | Set union is distributive over set intersection:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$ | From Intersection Distributes over Union:
:$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$
From the Duality Principle for Sets, exchanging $\cup$ for $\cap$ throughout, and vice versa, reveals the result:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$
{{qed}} | [[Definition:Set Union|Set union]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Intersection|set intersection]]:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$ | From [[Intersection Distributes over Union]]:
:$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$
From the [[Duality Principle for Sets]], exchanging $\cup$ for $\cap$ throughout, and vice versa, reveals the result:
:$R \cup \paren {S \cap T} = \paren {R \cup S} \cap \paren {R \cup T}$
{{qed}} | Union Distributes over Intersection/Proof 2 | https://proofwiki.org/wiki/Union_Distributes_over_Intersection | https://proofwiki.org/wiki/Union_Distributes_over_Intersection/Proof_2 | [
"Union Distributes over Intersection",
"Distributive Laws of Set Theory",
"Set Union",
"Set Intersection",
"Examples of Distributive Operations"
] | [
"Definition:Set Union",
"Definition:Distributive Operation",
"Definition:Set Intersection"
] | [
"Intersection Distributes over Union",
"Duality Principle for Sets"
] |
proofwiki-240 | Intersection Distributes over Union | Set intersection is distributive over set union:
:$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in R \cap \paren {S \cup T}
}}
{{eqn | o = \leadstoandfrom
| r = x \in R \land \paren {x \in S \lor x \in T}
| c = {{Defof|Set Union}} and {{Defof|Set Intersection}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in R \land x \in S} \lor \paren {x \in R \land... | [[Definition:Set Intersection|Set intersection]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Set Union|set union]]:
:$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$ | {{begin-eqn}}
{{eqn | o =
| r = x \in R \cap \paren {S \cup T}
}}
{{eqn | o = \leadstoandfrom
| r = x \in R \land \paren {x \in S \lor x \in T}
| c = {{Defof|Set Union}} and {{Defof|Set Intersection}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {x \in R \land x \in S} \lor \paren {x \in R \land... | Intersection Distributes over Union | https://proofwiki.org/wiki/Intersection_Distributes_over_Union | https://proofwiki.org/wiki/Intersection_Distributes_over_Union | [
"Intersection Distributes over Union",
"Distributive Laws of Set Theory",
"Set Intersection",
"Set Union",
"Examples of Distributive Operations"
] | [
"Definition:Set Intersection",
"Definition:Distributive Operation",
"Definition:Set Union"
] | [
"Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive"
] |
proofwiki-241 | Set Difference is Subset | :$S \setminus T \subseteq S$ | {{begin-eqn}}
{{eqn | l = x \in S \setminus T
| o = \leadsto
| r = x \in S \land x \notin T
| c = {{Defof|Set Difference}}
}}
{{eqn | o = \leadsto
| r = x \in S
| c = Rule of Simplification
}}
{{end-eqn}}
The result follows from the definition of subset.
{{Qed}} | :$S \setminus T \subseteq S$ | {{begin-eqn}}
{{eqn | l = x \in S \setminus T
| o = \leadsto
| r = x \in S \land x \notin T
| c = {{Defof|Set Difference}}
}}
{{eqn | o = \leadsto
| r = x \in S
| c = [[Rule of Simplification]]
}}
{{end-eqn}}
The result follows from the definition of [[Definition:Subset|subset]].
{{Qed}} | Set Difference is Subset/Proof 1 | https://proofwiki.org/wiki/Set_Difference_is_Subset | https://proofwiki.org/wiki/Set_Difference_is_Subset/Proof_1 | [
"Set Difference",
"Subsets",
"Set Difference is Subset"
] | [] | [
"Rule of Simplification",
"Definition:Subset"
] |
proofwiki-242 | Set Difference is Subset | :$S \setminus T \subseteq S$ | {{begin-eqn}}
{{eqn | l = S \setminus T
| r = S \cap \complement_S \left({T}\right)
| c = Set Difference as Intersection with Relative Complement
}}
{{eqn | o = \subseteq
| r = S
| c = Intersection is Subset
}}
{{end-eqn}}
{{Qed}} | :$S \setminus T \subseteq S$ | {{begin-eqn}}
{{eqn | l = S \setminus T
| r = S \cap \complement_S \left({T}\right)
| c = [[Set Difference as Intersection with Relative Complement]]
}}
{{eqn | o = \subseteq
| r = S
| c = [[Intersection is Subset]]
}}
{{end-eqn}}
{{Qed}} | Set Difference is Subset/Proof 2 | https://proofwiki.org/wiki/Set_Difference_is_Subset | https://proofwiki.org/wiki/Set_Difference_is_Subset/Proof_2 | [
"Set Difference",
"Subsets",
"Set Difference is Subset"
] | [] | [
"Set Difference as Intersection with Relative Complement",
"Intersection is Subset"
] |
proofwiki-243 | Set Difference with Empty Set is Self | The set difference between a set and the empty set is the set itself:
:$S \setminus \O = S$ | From Set Difference is Subset:
:$S \setminus \O \subseteq S$
From the definition of the empty set:
:$\forall x \in S: x \notin \O$
Let $x \in S$.
Thus:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S \land x \notin \O
| c =... | The [[Definition:Set Difference|set difference]] between a [[Definition:Set|set]] and the [[Definition:Empty Set|empty set]] is the [[Definition:Set|set]] itself:
:$S \setminus \O = S$ | From [[Set Difference is Subset]]:
:$S \setminus \O \subseteq S$
From the definition of the [[Definition:Empty Set|empty set]]:
:$\forall x \in S: x \notin \O$
Let $x \in S$.
Thus:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r... | Set Difference with Empty Set is Self | https://proofwiki.org/wiki/Set_Difference_with_Empty_Set_is_Self | https://proofwiki.org/wiki/Set_Difference_with_Empty_Set_is_Self | [
"Set Difference",
"Empty Set"
] | [
"Definition:Set Difference",
"Definition:Set",
"Definition:Empty Set",
"Definition:Set"
] | [
"Set Difference is Subset",
"Definition:Empty Set",
"Rule of Conjunction",
"Definition:Set Equality/Definition 2"
] |
proofwiki-244 | Set Difference with Self is Empty Set | The set difference of a set with itself is the empty set:
:$S \setminus S = \O$ | From Set is Subset of Itself:
:$S \subseteq S$
From Set Difference with Superset is Empty Set we have:
:$S \subseteq T \iff S \setminus T = \O$
Hence the result.
{{qed}} | The [[Definition:Set Difference|set difference]] of a [[Definition:Set|set]] with itself is the [[Definition:Empty Set|empty set]]:
:$S \setminus S = \O$ | From [[Set is Subset of Itself]]:
:$S \subseteq S$
From [[Set Difference with Superset is Empty Set]] we have:
:$S \subseteq T \iff S \setminus T = \O$
Hence the result.
{{qed}} | Set Difference with Self is Empty Set | https://proofwiki.org/wiki/Set_Difference_with_Self_is_Empty_Set | https://proofwiki.org/wiki/Set_Difference_with_Self_is_Empty_Set | [
"Set Difference",
"Empty Set"
] | [
"Definition:Set Difference",
"Definition:Set",
"Definition:Empty Set"
] | [
"Set is Subset of Itself",
"Set Difference with Superset is Empty Set"
] |
proofwiki-245 | Set Difference Equals First Set iff Sets are Disjoint | :$S \setminus T = S \iff S \cap T = \O$ | Assume $S, T \subseteq \Bbb U$ where $\Bbb U$ is a universal set.
{{begin-eqn}}
{{eqn | l = S \setminus T
| r = S
}}
{{eqn | ll= \leadstoandfrom
| l = S \cap \map \complement T
| r = S
| c = Set Difference as Intersection with Complement
}}
{{eqn | ll= \leadstoandfrom
| l = S
| o = \... | :$S \setminus T = S \iff S \cap T = \O$ | Assume $S, T \subseteq \Bbb U$ where $\Bbb U$ is a [[Definition:Universal Set|universal set]].
{{begin-eqn}}
{{eqn | l = S \setminus T
| r = S
}}
{{eqn | ll= \leadstoandfrom
| l = S \cap \map \complement T
| r = S
| c = [[Set Difference as Intersection with Complement]]
}}
{{eqn | ll= \leadstoa... | Set Difference Equals First Set iff Sets are Disjoint | https://proofwiki.org/wiki/Set_Difference_Equals_First_Set_iff_Sets_are_Disjoint | https://proofwiki.org/wiki/Set_Difference_Equals_First_Set_iff_Sets_are_Disjoint | [
"Set Difference",
"Set Intersection",
"Disjoint Sets"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Intersection with Subset is Subset",
"Intersection with Complement is Empty iff Subset",
"Complement of Complement"
] |
proofwiki-246 | Equal Set Differences iff Equal Intersections | :$R \setminus S = R \setminus T \iff R \cap S = R \cap T$ | {{begin-eqn}}
{{eqn | l = R \setminus S
| r = R \setminus T
}}
{{eqn | ll= \leadstoandfrom
| l = \set {x \in R: x \notin S}
| r = \set {x \in R: x \notin T}
| c = {{Defof|Set Difference}}
}}
{{eqn | ll= \leadstoandfrom
| q = \forall x \in R
| l = x \notin S
| o = \iff
| r... | :$R \setminus S = R \setminus T \iff R \cap S = R \cap T$ | {{begin-eqn}}
{{eqn | l = R \setminus S
| r = R \setminus T
}}
{{eqn | ll= \leadstoandfrom
| l = \set {x \in R: x \notin S}
| r = \set {x \in R: x \notin T}
| c = {{Defof|Set Difference}}
}}
{{eqn | ll= \leadstoandfrom
| q = \forall x \in R
| l = x \notin S
| o = \iff
| r... | Equal Set Differences iff Equal Intersections/Proof 1 | https://proofwiki.org/wiki/Equal_Set_Differences_iff_Equal_Intersections | https://proofwiki.org/wiki/Equal_Set_Differences_iff_Equal_Intersections/Proof_1 | [
"Set Difference",
"Set Intersection",
"Equal Set Differences iff Equal Intersections"
] | [] | [] |
proofwiki-247 | Equal Set Differences iff Equal Intersections | :$R \setminus S = R \setminus T \iff R \cap S = R \cap T$ | From Set Difference and Intersection form Partition:
: $\paren {R \setminus S} \cup \paren {R \cap S} = R = \paren {R \setminus T} \cup \paren {R \cap T}$
: $\paren {R \cap S} \cap \paren {R \setminus S} = \O = \paren {R \cap T} \cap \paren {R \setminus T}$
whatever $R, S, T$ might be.
Let $R \setminus S = R \setminus ... | :$R \setminus S = R \setminus T \iff R \cap S = R \cap T$ | From [[Set Difference and Intersection form Partition]]:
: $\paren {R \setminus S} \cup \paren {R \cap S} = R = \paren {R \setminus T} \cup \paren {R \cap T}$
: $\paren {R \cap S} \cap \paren {R \setminus S} = \O = \paren {R \cap T} \cap \paren {R \setminus T}$
whatever $R, S, T$ might be.
Let $R \setminus S = R \set... | Equal Set Differences iff Equal Intersections/Proof 2 | https://proofwiki.org/wiki/Equal_Set_Differences_iff_Equal_Intersections | https://proofwiki.org/wiki/Equal_Set_Differences_iff_Equal_Intersections/Proof_2 | [
"Set Difference",
"Set Intersection",
"Equal Set Differences iff Equal Intersections"
] | [] | [
"Set Difference and Intersection form Partition",
"Set Difference with Union is Set Difference",
"Set Difference with Disjoint Set",
"Set Difference with Union is Set Difference",
"Set Difference with Disjoint Set"
] |
proofwiki-248 | Euler's Number is Irrational | Euler's number $e$ is irrational. | {{AimForCont}} that $e$ is rational.
Then there exist coprime integers $m$ and $n$ (and we can choose $n$ to be positive) such that:
:$\dfrac m n = e = \ds \sum_{i \mathop = 0}^\infty \frac 1 {i!}$ from the definition of Euler's number.
Multiplying both sides by $n!$, observe that:
:$\dfrac m n n! = n! \ds \sum_{i \mat... | [[Definition:Euler's Number|Euler's number]] $e$ is [[Definition:Irrational Number|irrational]]. | {{AimForCont}} that $e$ is [[Definition:Rational Number|rational]].
Then there exist [[Definition:Coprime Integers|coprime integers]] $m$ and $n$ (and we can choose $n$ to be [[Definition:Positive Integer|positive]]) such that:
:$\dfrac m n = e = \ds \sum_{i \mathop = 0}^\infty \frac 1 {i!}$ from the definition of [[... | Euler's Number is Irrational | https://proofwiki.org/wiki/Euler's_Number_is_Irrational | https://proofwiki.org/wiki/Euler's_Number_is_Irrational | [
"Irrationality Proofs",
"Euler's Number"
] | [
"Definition:Euler's Number",
"Definition:Irrational Number"
] | [
"Definition:Rational Number",
"Definition:Coprime/Integers",
"Definition:Positive/Integer",
"Definition:Euler's Number/Limit of Series",
"Sum of Infinite Geometric Sequence",
"Definition:Integer",
"Definition:Addition/Integers",
"Definition:Subtraction/Integers",
"Definition:Integer",
"Definition:... |
proofwiki-249 | Set Difference with Intersection | Let $S$ and $T$ be sets. | 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... | Let $S$ and $T$ be [[Definition:Set|sets]]. | 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/Set_Difference_with_Intersection | https://proofwiki.org/wiki/Intersection_with_Set_Difference_is_Set_Difference_with_Intersection/Proof_1 | [
"Set Difference",
"Set Intersection",
"Empty Set"
] | [
"Definition:Set"
] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Intersection is Commutative",
"Intersection is Associative",
"Set Difference as Intersection with Complement"
] |
proofwiki-250 | Set Difference with Intersection | Let $S$ and $T$ be sets. | {{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... | Let $S$ and $T$ be [[Definition:Set|sets]]. | {{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/Set_Difference_with_Intersection | https://proofwiki.org/wiki/Intersection_with_Set_Difference_is_Set_Difference_with_Intersection/Proof_2 | [
"Set Difference",
"Set Intersection",
"Empty Set"
] | [
"Definition:Set"
] | [
"Rule of Commutation",
"Rule of Association"
] |
proofwiki-251 | Pascal's Rule | For positive integers $n, k$ with $1 \le k \le n$:
:$\dbinom n {k - 1} + \dbinom n k = \dbinom {n + 1} k$
This is also valid for the real number definition:
:$\forall r \in \R, k \in \Z: \dbinom r {k - 1} + \dbinom r k = \dbinom {r + 1} k$ | Suppose you were a member of a club with $n + 1$ members (including you).
Suppose it were time to elect a committee of $k$ members from that club.
From Cardinality of Set of Subsets, there are $\dbinom {n + 1} k$ ways to select the members to form this committee.
Now, you yourself may or may not be elected a member of ... | For [[Definition:Positive Integer|positive integers]] $n, k$ with $1 \le k \le n$:
:$\dbinom n {k - 1} + \dbinom n k = \dbinom {n + 1} k$
This is also valid for the [[Definition:Binomial Coefficient/Real Numbers|real number definition]]:
:$\forall r \in \R, k \in \Z: \dbinom r {k - 1} + \dbinom r k = \dbinom {r + 1} ... | Suppose you were a member of a club with $n + 1$ members (including you).
Suppose it were time to elect a committee of $k$ members from that club.
From [[Cardinality of Set of Subsets]], there are $\dbinom {n + 1} k$ ways to select the members to form this committee.
Now, you yourself may or may not be elected a me... | Pascal's Rule/Combinatorial Proof | https://proofwiki.org/wiki/Pascal's_Rule | https://proofwiki.org/wiki/Pascal's_Rule/Combinatorial_Proof | [
"Pascal's Rule",
"Binomial Coefficients"
] | [
"Definition:Positive/Integer",
"Definition:Binomial Coefficient/Real Numbers"
] | [
"Cardinality of Set of Subsets",
"Cardinality of Set of Subsets",
"Cardinality of Set of Subsets"
] |
proofwiki-252 | Pascal's Rule | For positive integers $n, k$ with $1 \le k \le n$:
:$\dbinom n {k - 1} + \dbinom n k = \dbinom {n + 1} k$
This is also valid for the real number definition:
:$\forall r \in \R, k \in \Z: \dbinom r {k - 1} + \dbinom r k = \dbinom {r + 1} k$ | Let $n, k \in \N$ with $1 \le k \le n$.
{{begin-eqn}}
{{eqn | l = \binom n k + \binom n {k - 1}
| r = \frac {n!} {k! \, \paren {n - k}!} + \frac {n!} {\paren {k - 1}! \, \paren {n - \paren {k - 1} }!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac {n! \, \paren {n - \paren {k - 1} } } {k! \, \par... | For [[Definition:Positive Integer|positive integers]] $n, k$ with $1 \le k \le n$:
:$\dbinom n {k - 1} + \dbinom n k = \dbinom {n + 1} k$
This is also valid for the [[Definition:Binomial Coefficient/Real Numbers|real number definition]]:
:$\forall r \in \R, k \in \Z: \dbinom r {k - 1} + \dbinom r k = \dbinom {r + 1} ... | Let $n, k \in \N$ with $1 \le k \le n$.
{{begin-eqn}}
{{eqn | l = \binom n k + \binom n {k - 1}
| r = \frac {n!} {k! \, \paren {n - k}!} + \frac {n!} {\paren {k - 1}! \, \paren {n - \paren {k - 1} }!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac {n! \, \paren {n - \paren {k - 1} } } {k! \, \pa... | Pascal's Rule/Direct Proof | https://proofwiki.org/wiki/Pascal's_Rule | https://proofwiki.org/wiki/Pascal's_Rule/Direct_Proof | [
"Pascal's Rule",
"Binomial Coefficients"
] | [
"Definition:Positive/Integer",
"Definition:Binomial Coefficient/Real Numbers"
] | [
"Addition of Fractions"
] |
proofwiki-253 | Set Difference Union Intersection | :$S = \paren {S \setminus T} \cup \paren {S \cap T}$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cup \paren {S \cap T}
| r = \paren {\paren {S \setminus T} \cup S} \cap \paren {\paren {S \setminus T} \cup T}
| c = Union Distributes over Intersection
}}
{{eqn | r = S \cap \paren {\paren {S \setminus T} \cup T}
| c = Set Difference Union First Set i... | :$S = \paren {S \setminus T} \cup \paren {S \cap T}$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cup \paren {S \cap T}
| r = \paren {\paren {S \setminus T} \cup S} \cap \paren {\paren {S \setminus T} \cup T}
| c = [[Union Distributes over Intersection]]
}}
{{eqn | r = S \cap \paren {\paren {S \setminus T} \cup T}
| c = [[Set Difference Union First... | Set Difference Union Intersection/Proof 1 | https://proofwiki.org/wiki/Set_Difference_Union_Intersection | https://proofwiki.org/wiki/Set_Difference_Union_Intersection/Proof_1 | [
"Set Difference",
"Set Union",
"Set Intersection",
"Set Difference Union Intersection"
] | [] | [
"Union Distributes over Intersection",
"Set Difference Union First Set is First Set",
"Set Difference Union Second Set is Union",
"Absorption Laws (Set Theory)/Intersection Absorbs Union"
] |
proofwiki-254 | Set Difference Union Intersection | :$S = \paren {S \setminus T} \cup \paren {S \cap T}$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cup \paren {S \cap T}
| r = S \setminus \paren {T \setminus T}
| c = Set Difference with Set Difference is Union of Set Difference with Intersection
}}
{{eqn | r = S \setminus \O
| c = Set Difference with Self is Empty Set
}}
{{eqn | r = S
| c = ... | :$S = \paren {S \setminus T} \cup \paren {S \cap T}$ | {{begin-eqn}}
{{eqn | l = \paren {S \setminus T} \cup \paren {S \cap T}
| r = S \setminus \paren {T \setminus T}
| c = [[Set Difference with Set Difference is Union of Set Difference with Intersection]]
}}
{{eqn | r = S \setminus \O
| c = [[Set Difference with Self is Empty Set]]
}}
{{eqn | r = S
... | Set Difference Union Intersection/Proof 2 | https://proofwiki.org/wiki/Set_Difference_Union_Intersection | https://proofwiki.org/wiki/Set_Difference_Union_Intersection/Proof_2 | [
"Set Difference",
"Set Union",
"Set Intersection",
"Set Difference Union Intersection"
] | [] | [
"Set Difference with Set Difference is Union of Set Difference with Intersection",
"Set Difference with Self is Empty Set",
"Set Difference with Empty Set is Self"
] |
proofwiki-255 | Set Difference Union Intersection | :$S = \paren {S \setminus T} \cup \paren {S \cap T}$ | By Set Difference is Subset:
:$S \setminus T \subseteq S$
By Intersection is Subset:
:$S \cap T \subseteq S$
Hence from Union is Smallest Superset:
:$\paren {S \setminus T} \cup \paren {S \cap T} \subseteq S$
Let $s \in S$.
Either:
:$s \in T$, in which case $s \in S \cap T$ by definition of set intersection
or
:$s \not... | :$S = \paren {S \setminus T} \cup \paren {S \cap T}$ | By [[Set Difference is Subset]]:
:$S \setminus T \subseteq S$
By [[Intersection is Subset]]:
:$S \cap T \subseteq S$
Hence from [[Union is Smallest Superset]]:
:$\paren {S \setminus T} \cup \paren {S \cap T} \subseteq S$
Let $s \in S$.
Either:
:$s \in T$, in which case $s \in S \cap T$ by definition of [[Definitio... | Set Difference Union Intersection/Proof 3 | https://proofwiki.org/wiki/Set_Difference_Union_Intersection | https://proofwiki.org/wiki/Set_Difference_Union_Intersection/Proof_3 | [
"Set Difference",
"Set Union",
"Set Intersection",
"Set Difference Union Intersection"
] | [] | [
"Set Difference is Subset",
"Intersection is Subset",
"Union is Smallest Superset",
"Definition:Set Intersection",
"Definition:Set Difference",
"Definition:Set Union",
"Definition:Subset",
"Definition:Set Equality/Definition 2"
] |
proofwiki-256 | Absorption Laws (Logic) | For any two propositions $p$ and $q$, we have: | By calculation:
{{begin-eqn}}
{{eqn | l = p \land \paren {p \lor q}
| r = \paren {p \lor \bot} \land \paren {p \lor q}
| c = Disjunction with Contradiction
}}
{{eqn | r = p \lor \paren {\bot \land q}
| c = Disjunction is Left Distributive over Conjunction
}}
{{eqn | r = p \lor \bot
| c = Conjunc... | For any two [[Definition:Proposition|propositions]] $p$ and $q$, we have: | By calculation:
{{begin-eqn}}
{{eqn | l = p \land \paren {p \lor q}
| r = \paren {p \lor \bot} \land \paren {p \lor q}
| c = [[Disjunction with Contradiction]]
}}
{{eqn | r = p \lor \paren {\bot \land q}
| c = [[Disjunction is Left Distributive over Conjunction]]
}}
{{eqn | r = p \lor \bot
| c ... | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof 2 | https://proofwiki.org/wiki/Absorption_Laws_(Logic) | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Proof_2 | [
"Conjunction",
"Disjunction"
] | [
"Definition:Proposition"
] | [
"Disjunction with Contradiction",
"Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive",
"Conjunction with Contradiction",
"Disjunction with Contradiction"
] |
proofwiki-257 | Absorption Laws (Logic) | For any two propositions $p$ and $q$, we have: | We apply the Method of Truth Tables.
As can be seen by inspection, the appropriate truth values match for all boolean interpretations.
$\begin{array}{|ccccc||c|} \hline
p & \land & (p & \lor & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F & \F & \T & \T & \F \\
\T & \T & \T & \T & \F & \T \\
\T & \T & \T & \T... | For any two [[Definition:Proposition|propositions]] $p$ and $q$, we have: | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the appropriate [[Definition:Truth Value|truth values]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|ccccc||c|} \hline
p & \land & (p & \lor & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F ... | Absorption Laws (Logic)/Conjunction Absorbs Disjunction/Proof by Truth Table | https://proofwiki.org/wiki/Absorption_Laws_(Logic) | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Conjunction_Absorbs_Disjunction/Proof_by_Truth_Table | [
"Conjunction",
"Disjunction"
] | [
"Definition:Proposition"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Boolean Interpretation"
] |
proofwiki-258 | Absorption Laws (Logic) | For any two propositions $p$ and $q$, we have: | {{begin-eqn}}
{{eqn | l = p \lor \paren {p \land q}
| r = \paren {p \land \top} \lor \paren {p \land q}
| c = Conjunction with Tautology
}}
{{eqn | r = p \land \paren {\top \lor q}
| c = Conjunction is Left Distributive over Disjunction
}}
{{eqn | r = p \land \top
| c = Disjunction with Tautolog... | For any two [[Definition:Proposition|propositions]] $p$ and $q$, we have: | {{begin-eqn}}
{{eqn | l = p \lor \paren {p \land q}
| r = \paren {p \land \top} \lor \paren {p \land q}
| c = [[Conjunction with Tautology]]
}}
{{eqn | r = p \land \paren {\top \lor q}
| c = [[Conjunction is Left Distributive over Disjunction]]
}}
{{eqn | r = p \land \top
| c = [[Disjunction wit... | Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Proof 2 | https://proofwiki.org/wiki/Absorption_Laws_(Logic) | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction/Proof_2 | [
"Conjunction",
"Disjunction"
] | [
"Definition:Proposition"
] | [
"Conjunction with Tautology",
"Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive",
"Disjunction with Tautology",
"Conjunction with Tautology"
] |
proofwiki-259 | Absorption Laws (Logic) | For any two propositions $p$ and $q$, we have: | We apply the Method of Truth Tables.
As can be seen by inspection, the appropriate truth values match for all boolean interpretations.
$\begin{array}{|ccccc||c|} \hline
p & \lor & (p & \land & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F & \F & \F & \T & \F \\
\T & \T & \T & \F & \F & \T \\
\T & \T & \T & \T... | For any two [[Definition:Proposition|propositions]] $p$ and $q$, we have: | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the appropriate [[Definition:Truth Value|truth values]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|ccccc||c|} \hline
p & \lor & (p & \land & q) & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F ... | Absorption Laws (Logic)/Disjunction Absorbs Conjunction/Proof by Truth Table | https://proofwiki.org/wiki/Absorption_Laws_(Logic) | https://proofwiki.org/wiki/Absorption_Laws_(Logic)/Disjunction_Absorbs_Conjunction/Proof_by_Truth_Table | [
"Conjunction",
"Disjunction"
] | [
"Definition:Proposition"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Boolean Interpretation"
] |
proofwiki-260 | Set Difference with Set Difference | :$S \setminus \paren {S \setminus T} = S \cap T = T \setminus \paren {T \setminus S}$ | From the {{axiom-link|Transitivity}}, all sets are classes.
The result then follows from Class Difference with Class Difference.
{{finish|Find the result that demonstrate the set difference of two sets is also a set, intersection as well. Probably via subset of set is set or something.}} | :$S \setminus \paren {S \setminus T} = S \cap T = T \setminus \paren {T \setminus S}$ | From the {{axiom-link|Transitivity}}, all [[Definition:Set|sets]] are [[Definition:Class (Class Theory)|classes]].
The result then follows from [[Class Difference with Class Difference]].
{{finish|Find the result that demonstrate the set difference of two sets is also a set, intersection as well. Probably via subset ... | Set Difference with Set Difference/Proof 2 | https://proofwiki.org/wiki/Set_Difference_with_Set_Difference | https://proofwiki.org/wiki/Set_Difference_with_Set_Difference/Proof_2 | [
"Set Difference with Set Difference",
"Set Intersection",
"Set Difference"
] | [] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Class Difference with Class Difference"
] |
proofwiki-261 | Set Difference is Anticommutative | Set difference is an anticommutative operation:
:$S = T \iff S \setminus T = T \setminus S = \O$ | From Set Difference with Superset is Empty Set we have:
:$S \subseteq T \iff S \setminus T = \O$
:$T \subseteq S \iff T \setminus S = \O$
The result follows from definition of set equality:
:$S = T \iff \paren {S \subseteq T} \land \paren {T \subseteq S}$
{{qed}} | [[Definition:Set Difference|Set difference]] is an [[Definition:Anticommutative|anticommutative]] operation:
:$S = T \iff S \setminus T = T \setminus S = \O$ | From [[Set Difference with Superset is Empty Set]] we have:
:$S \subseteq T \iff S \setminus T = \O$
:$T \subseteq S \iff T \setminus S = \O$
The result follows from definition of [[Definition:Set Equality/Definition 2|set equality]]:
:$S = T \iff \paren {S \subseteq T} \land \paren {T \subseteq S}$
{{qed}} | Set Difference is Anticommutative | https://proofwiki.org/wiki/Set_Difference_is_Anticommutative | https://proofwiki.org/wiki/Set_Difference_is_Anticommutative | [
"Set Difference",
"Empty Set"
] | [
"Definition:Set Difference",
"Definition:Anticommutative"
] | [
"Set Difference with Superset is Empty Set",
"Definition:Set Equality/Definition 2"
] |
proofwiki-262 | Dirichlet's Test for Uniform Convergence | Let $D$ be a set.
Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space.
Let $a_i, b_i$ be mappings from $D \to M$.
{{explain|What is $M$? Should be $V$?}}
Let the following conditions be satisfied:
:$(1): \quad$ The sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty \map {a_n} x$ be bounded on $D$
:... | Suppose $\map {b_n} x \ge \map {b_{n + 1} } x$ for each $x \in D$.
All we need to show is that:
:$\ds \sum_{n \mathop = 1}^\infty \size {\map {b_n} x - \map {b_{n + 1} } x}$
converges uniformly on $D$.
{{explain|Why is the above sufficient? On the surface of it, no reference has been made to $\map {a_n} x$. Its connect... | Let $D$ be a [[Definition:Set|set]].
Let $\struct {V, \norm {\,\cdot\,} }$ be a [[Definition:Normed Vector Space|normed vector space]].
Let $a_i, b_i$ be [[Definition:Mapping|mappings]] from $D \to M$.
{{explain|What is $M$? Should be $V$?}}
Let the following conditions be satisfied:
:$(1): \quad$ The [[Definition... | Suppose $\map {b_n} x \ge \map {b_{n + 1} } x$ for each $x \in D$.
All we need to show is that:
:$\ds \sum_{n \mathop = 1}^\infty \size {\map {b_n} x - \map {b_{n + 1} } x}$
[[Definition:Uniform Convergence|converges uniformly]] on $D$.
{{explain|Why is the above sufficient? On the surface of it, no reference has bee... | Dirichlet's Test for Uniform Convergence | https://proofwiki.org/wiki/Dirichlet's_Test_for_Uniform_Convergence | https://proofwiki.org/wiki/Dirichlet's_Test_for_Uniform_Convergence | [
"Dirichlet's Test for Uniform Convergence",
"Convergence Tests",
"Convergence",
"Direct Proofs",
"Analysis"
] | [
"Definition:Set",
"Definition:Normed Vector Space",
"Definition:Mapping",
"Definition:Series/Sequence of Partial Sums",
"Definition:Bounded Sequence",
"Definition:Monotone (Order Theory)/Sequence",
"Definition:Uniform Convergence",
"Definition:Uniform Convergence"
] | [
"Definition:Uniform Convergence",
"Definition:Cauchy Sequence/Cauchy Criterion",
"Definition:Uniform Convergence"
] |
proofwiki-263 | Relative Complement of Empty Set | The relative complement of the empty set is the set itself:
:$\relcomp S \O = S$ | {{begin-eqn}}
{{eqn | l = \relcomp S \O
| r = S \setminus \O
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = S
| c = Set Difference with Empty Set is Self
}}
{{end-eqn}}
{{qed}} | The [[Definition:Relative Complement|relative complement]] of the [[Definition:Empty Set|empty set]] is the [[Definition:Set|set]] itself:
:$\relcomp S \O = S$ | {{begin-eqn}}
{{eqn | l = \relcomp S \O
| r = S \setminus \O
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = S
| c = [[Set Difference with Empty Set is Self]]
}}
{{end-eqn}}
{{qed}} | Relative Complement of Empty Set | https://proofwiki.org/wiki/Relative_Complement_of_Empty_Set | https://proofwiki.org/wiki/Relative_Complement_of_Empty_Set | [
"Relative Complement",
"Empty Set"
] | [
"Definition:Relative Complement",
"Definition:Empty Set",
"Definition:Set"
] | [
"Set Difference with Empty Set is Self"
] |
proofwiki-264 | Relative Complement with Self is Empty Set | The relative complement of a set in itself is the empty set:
:$\relcomp S S = \O$ | {{begin-eqn}}
{{eqn | l = \relcomp S S
| r = S \setminus S
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \O
| c = Set Difference with Self is Empty Set
}}
{{end-eqn}}
{{qed}} | The [[Definition:Relative Complement|relative complement]] of a [[Definition:Set|set]] in itself is the [[Definition:Empty Set|empty set]]:
:$\relcomp S S = \O$ | {{begin-eqn}}
{{eqn | l = \relcomp S S
| r = S \setminus S
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \O
| c = [[Set Difference with Self is Empty Set]]
}}
{{end-eqn}}
{{qed}} | Relative Complement with Self is Empty Set | https://proofwiki.org/wiki/Relative_Complement_with_Self_is_Empty_Set | https://proofwiki.org/wiki/Relative_Complement_with_Self_is_Empty_Set | [
"Relative Complement",
"Empty Set"
] | [
"Definition:Relative Complement",
"Definition:Set",
"Definition:Empty Set"
] | [
"Set Difference with Self is Empty Set"
] |
proofwiki-265 | Relative Complement of Relative Complement | :$\relcomp S {\relcomp S T} = T$ | By the definition of relative complement:
:$\relcomp S {\relcomp S T} = S \setminus \paren {S \setminus T}$
Let $t \in T$.
Then by the definition of set difference:
:$t \notin S \setminus T$
Since $t \in T$ and $T \subseteq S$, by the definition of subset:
:$t \in S$
Thus:
:$t \in \paren {S \setminus \paren {S \setminu... | :$\relcomp S {\relcomp S T} = T$ | By the definition of [[Definition:Relative Complement|relative complement]]:
:$\relcomp S {\relcomp S T} = S \setminus \paren {S \setminus T}$
Let $t \in T$.
Then by the definition of [[Definition:Set Difference|set difference]]:
:$t \notin S \setminus T$
Since $t \in T$ and $T \subseteq S$, by the definition of [[D... | Relative Complement of Relative Complement/Proof 1 | https://proofwiki.org/wiki/Relative_Complement_of_Relative_Complement | https://proofwiki.org/wiki/Relative_Complement_of_Relative_Complement/Proof_1 | [
"Relative Complement of Relative Complement",
"Relative Complement"
] | [] | [
"Definition:Relative Complement",
"Definition:Set Difference",
"Definition:subset",
"Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication",
"Modus Ponendo Ponens"
] |
proofwiki-266 | Relative Complement of Relative Complement | :$\relcomp S {\relcomp S T} = T$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\relcomp S T}
| r = S \setminus \paren {S \setminus T}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = S \cap T
| c = Set Difference with Set Difference
}}
{{end-eqn}}
The definition of the relative complement requires that:
:$T \subseteq S$
But from Intersecti... | :$\relcomp S {\relcomp S T} = T$ | {{begin-eqn}}
{{eqn | l = \relcomp S {\relcomp S T}
| r = S \setminus \paren {S \setminus T}
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = S \cap T
| c = [[Set Difference with Set Difference]]
}}
{{end-eqn}}
The definition of the [[Definition:Relative Complement|relative complement]] requires t... | Relative Complement of Relative Complement/Proof 2 | https://proofwiki.org/wiki/Relative_Complement_of_Relative_Complement | https://proofwiki.org/wiki/Relative_Complement_of_Relative_Complement/Proof_2 | [
"Relative Complement of Relative Complement",
"Relative Complement"
] | [] | [
"Set Difference with Set Difference",
"Definition:Relative Complement",
"Intersection with Subset is Subset"
] |
proofwiki-267 | Intersection with Relative Complement is Empty | The intersection of a set and its relative complement is the empty set:
:$T \cap \relcomp S T = \O$ | {{begin-eqn}}
{{eqn | l = T \cap \relcomp S T
| r = \paren {S \setminus T} \cap T
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \O
| c = Set Difference Intersection with Second Set is Empty Set
}}
{{end-eqn}}
{{qed}} | The [[Definition:Set Intersection|intersection]] of a [[Definition:Set|set]] and its [[Definition:Relative Complement|relative complement]] is the [[Definition:Empty Set|empty set]]:
:$T \cap \relcomp S T = \O$ | {{begin-eqn}}
{{eqn | l = T \cap \relcomp S T
| r = \paren {S \setminus T} \cap T
| c = {{Defof|Relative Complement}}
}}
{{eqn | r = \O
| c = [[Set Difference Intersection with Second Set is Empty Set]]
}}
{{end-eqn}}
{{qed}} | Intersection with Relative Complement is Empty | https://proofwiki.org/wiki/Intersection_with_Relative_Complement_is_Empty | https://proofwiki.org/wiki/Intersection_with_Relative_Complement_is_Empty | [
"Relative Complement",
"Set Intersection"
] | [
"Definition:Set Intersection",
"Definition:Set",
"Definition:Relative Complement",
"Definition:Empty Set"
] | [
"Set Difference Intersection with Second Set is Empty Set"
] |
proofwiki-268 | Union with Relative Complement | The union of a set $T$ and its relative complement in $S$ is the set $S$:
:$\relcomp S T \cup T = S$ | === Step 1 ===
By the definition of relative complement, we have that:
:$\relcomp S T \subseteq S$
and:
:$T \subseteq S$
Hence by Union is Smallest Superset:
:$\relcomp S T \cup T\subseteq S$
{{qed|lemma}} | The [[Definition:Set Union|union]] of a [[Definition:Set|set]] $T$ and its [[Definition:Relative Complement|relative complement]] in $S$ is the set $S$:
:$\relcomp S T \cup T = S$ | === Step 1 ===
By the definition of [[Definition:Relative Complement|relative complement]], we have that:
:$\relcomp S T \subseteq S$
and:
:$T \subseteq S$
Hence by [[Union is Smallest Superset]]:
:$\relcomp S T \cup T\subseteq S$
{{qed|lemma}} | Union with Relative Complement | https://proofwiki.org/wiki/Union_with_Relative_Complement | https://proofwiki.org/wiki/Union_with_Relative_Complement | [
"Relative Complement",
"Set Union"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Relative Complement"
] | [
"Definition:Relative Complement",
"Union is Smallest Superset",
"Definition:Relative Complement"
] |
proofwiki-269 | Set Difference as Intersection with Relative Complement | Let $A, B \subseteq S$.
Then the set difference between $A$ and $B$ can be expressed as the intersection with the relative complement with respect to $S$:
:$A \setminus B = A \cap \relcomp S B$ | {{begin-eqn}}
{{eqn | l = A \setminus B
| r = \set {x: x \in A \land x \notin B}
| c = {{Defof|Set Difference}}
}}
{{eqn | r = \set {x: \paren {x \in A \land x \in X} \land x \notin B}
| c = {{Defof|Subset}}, Modus Ponens and Rule of Conjunction
}}
{{eqn | r = \set {x: x \in A \land \paren {x \in X \l... | Let $A, B \subseteq S$.
Then the [[Definition:Set Difference|set difference]] between $A$ and $B$ can be expressed as the [[Definition:Set Intersection|intersection]] with the [[Definition:Relative Complement|relative complement]] with respect to $S$:
:$A \setminus B = A \cap \relcomp S B$ | {{begin-eqn}}
{{eqn | l = A \setminus B
| r = \set {x: x \in A \land x \notin B}
| c = {{Defof|Set Difference}}
}}
{{eqn | r = \set {x: \paren {x \in A \land x \in X} \land x \notin B}
| c = {{Defof|Subset}}, [[Modus Ponens]] and [[Rule of Conjunction]]
}}
{{eqn | r = \set {x: x \in A \land \paren {x ... | Set Difference as Intersection with Relative Complement | https://proofwiki.org/wiki/Set_Difference_as_Intersection_with_Relative_Complement | https://proofwiki.org/wiki/Set_Difference_as_Intersection_with_Relative_Complement | [
"Set Difference",
"Relative Complement",
"Set Intersection"
] | [
"Definition:Set Difference",
"Definition:Set Intersection",
"Definition:Relative Complement"
] | [
"Modus Ponendo Ponens",
"Rule of Conjunction",
"Rule of Association/Conjunction"
] |
proofwiki-270 | Equivalence of Axiom Schemata for Groups | In the definition of a group, the axioms for the existence of an identity element and for closure under taking inverses can be replaced by the following two axioms:
:Given a group $G$, there exists at least one element $e \in G$ such that $e$ is a '''left identity''';
:For any element $g$ in a group $G$, there exists a... | Suppose we define a group $G$ in the usual way, but make the first pair of axiom replacements listed above:
:the existence of a left identity
:every element has a left inverse.
Let $e \in G$ be a left identity and $g \in G$.
Then, from Left Inverse for All is Right Inverse, each left inverse is also a right inverse wit... | In the definition of a [[Definition:Group|group]], the [[Axiom:Group Axioms|axioms]] for the existence of an [[Definition:Identity Element|identity element]] and for [[Definition:Closed Algebraic Structure|closure]] under taking [[Definition:Inverse Element|inverses]] can be replaced by the following two axioms:
:Give... | Suppose we define a [[Definition:Group|group]] $G$ in the usual way, but make the first pair of axiom replacements listed above:
:the existence of a [[Definition:Left Identity|left identity]]
:every element has a [[Definition:Left Inverse Element|left inverse]].
Let $e \in G$ be a [[Definition:Left Identity|left iden... | Equivalence of Axiom Schemata for Groups | https://proofwiki.org/wiki/Equivalence_of_Axiom_Schemata_for_Groups | https://proofwiki.org/wiki/Equivalence_of_Axiom_Schemata_for_Groups | [
"Group Theory"
] | [
"Definition:Group",
"Axiom:Group Axioms",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Element",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition... | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Left Inverse for All is Right Inverse",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Inverse (Ab... |
proofwiki-271 | Group has Latin Square Property | Let $\struct {G, \circ}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.
Similarly, there exists a unique $h \in G$ such that $h \circ a = b$. | Follows directly from the definition of both a Cayley table and a Latin square. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then $G$ satisfies the [[Definition:Latin Square Property|Latin square property]].
That is, for all $a, b \in G$, there exists a [[Definition:Unique|unique]] $g \in G$ such that $a \circ g = b$.
Similarly, there exists a [[Definition:Unique|unique]] $h \in G$... | Follows directly from the definition of both a [[Definition:Cayley Table|Cayley table]] and a [[Definition:Latin Square|Latin square]]. | Group has Latin Square Property/Corollary/Proof 1 | https://proofwiki.org/wiki/Group_has_Latin_Square_Property | https://proofwiki.org/wiki/Group_has_Latin_Square_Property/Corollary/Proof_1 | [
"Group has Latin Square Property",
"Latin Square Property",
"Latin Squares",
"Group Theory"
] | [
"Definition:Group",
"Definition:Latin Square Property",
"Definition:Unique",
"Definition:Unique"
] | [
"Definition:Cayley Table",
"Definition:Latin Square"
] |
proofwiki-272 | Group has Latin Square Property | Let $\struct {G, \circ}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.
Similarly, there exists a unique $h \in G$ such that $h \circ a = b$. | Let $G$ be a finite group whose order is $n$.
Let $\tuple {x_1, x_2, \ldots, x_n}$ be the elements of the underlying set of $G$ in the order they appear in the headings of the Cayley table of $G$.
Consider the row of the Cayley table headed with $a$.
The elements of that row are:
:$\tuple {a x_1, a x_2, \ldots, a x_n}$... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then $G$ satisfies the [[Definition:Latin Square Property|Latin square property]].
That is, for all $a, b \in G$, there exists a [[Definition:Unique|unique]] $g \in G$ such that $a \circ g = b$.
Similarly, there exists a [[Definition:Unique|unique]] $h \in G$... | Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Order of Structure|order]] is $n$.
Let $\tuple {x_1, x_2, \ldots, x_n}$ be the [[Definition:Element|elements]] of the [[Definition:Underlying Set of Structure|underlying set]] of $G$ in the order they appear in the headings of the [[Definition:Ca... | Group has Latin Square Property/Corollary/Proof 2 | https://proofwiki.org/wiki/Group_has_Latin_Square_Property | https://proofwiki.org/wiki/Group_has_Latin_Square_Property/Corollary/Proof_2 | [
"Group has Latin Square Property",
"Latin Square Property",
"Latin Squares",
"Group Theory"
] | [
"Definition:Group",
"Definition:Latin Square Property",
"Definition:Unique",
"Definition:Unique"
] | [
"Definition:Finite Group",
"Definition:Order of Structure",
"Definition:Element",
"Definition:Underlying Set/Abstract Algebra",
"Definition:Cayley Table",
"Definition:Array/Row",
"Definition:Cayley Table",
"Definition:Array/Element",
"Definition:Array/Row",
"Definition:Regular Representations/Left... |
proofwiki-273 | Group has Latin Square Property | Let $\struct {G, \circ}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.
Similarly, there exists a unique $h \in G$ such that $h \circ a = b$. | {{begin-eqn}}
{{eqn | l = g
| r = a^{-1} \circ b
}}
{{eqn | ll= \leadsto
| l = a \circ g
| r = a \circ \paren {a^{-1} \circ b}
}}
{{eqn | ll= \leadsto
| l = a \circ g
| r = \paren {a \circ a^{-1} } \circ b
| c = {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
| l = a \circ g
| ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then $G$ satisfies the [[Definition:Latin Square Property|Latin square property]].
That is, for all $a, b \in G$, there exists a [[Definition:Unique|unique]] $g \in G$ such that $a \circ g = b$.
Similarly, there exists a [[Definition:Unique|unique]] $h \in G$... | {{begin-eqn}}
{{eqn | l = g
| r = a^{-1} \circ b
}}
{{eqn | ll= \leadsto
| l = a \circ g
| r = a \circ \paren {a^{-1} \circ b}
}}
{{eqn | ll= \leadsto
| l = a \circ g
| r = \paren {a \circ a^{-1} } \circ b
| c = {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
| l = a \circ g
| ... | Group has Latin Square Property/Proof 1 | https://proofwiki.org/wiki/Group_has_Latin_Square_Property | https://proofwiki.org/wiki/Group_has_Latin_Square_Property/Proof_1 | [
"Group has Latin Square Property",
"Latin Square Property",
"Latin Squares",
"Group Theory"
] | [
"Definition:Group",
"Definition:Latin Square Property",
"Definition:Unique",
"Definition:Unique"
] | [] |
proofwiki-274 | Group has Latin Square Property | Let $\struct {G, \circ}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.
Similarly, there exists a unique $h \in G$ such that $h \circ a = b$. | We shall prove that this is true for the first equation:
{{begin-eqn}}
{{eqn | l = a \circ g
| r = b
}}
{{eqn | ll= \leadstoandfrom
| l = a^{-1} \circ \paren {a \circ g}
| r = a^{-1} \circ b
| c = $\circ$ is a Cancellable Binary Operation
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {a^{-1}... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then $G$ satisfies the [[Definition:Latin Square Property|Latin square property]].
That is, for all $a, b \in G$, there exists a [[Definition:Unique|unique]] $g \in G$ such that $a \circ g = b$.
Similarly, there exists a [[Definition:Unique|unique]] $h \in G$... | We shall prove that this is true for the first equation:
{{begin-eqn}}
{{eqn | l = a \circ g
| r = b
}}
{{eqn | ll= \leadstoandfrom
| l = a^{-1} \circ \paren {a \circ g}
| r = a^{-1} \circ b
| c = $\circ$ is a [[Cancellation Laws|Cancellable Binary Operation]]
}}
{{eqn | ll= \leadstoandfrom
... | Group has Latin Square Property/Proof 2 | https://proofwiki.org/wiki/Group_has_Latin_Square_Property | https://proofwiki.org/wiki/Group_has_Latin_Square_Property/Proof_2 | [
"Group has Latin Square Property",
"Latin Square Property",
"Latin Squares",
"Group Theory"
] | [
"Definition:Group",
"Definition:Latin Square Property",
"Definition:Unique",
"Definition:Unique"
] | [
"Cancellation Laws",
"Definition:Logical Equivalence"
] |
proofwiki-275 | Group has Latin Square Property | Let $\struct {G, \circ}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.
Similarly, there exists a unique $h \in G$ such that $h \circ a = b$. | Suppose that $\exists x, y \in G: a \circ x = b = a \circ y$.
{{begin-eqn}}
{{eqn | l = a \circ x
| r = a \circ y
| c =
}}
{{eqn | ll= \leadsto
| l = a^{-1} \circ \paren {a \circ x}
| r = a^{-1} \circ \paren {a \circ y}
| c = {{Group-axiom|3}}
}}
{{eqn | ll= \leadsto
| l = \paren {a... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then $G$ satisfies the [[Definition:Latin Square Property|Latin square property]].
That is, for all $a, b \in G$, there exists a [[Definition:Unique|unique]] $g \in G$ such that $a \circ g = b$.
Similarly, there exists a [[Definition:Unique|unique]] $h \in G$... | Suppose that $\exists x, y \in G: a \circ x = b = a \circ y$.
{{begin-eqn}}
{{eqn | l = a \circ x
| r = a \circ y
| c =
}}
{{eqn | ll= \leadsto
| l = a^{-1} \circ \paren {a \circ x}
| r = a^{-1} \circ \paren {a \circ y}
| c = {{Group-axiom|3}}
}}
{{eqn | ll= \leadsto
| l = \paren {... | Group has Latin Square Property/Proof 3 | https://proofwiki.org/wiki/Group_has_Latin_Square_Property | https://proofwiki.org/wiki/Group_has_Latin_Square_Property/Proof_3 | [
"Group has Latin Square Property",
"Latin Square Property",
"Latin Squares",
"Group Theory"
] | [
"Definition:Group",
"Definition:Latin Square Property",
"Definition:Unique",
"Definition:Unique"
] | [
"Definition:Unique",
"Axiom:Group Axioms",
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-276 | Group has Latin Square Property | Let $\struct {G, \circ}$ be a group.
Then $G$ satisfies the Latin square property.
That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.
Similarly, there exists a unique $h \in G$ such that $h \circ a = b$. | We shall prove that this is true for the first equation:
{{begin-eqn}}
{{eqn | l = b
| r = a \circ g
}}
{{eqn | ll= \leadsto
| l = a^{-1} \circ b
| r = a^{-1} \circ \paren {a \circ g}
| c = {{Group-axiom|3}}
}}
{{eqn | r = \paren {a^{-1} \circ a} \circ g
| c = {{Group-axiom|1}}
}}
{{eqn | ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then $G$ satisfies the [[Definition:Latin Square Property|Latin square property]].
That is, for all $a, b \in G$, there exists a [[Definition:Unique|unique]] $g \in G$ such that $a \circ g = b$.
Similarly, there exists a [[Definition:Unique|unique]] $h \in G$... | We shall prove that this is true for the first equation:
{{begin-eqn}}
{{eqn | l = b
| r = a \circ g
}}
{{eqn | ll= \leadsto
| l = a^{-1} \circ b
| r = a^{-1} \circ \paren {a \circ g}
| c = {{Group-axiom|3}}
}}
{{eqn | r = \paren {a^{-1} \circ a} \circ g
| c = {{Group-axiom|1}}
}}
{{eqn ... | Group has Latin Square Property/Proof 4 | https://proofwiki.org/wiki/Group_has_Latin_Square_Property | https://proofwiki.org/wiki/Group_has_Latin_Square_Property/Proof_4 | [
"Group has Latin Square Property",
"Latin Square Property",
"Latin Squares",
"Group Theory"
] | [
"Definition:Group",
"Definition:Latin Square Property",
"Definition:Unique",
"Definition:Unique"
] | [] |
proofwiki-277 | Symmetric Difference is Commutative | Symmetric difference is commutative:
:$S \symdif T = T \symdif S$ | {{begin-eqn}}
{{eqn | l = S \symdif T
| r = \paren {S \setminus T} \cup \paren {T \setminus S}
| c = {{Defof|Symmetric Difference}}
}}
{{eqn | r = \paren {T \setminus S} \cup \paren {S \setminus T}
| c = Union is Commutative
}}
{{eqn | r = T \symdif S
| c = {{Defof|Symmetric Difference}}
}}
{{en... | [[Definition:Symmetric Difference|Symmetric difference]] is [[Definition:Commutative Operation|commutative]]:
:$S \symdif T = T \symdif S$ | {{begin-eqn}}
{{eqn | l = S \symdif T
| r = \paren {S \setminus T} \cup \paren {T \setminus S}
| c = {{Defof|Symmetric Difference}}
}}
{{eqn | r = \paren {T \setminus S} \cup \paren {S \setminus T}
| c = [[Union is Commutative]]
}}
{{eqn | r = T \symdif S
| c = {{Defof|Symmetric Difference}}
}}
... | Symmetric Difference is Commutative | https://proofwiki.org/wiki/Symmetric_Difference_is_Commutative | https://proofwiki.org/wiki/Symmetric_Difference_is_Commutative | [
"Symmetric Difference",
"Commutative Laws of Set Theory",
"Examples of Commutative Operations"
] | [
"Definition:Symmetric Difference",
"Definition:Commutative/Operation"
] | [
"Union is Commutative"
] |
proofwiki-278 | Equivalence of Definitions of Symmetric Difference | Let $S$ and $T$ be sets.
{{TFAE|def = Symmetric Difference|view = symmetric difference $S \symdif T$ between $S$ and $T$}}
=== Definition 1 ===
{{:Definition:Symmetric Difference/Definition 1}}
=== Definition 2 ===
{{:Definition:Symmetric Difference/Definition 2}}
=== Definition 3 ===
{{:Definition:Symmetric Difference... | === $(1)$ iff $(2)$ ===
{{:Equivalence of Definitions of Symmetric Difference/(1) iff (2)}}
{{qed|lemma}} | Let $S$ and $T$ be [[Definition:Set|sets]].
{{TFAE|def = Symmetric Difference|view = symmetric difference $S \symdif T$ between $S$ and $T$}}
=== [[Definition:Symmetric Difference/Definition 1|Definition 1]] ===
{{:Definition:Symmetric Difference/Definition 1}}
=== [[Definition:Symmetric Difference/Definition 2|Defin... | === [[Equivalence of Definitions of Symmetric Difference/(1) iff (2)|$(1)$ iff $(2)$]] ===
{{:Equivalence of Definitions of Symmetric Difference/(1) iff (2)}}
{{qed|lemma}} | Equivalence of Definitions of Symmetric Difference | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Symmetric_Difference | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Symmetric_Difference | [
"Equivalence of Definitions of Symmetric Difference",
"Symmetric Difference",
"Set Difference",
"Set Intersection",
"Set Union"
] | [
"Definition:Set",
"Definition:Symmetric Difference/Definition 1",
"Definition:Symmetric Difference/Definition 2",
"Definition:Symmetric Difference/Definition 3",
"Definition:Symmetric Difference/Definition 4",
"Definition:Symmetric Difference/Definition 5"
] | [
"Equivalence of Definitions of Symmetric Difference/(1) iff (2)"
] |
proofwiki-279 | Symmetric Difference of Equal Sets | The symmetric difference of two equal sets is the empty set:
:$S = T \iff S \symdif T = \O$ | {{begin-eqn}}
{{eqn | l = S = T
| o = \leadstoandfrom
| r = S \subseteq T \land T \subseteq S
| c = {{Defof|Set Equality}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {S \setminus T = \O} \land \paren {T \setminus S = \O}
| c = Set Difference with Superset is Empty Set
}}
{{eqn | o = \lead... | The [[Definition:Symmetric Difference|symmetric difference]] of two [[Definition:Set Equality|equal sets]] is the [[Definition:Empty Set|empty set]]:
:$S = T \iff S \symdif T = \O$ | {{begin-eqn}}
{{eqn | l = S = T
| o = \leadstoandfrom
| r = S \subseteq T \land T \subseteq S
| c = {{Defof|Set Equality}}
}}
{{eqn | o = \leadstoandfrom
| r = \paren {S \setminus T = \O} \land \paren {T \setminus S = \O}
| c = [[Set Difference with Superset is Empty Set]]
}}
{{eqn | o = \... | Symmetric Difference of Equal Sets | https://proofwiki.org/wiki/Symmetric_Difference_of_Equal_Sets | https://proofwiki.org/wiki/Symmetric_Difference_of_Equal_Sets | [
"Symmetric Difference",
"Empty Set",
"Set Equality"
] | [
"Definition:Symmetric Difference",
"Definition:Set Equality",
"Definition:Empty Set"
] | [
"Set Difference with Superset is Empty Set",
"Union is Empty iff Sets are Empty"
] |
proofwiki-280 | Union is Empty iff Sets are Empty | If the union of two sets is the empty set, then both are themselves empty:
:$S \cup T = \O \iff S = \O \land T = \O$ | {{begin-eqn}}
{{eqn | r = S \cup T = \O
| o =
}}
{{eqn | ll= \leadstoandfrom
| o =
| q = \neg \exists x
| r = x \in \paren {S \cup T}
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| q = \forall x
| r = \neg \paren {x \in \paren {S \cup T} }
| c =... | If the [[Definition:Set Union|union]] of two [[Definition:Set|sets]] is the [[Definition:Empty Set|empty set]], then both are themselves [[Definition:Empty Set|empty]]:
:$S \cup T = \O \iff S = \O \land T = \O$ | {{begin-eqn}}
{{eqn | r = S \cup T = \O
| o =
}}
{{eqn | ll= \leadstoandfrom
| o =
| q = \neg \exists x
| r = x \in \paren {S \cup T}
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadstoandfrom
| o =
| q = \forall x
| r = \neg \paren {x \in \paren {S \cup T} }
| c =... | Union is Empty iff Sets are Empty/Proof 1 | https://proofwiki.org/wiki/Union_is_Empty_iff_Sets_are_Empty | https://proofwiki.org/wiki/Union_is_Empty_iff_Sets_are_Empty/Proof_1 | [
"Union is Empty iff Sets are Empty",
"Set Union",
"Empty Set"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Empty Set",
"Definition:Empty Set"
] | [
"De Morgan's Laws (Predicate Logic)",
"De Morgan's Laws (Logic)/Conjunction of Negations"
] |
proofwiki-281 | Union is Empty iff Sets are Empty | If the union of two sets is the empty set, then both are themselves empty:
:$S \cup T = \O \iff S = \O \land T = \O$ | Let $S \cup T = \O$.
We have:
{{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = S \cup T
| c = Set is Subset of Union
}}
{{eqn | ll= \leadsto
| l = S
| o = \subseteq
| r = \O
| c = {{hypothesis}}
}}
{{end-eqn}}
From Empty Set is Subset of All Sets:
:$\O \subseteq S$
So it follo... | If the [[Definition:Set Union|union]] of two [[Definition:Set|sets]] is the [[Definition:Empty Set|empty set]], then both are themselves [[Definition:Empty Set|empty]]:
:$S \cup T = \O \iff S = \O \land T = \O$ | Let $S \cup T = \O$.
We have:
{{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = S \cup T
| c = [[Set is Subset of Union]]
}}
{{eqn | ll= \leadsto
| l = S
| o = \subseteq
| r = \O
| c = {{hypothesis}}
}}
{{end-eqn}}
From [[Empty Set is Subset of All Sets]]:
:$\O \subseteq S$
... | Union is Empty iff Sets are Empty/Proof 2 | https://proofwiki.org/wiki/Union_is_Empty_iff_Sets_are_Empty | https://proofwiki.org/wiki/Union_is_Empty_iff_Sets_are_Empty/Proof_2 | [
"Union is Empty iff Sets are Empty",
"Set Union",
"Empty Set"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Empty Set",
"Definition:Empty Set"
] | [
"Set is Subset of Union",
"Empty Set is Subset of All Sets",
"Definition:Set Equality/Definition 2"
] |
proofwiki-282 | Symmetric Difference with Empty Set | :$S \symdif \O = S$
where $\symdif$ denotes the symmetric difference. | {{begin-eqn}}
{{eqn | l = S \symdif \O
| r = \paren {S \cup \O} \setminus \paren {S \cap \O}
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | r = S \setminus \paren {S \cap \O}
| c = Union with Empty Set
}}
{{eqn | r = S \setminus \O
| c = Intersection with Empty Set
}}
{{eqn | r = S
... | :$S \symdif \O = S$
where $\symdif$ denotes the [[Definition:Symmetric Difference|symmetric difference]]. | {{begin-eqn}}
{{eqn | l = S \symdif \O
| r = \paren {S \cup \O} \setminus \paren {S \cap \O}
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | r = S \setminus \paren {S \cap \O}
| c = [[Union with Empty Set]]
}}
{{eqn | r = S \setminus \O
| c = [[Intersection with Empty Set]]
}}
{{eqn | ... | Symmetric Difference with Empty Set | https://proofwiki.org/wiki/Symmetric_Difference_with_Empty_Set | https://proofwiki.org/wiki/Symmetric_Difference_with_Empty_Set | [
"Symmetric Difference",
"Empty Set"
] | [
"Definition:Symmetric Difference"
] | [
"Union with Empty Set",
"Intersection with Empty Set",
"Set Difference with Empty Set is Self"
] |
proofwiki-283 | Intersection Distributes over Symmetric Difference | Intersection is distributive over symmetric difference:
{{begin-eqn}}
{{eqn | l = \paren {R \symdif S} \cap T
| r = \paren {R \cap T} \symdif \paren {S \cap T}
}}
{{eqn | l = T \cap \paren {R \symdif S}
| r = \paren {T \cap R} \symdif \paren {T \cap S}
}}
{{end-eqn}} | From Set Intersection Distributes over Set Difference, we have:
:$\paren {R \setminus S} \cap T = \paren {R \cap T} \setminus \paren {S \cap T}$
So:
{{begin-eqn}}
{{eqn | l = \paren {R \cap T} \symdif \paren {S \cap T}
| r = \paren {\paren {R \cap T} \setminus \paren {S \cap T} } \cup \paren {\paren {S \cap T} \s... | [[Definition:Set Intersection|Intersection]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Symmetric Difference|symmetric difference]]:
{{begin-eqn}}
{{eqn | l = \paren {R \symdif S} \cap T
| r = \paren {R \cap T} \symdif \paren {S \cap T}
}}
{{eqn | l = T \cap \paren {R \symdif S}
... | From [[Set Intersection Distributes over Set Difference]], we have:
:$\paren {R \setminus S} \cap T = \paren {R \cap T} \setminus \paren {S \cap T}$
So:
{{begin-eqn}}
{{eqn | l = \paren {R \cap T} \symdif \paren {S \cap T}
| r = \paren {\paren {R \cap T} \setminus \paren {S \cap T} } \cup \paren {\paren {S \ca... | Intersection Distributes over Symmetric Difference | https://proofwiki.org/wiki/Intersection_Distributes_over_Symmetric_Difference | https://proofwiki.org/wiki/Intersection_Distributes_over_Symmetric_Difference | [
"Set Intersection",
"Symmetric Difference",
"Examples of Distributive Operations"
] | [
"Definition:Set Intersection",
"Definition:Distributive Operation",
"Definition:Symmetric Difference"
] | [
"Set Intersection Distributes over Set Difference",
"Set Intersection Distributes over Set Difference",
"Intersection Distributes over Union",
"Intersection is Commutative"
] |
proofwiki-284 | Symmetric Difference of Unions | Let $R$, $S$ and $T$ be sets.
Then:
:$\paren {R \cup T} \symdif \paren {S \cup T} = \paren {R \symdif S} \setminus T$
where:
:$\symdif$ denotes the symmetric difference
:$\setminus$ denotes set difference
:$\cup$ denotes set union | {{begin-eqn}}
{{eqn | l = \paren {R \cup T} \symdif \paren {S \cup T}
| r = \paren {\paren {R \cup T} \setminus \paren {S \cup T} } \cup \paren {\paren {S \cup T} \setminus \paren {R \cup T} }
| c = {{Defof|Symmetric Difference|index = 1}}
}}
{{eqn | r = \paren {\paren {R \setminus S} \setminus T} \cup \par... | Let $R$, $S$ and $T$ be [[Definition:Set|sets]].
Then:
:$\paren {R \cup T} \symdif \paren {S \cup T} = \paren {R \symdif S} \setminus T$
where:
:$\symdif$ denotes the [[Definition:Symmetric Difference|symmetric difference]]
:$\setminus$ denotes [[Definition:Set Difference|set difference]]
:$\cup$ denotes [[Definition:... | {{begin-eqn}}
{{eqn | l = \paren {R \cup T} \symdif \paren {S \cup T}
| r = \paren {\paren {R \cup T} \setminus \paren {S \cup T} } \cup \paren {\paren {S \cup T} \setminus \paren {R \cup T} }
| c = {{Defof|Symmetric Difference|index = 1}}
}}
{{eqn | r = \paren {\paren {R \setminus S} \setminus T} \cup \par... | Symmetric Difference of Unions | https://proofwiki.org/wiki/Symmetric_Difference_of_Unions | https://proofwiki.org/wiki/Symmetric_Difference_of_Unions | [
"Symmetric Difference",
"Set Union"
] | [
"Definition:Set",
"Definition:Symmetric Difference",
"Definition:Set Difference",
"Definition:Set Union"
] | [
"Set Difference with Union",
"Set Difference is Right Distributive over Union",
"Category:Symmetric Difference",
"Category:Set Union"
] |
proofwiki-285 | Intersection with Universal Set | The intersection of a set with the universal set is the set itself:
:$\mathbb U \cap S = S$ | {{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = \mathbb U
| c = {{Defof|Universal Set}}
}}
{{eqn | ll= \leadstoandfrom
| l = \mathbb U \cap S
| r = S
| c = Intersection with Subset is Subset
}}
{{end-eqn}}
{{qed}} | The [[Definition:Set Intersection|intersection]] of a [[Definition:Set|set]] with the [[Definition:Universal Set|universal set]] is the set itself:
:$\mathbb U \cap S = S$ | {{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = \mathbb U
| c = {{Defof|Universal Set}}
}}
{{eqn | ll= \leadstoandfrom
| l = \mathbb U \cap S
| r = S
| c = [[Intersection with Subset is Subset]]
}}
{{end-eqn}}
{{qed}} | Intersection with Universal Set | https://proofwiki.org/wiki/Intersection_with_Universal_Set | https://proofwiki.org/wiki/Intersection_with_Universal_Set | [
"Universal Set",
"Set Intersection"
] | [
"Definition:Set Intersection",
"Definition:Set",
"Definition:Universal Set"
] | [
"Intersection with Subset is Subset"
] |
proofwiki-286 | Union with Universal Set | The union of a set with the universal set is the universal set:
:$\mathbb U \cup S = \mathbb U$ | {{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = \mathbb U
| c = {{Defof|Universal Set}}
}}
{{eqn | ll= \leadstoandfrom
| l = \mathbb U \cup S
| r = \mathbb U
| c = Union with Superset is Superset
}}
{{end-eqn}}
{{qed}} | The [[Definition:Set Union|union]] of a [[Definition:Set|set]] with the [[Definition:Universal Set|universal set]] is the [[Definition:Universal Set|universal set]]:
:$\mathbb U \cup S = \mathbb U$ | {{begin-eqn}}
{{eqn | l = S
| o = \subseteq
| r = \mathbb U
| c = {{Defof|Universal Set}}
}}
{{eqn | ll= \leadstoandfrom
| l = \mathbb U \cup S
| r = \mathbb U
| c = [[Union with Superset is Superset]]
}}
{{end-eqn}}
{{qed}} | Union with Universal Set | https://proofwiki.org/wiki/Union_with_Universal_Set | https://proofwiki.org/wiki/Union_with_Universal_Set | [
"Universal Set",
"Set Union"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Universal Set",
"Definition:Universal Set"
] | [
"Union with Superset is Superset"
] |
proofwiki-287 | Complement of Empty Set is Universal Set | The complement of the empty set is the universal set:
:$\map \complement \O = \mathbb U$ | Substitute $\mathbb U$ for $S$ in $\relcomp S \O = S$ from Relative Complement of Empty Set.
{{qed}} | The [[Definition:Set Complement|complement]] of the [[Definition:Empty Set|empty set]] is the [[Definition:Universal Set|universal set]]:
:$\map \complement \O = \mathbb U$ | Substitute $\mathbb U$ for $S$ in $\relcomp S \O = S$ from [[Relative Complement of Empty Set]].
{{qed}} | Complement of Empty Set is Universal Set | https://proofwiki.org/wiki/Complement_of_Empty_Set_is_Universal_Set | https://proofwiki.org/wiki/Complement_of_Empty_Set_is_Universal_Set | [
"Set Complement",
"Empty Set",
"Universal Set"
] | [
"Definition:Set Complement",
"Definition:Empty Set",
"Definition:Universal Set"
] | [
"Relative Complement of Empty Set"
] |
proofwiki-288 | Complement of Universal Set is Empty Set | The complement of the universal set is the empty set:
:$\map \complement {\mathbb U} = \O$ | Substitute $\mathbb U$ for $S$ in $\relcomp S S = \O$ from Relative Complement with Self is Empty Set.
{{Qed}} | The [[Definition:Set Complement|complement]] of the [[Definition:Universal Set|universal set]] is the [[Definition:Empty Set|empty set]]:
:$\map \complement {\mathbb U} = \O$ | Substitute $\mathbb U$ for $S$ in $\relcomp S S = \O$ from [[Relative Complement with Self is Empty Set]].
{{Qed}} | Complement of Universal Set is Empty Set | https://proofwiki.org/wiki/Complement_of_Universal_Set_is_Empty_Set | https://proofwiki.org/wiki/Complement_of_Universal_Set_is_Empty_Set | [
"Set Complement",
"Universal Set",
"Empty Set"
] | [
"Definition:Set Complement",
"Definition:Universal Set",
"Definition:Empty Set"
] | [
"Relative Complement with Self is Empty Set"
] |
proofwiki-289 | Complement of Complement | The complement of the complement of a set is the set itself:
:$\map \complement {\map \complement S} = S$ | Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $\relcomp S {\relcomp S T} = T$ from Relative Complement of Relative Complement.
{{qed}} | The [[Definition:Set Complement|complement]] of the [[Definition:Set Complement|complement]] of a [[Definition:Set|set]] is the set itself:
:$\map \complement {\map \complement S} = S$ | Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $\relcomp S {\relcomp S T} = T$ from [[Relative Complement of Relative Complement]].
{{qed}} | Complement of Complement | https://proofwiki.org/wiki/Complement_of_Complement | https://proofwiki.org/wiki/Complement_of_Complement | [
"Set Complement"
] | [
"Definition:Set Complement",
"Definition:Set Complement",
"Definition:Set"
] | [
"Relative Complement of Relative Complement"
] |
proofwiki-290 | Intersection with Complement | The intersection of a set and its complement is the empty set:
:$S \cap \map \complement S = \O$ | Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cap \relcomp S T = \O$ from Intersection with Relative Complement is Empty.
{{qed}} | The [[Definition:Set Intersection|intersection]] of a [[Definition:Set|set]] and its [[Definition:Set Complement|complement]] is the [[Definition:Empty Set|empty set]]:
:$S \cap \map \complement S = \O$ | Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cap \relcomp S T = \O$ from [[Intersection with Relative Complement is Empty]].
{{qed}} | Intersection with Complement | https://proofwiki.org/wiki/Intersection_with_Complement | https://proofwiki.org/wiki/Intersection_with_Complement | [
"Set Complement",
"Set Intersection",
"Empty Set"
] | [
"Definition:Set Intersection",
"Definition:Set",
"Definition:Set Complement",
"Definition:Empty Set"
] | [
"Intersection with Relative Complement is Empty"
] |
proofwiki-291 | Union with Complement | The union of a set and its complement is the universal set:
:$S \cup \map \complement S = \mathbb U$ | Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cup \relcomp S T = S$ from Union with Relative Complement.
{{qed}}
{{LEM|Union with Relative Complement}} | The [[Definition:Set Union|union]] of a [[Definition:Set|set]] and its [[Definition:Set Complement|complement]] is the [[Definition:Universal Set|universal set]]:
:$S \cup \map \complement S = \mathbb U$ | Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cup \relcomp S T = S$ from [[Union with Relative Complement]].
{{qed}}
{{LEM|Union with Relative Complement}} | Union with Complement | https://proofwiki.org/wiki/Union_with_Complement | https://proofwiki.org/wiki/Union_with_Complement | [
"Set Union",
"Set Complement",
"Universal Set"
] | [
"Definition:Set Union",
"Definition:Set",
"Definition:Set Complement",
"Definition:Universal Set"
] | [
"Union with Relative Complement"
] |
proofwiki-292 | Set with Complement forms Partition | Let $\O \subset S \subset \mathbb U$.
Then $S$ and its complement $\map \complement S$ form a partition of the universal set $\mathbb U$. | Follows directly from Set with Relative Complement forms Partition:
If $\O \subset T \subset S$, then $\set {T, \relcomp S T}$ is a partition of $S$.
{{Qed}}
Category:Set Complement
Category:Universal Set
Category:Set Partitions
6sxbs5qe78y1wegwid1kahz3yitbtoj | Let $\O \subset S \subset \mathbb U$.
Then $S$ and its [[Definition:Set Complement|complement]] $\map \complement S$ form a [[Definition:Set Partition|partition]] of the [[Definition:Universal Set|universal set]] $\mathbb U$. | Follows directly from [[Set with Relative Complement forms Partition]]:
If $\O \subset T \subset S$, then $\set {T, \relcomp S T}$ is a [[Definition:Set Partition|partition]] of $S$.
{{Qed}}
[[Category:Set Complement]]
[[Category:Universal Set]]
[[Category:Set Partitions]]
6sxbs5qe78y1wegwid1kahz3yitbtoj | Set with Complement forms Partition | https://proofwiki.org/wiki/Set_with_Complement_forms_Partition | https://proofwiki.org/wiki/Set_with_Complement_forms_Partition | [
"Set Complement",
"Universal Set",
"Set Partitions"
] | [
"Definition:Set Complement",
"Definition:Set Partition",
"Definition:Universal Set"
] | [
"Set Difference and Intersection form Partition/Corollary 2",
"Definition:Set Partition",
"Category:Set Complement",
"Category:Universal Set",
"Category:Set Partitions"
] |
proofwiki-293 | Set Difference as Intersection with Complement | Set difference can be expressed as the intersection with the set complement:
:$A \setminus B = A \cap \map \complement B$ | This follows directly from Set Difference as Intersection with Relative Complement:
:$A \setminus B = A \cap \relcomp S B$
Let $S = \Bbb U$.
Since $A, B \subseteq \Bbb U$ by definition of the universal set, the result follows.
{{qed}} | [[Definition:Set Difference|Set difference]] can be expressed as the [[Definition:Set Intersection|intersection]] with the [[Definition:Set Complement|set complement]]:
:$A \setminus B = A \cap \map \complement B$ | This follows directly from [[Set Difference as Intersection with Relative Complement]]:
:$A \setminus B = A \cap \relcomp S B$
Let $S = \Bbb U$.
Since $A, B \subseteq \Bbb U$ by definition of [[Definition:Universal Set|the universal set]], the result follows.
{{qed}} | Set Difference as Intersection with Complement | https://proofwiki.org/wiki/Set_Difference_as_Intersection_with_Complement | https://proofwiki.org/wiki/Set_Difference_as_Intersection_with_Complement | [
"Set Complement",
"Set Difference",
"Set Intersection"
] | [
"Definition:Set Difference",
"Definition:Set Intersection",
"Definition:Set Complement"
] | [
"Set Difference as Intersection with Relative Complement",
"Definition:Universal Set"
] |
proofwiki-294 | Set Difference of Complements | :$\map \complement S \setminus \map \complement T = T \setminus S$ | {{begin-eqn}}
{{eqn | l = \map \complement S \setminus \map \complement T
| r = \set {x: x \in \map \complement S \land x \notin \map \complement T}
| c = {{Defof|Set Difference}}
}}
{{eqn | r = \set {x: x \notin S \land x \in T}
| c = {{Defof|Set Complement}}
}}
{{eqn | r = \set {x: x \in T \land x \... | :$\map \complement S \setminus \map \complement T = T \setminus S$ | {{begin-eqn}}
{{eqn | l = \map \complement S \setminus \map \complement T
| r = \set {x: x \in \map \complement S \land x \notin \map \complement T}
| c = {{Defof|Set Difference}}
}}
{{eqn | r = \set {x: x \notin S \land x \in T}
| c = {{Defof|Set Complement}}
}}
{{eqn | r = \set {x: x \in T \land x \... | Set Difference of Complements | https://proofwiki.org/wiki/Set_Difference_of_Complements | https://proofwiki.org/wiki/Set_Difference_of_Complements | [
"Set Difference",
"Set Complement"
] | [] | [
"Rule of Commutation",
"Category:Set Difference",
"Category:Set Complement"
] |
proofwiki-295 | Symmetric Difference of Complements | The symmetric difference of two sets equals the symmetric difference of their complements:
:$\map \complement S \symdif \map \complement T = S \symdif T$ | {{begin-eqn}}
{{eqn | l = \map \complement S \symdif \map \complement T
| r = \paren {\map \complement S \setminus \map \complement T} \cup \paren {\map \complement T \setminus \map \complement S}
| c = {{Defof|Symmetric Difference}}
}}
{{eqn | r = \paren {T \setminus S} \cup \paren {S \setminus T}
| ... | The [[Definition:Symmetric Difference|symmetric difference]] of two [[Definition:Set|sets]] equals the [[Definition:Symmetric Difference|symmetric difference]] of their [[Definition:Set Complement|complements]]:
:$\map \complement S \symdif \map \complement T = S \symdif T$ | {{begin-eqn}}
{{eqn | l = \map \complement S \symdif \map \complement T
| r = \paren {\map \complement S \setminus \map \complement T} \cup \paren {\map \complement T \setminus \map \complement S}
| c = {{Defof|Symmetric Difference}}
}}
{{eqn | r = \paren {T \setminus S} \cup \paren {S \setminus T}
| ... | Symmetric Difference of Complements | https://proofwiki.org/wiki/Symmetric_Difference_of_Complements | https://proofwiki.org/wiki/Symmetric_Difference_of_Complements | [
"Symmetric Difference",
"Set Complement"
] | [
"Definition:Symmetric Difference",
"Definition:Set",
"Definition:Symmetric Difference",
"Definition:Set Complement"
] | [
"Set Difference of Complements"
] |
proofwiki-296 | Symmetric Difference with Universal Set | :$\mathbb U \symdif S = \map \complement S$
where:
:$\mathbb U$ denotes the universal set
:$\symdif$ denotes symmetric difference. | {{begin-eqn}}
{{eqn | l = \mathbb U \symdif S
| r = \mathbb U \cup S \setminus \mathbb U \cap S
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | r = \mathbb U \cup S \setminus S
| c = Intersection with Universal Set
}}
{{eqn | r = \mathbb U \setminus S
| c = Union with Universal Set
}}
... | :$\mathbb U \symdif S = \map \complement S$
where:
:$\mathbb U$ denotes the [[Definition:Universal Set|universal set]]
:$\symdif$ denotes [[Definition:Symmetric Difference|symmetric difference]]. | {{begin-eqn}}
{{eqn | l = \mathbb U \symdif S
| r = \mathbb U \cup S \setminus \mathbb U \cap S
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | r = \mathbb U \cup S \setminus S
| c = [[Intersection with Universal Set]]
}}
{{eqn | r = \mathbb U \setminus S
| c = [[Union with Universal S... | Symmetric Difference with Universal Set | https://proofwiki.org/wiki/Symmetric_Difference_with_Universal_Set | https://proofwiki.org/wiki/Symmetric_Difference_with_Universal_Set | [
"Symmetric Difference",
"Universal Set",
"Set Complement"
] | [
"Definition:Universal Set",
"Definition:Symmetric Difference"
] | [
"Intersection with Universal Set",
"Union with Universal Set"
] |
proofwiki-297 | Symmetric Difference with Self is Empty Set | The symmetric difference of a set with itself is the empty set:
:$S \symdif S = \O$ | This follows directly from Symmetric Difference of Equal Sets:
:$S \symdif T = \O \iff S = T$
substituting $S$ for $T$.
{{Qed}} | The [[Definition:Symmetric Difference|symmetric difference]] of a [[Definition:Set|set]] with itself is the [[Definition:Empty Set|empty set]]:
:$S \symdif S = \O$ | This follows directly from [[Symmetric Difference of Equal Sets]]:
:$S \symdif T = \O \iff S = T$
substituting $S$ for $T$.
{{Qed}} | Symmetric Difference with Self is Empty Set | https://proofwiki.org/wiki/Symmetric_Difference_with_Self_is_Empty_Set | https://proofwiki.org/wiki/Symmetric_Difference_with_Self_is_Empty_Set | [
"Symmetric Difference",
"Empty Set"
] | [
"Definition:Symmetric Difference",
"Definition:Set",
"Definition:Empty Set"
] | [
"Symmetric Difference of Equal Sets"
] |
proofwiki-298 | Symmetric Difference with Complement | The symmetric difference of a set with its complement is the universal set:
:$S \symdif \relcomp {} S = \mathbb U$ | {{begin-eqn}}
{{eqn | l = S \symdif \relcomp {} S
| r = \paren {S \cup \relcomp {} S} \setminus \paren {S \cap \relcomp {} S}
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | r = \paren {S \cup \relcomp {} S} \setminus \O
| c = Intersection with Complement
}}
{{eqn | r = \mathbb U \setminus \... | The [[Definition:Symmetric Difference|symmetric difference]] of a [[Definition:Set|set]] with its [[Definition:Set Complement|complement]] is the [[Definition:Universal Set|universal set]]:
:$S \symdif \relcomp {} S = \mathbb U$ | {{begin-eqn}}
{{eqn | l = S \symdif \relcomp {} S
| r = \paren {S \cup \relcomp {} S} \setminus \paren {S \cap \relcomp {} S}
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | r = \paren {S \cup \relcomp {} S} \setminus \O
| c = [[Intersection with Complement]]
}}
{{eqn | r = \mathbb U \setmin... | Symmetric Difference with Complement | https://proofwiki.org/wiki/Symmetric_Difference_with_Complement | https://proofwiki.org/wiki/Symmetric_Difference_with_Complement | [
"Symmetric Difference",
"Set Complement",
"Universal Set"
] | [
"Definition:Symmetric Difference",
"Definition:Set",
"Definition:Set Complement",
"Definition:Universal Set"
] | [
"Intersection with Complement",
"Union with Complement",
"Set Difference with Empty Set is Self"
] |
proofwiki-299 | Symmetric Difference is Associative | Symmetric difference is associative:
:$R \symdif \paren {S \symdif T} = \paren {R \symdif S} \symdif T$ | We can directly expand the expressions for $R \symdif \paren {S \symdif T}$ and $\paren {R \symdif S} \symdif T$, and see that they come to the same thing.
Expanding the {{RHS}}:
{{begin-eqn}}
{{eqn | o =
| r = \paren {R \symdif S} \symdif T
| c =
}}
{{eqn | r = \paren {\paren {\paren {R \cap \overline S}... | [[Definition:Symmetric Difference|Symmetric difference]] is [[Definition:Associative Operation|associative]]:
:$R \symdif \paren {S \symdif T} = \paren {R \symdif S} \symdif T$ | We can directly expand the expressions for $R \symdif \paren {S \symdif T}$ and $\paren {R \symdif S} \symdif T$, and see that they come to the same thing.
Expanding the {{RHS}}:
{{begin-eqn}}
{{eqn | o =
| r = \paren {R \symdif S} \symdif T
| c =
}}
{{eqn | r = \paren {\paren {\paren {R \cap \overline ... | Symmetric Difference is Associative/Proof 1 | https://proofwiki.org/wiki/Symmetric_Difference_is_Associative | https://proofwiki.org/wiki/Symmetric_Difference_is_Associative/Proof_1 | [
"Symmetric Difference is Associative",
"Symmetric Difference",
"Associative Laws of Set Theory",
"Examples of Associative Operations"
] | [
"Definition:Symmetric Difference",
"Definition:Associative Operation"
] | [
"Intersection Distributes over Union",
"De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection",
"De Morgan's Laws (Set Theory)/Set Complement/Complement of Union",
"Intersection Distributes over Union",
"Intersection Distributes over Union",
"De Morgan's Laws (Set Theory)/Set Complement/... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.