hcju/proofwikiqa huggingface dataset:
First we show that $A \subseteq B \land S \subseteq T \implies A \times S \subseteq B \times T$. First, let $A = \O$ or $S = \O$. Then from Cartesian Product is Empty iff Factor is Empty: :$A \times S = \O \subseteq B \times T$ so the result holds. Next, let $A, S \ne \O$. Then from Cartesian Product is Empty iff Factor is Empty: :$A \times S \ne \O$ and we can use the following argument: {{begin-eqn}} {{eqn | o =       | r = \tuple {x, y} \in A \times S }} {{eqn | o = \leadsto       | r = x \in A, y \in S       | c = {{Defof|Cartesian Product}} }} {{eqn | o = \leadsto       | r = x \in B, y \in T       | c = {{Defof|Subset}} }} {{eqn | o = \leadsto       | r = \tuple {x, y} \in B \times T       | c = {{Defof|Cartesian Product}} }} {{end-eqn}} Thus $A \times S \subseteq B \times T$ as we were to prove. Now we show that if $A, S \ne \O$, then: :$A \times S \subseteq B \times T \implies A \subseteq B \land S \subseteq T$ So suppose that $A \times S \subseteq B \times T$. First note that if $A = \O$, then $A \times S = \O \subseteq B \times T$, whatever $S$ is, so it is not necessarily the case that $S \subseteq T$. Similarly if $S = \O$; it is not necessarily the case that $A \subseteq B$. So that explains the restriction $A, S \ne \O$. Now, as $A, S \ne \O$, $\exists x \in A, y \in S$. Thus: {{begin-eqn}} {{eqn | o =       | r = x \in A, y \in S }} {{eqn | o = \leadsto       | r = \tuple {x, y} \in A \times S       | c = {{Defof|Cartesian Product}} }} {{eqn | o = \leadsto       | r = \tuple {x, y} \in B \times T       | c = {{Defof|Subset}} }} {{eqn | o = \leadsto       | r = x \in B, y \in T       | c = {{Defof|Cartesian Product}} }} {{end-eqn}} So when $A, S \ne \O$, we have: :$A \subseteq S \land B \subseteq T \implies A \times S \subseteq B \times T$ :$A \times S \subseteq B \times T \implies A \subseteq B \land S \subseteq T$ from which: :$A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$ {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \tuple {x, y} \in \paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S_1 \land x \in S_2} \land \paren {y \in T_1 \land y \in T_2}       | c = {{Defof|Cartesian Product}} and {{Defof|Set Intersection}} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S_1 \land y \in T_1} \land \paren {x \in S_2 \land y \in T_2}       | c = Rule of Commutation, Rule of Association }} {{eqn | o = \leadstoandfrom       | r = \tuple {x, y} \in S_1 \times T_1 \land \tuple {x, y} \in S_2 \times T_2       | c = {{Defof|Cartesian Product}} }} {{eqn | o = \leadstoandfrom       | r = \tuple {x, y} \in \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2}       | c = {{Defof|Set Intersection}} }} {{end-eqn}} {{qed}}
Take the result Cartesian Product of Unions: :$\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2} = \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \cup \paren {S_1 \times T_2} \cup \paren {S_2 \times T_1}$ Put $S_1 = S_2 = A, T_1 = B, T_2 = C$: {{begin-eqn}} {{eqn | o =        | r = A \times \paren {B \cup C}       | c =  }} {{eqn | r = \paren {A \cup A} \times \paren {B \cup C}       | c = Union is Idempotent }} {{eqn | r = \paren {A \times B} \cup \paren {A \times C} \cup \paren {A \times C} \cup \paren {A \times B}       | c = Cartesian Product of Unions }} {{eqn | r = \paren {A \times B} \cup \paren {A \times C}       | c = Union is Idempotent }} {{end-eqn}} Thus: :$A \times \paren {B \cup C} = \paren {A \times B} \cup \paren {A \times C}$ The other result is proved similarly. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \tuple {x, y} \in \paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2} }} {{eqn | ll= \leadstoandfrom       | o =        | r = \paren {x \in S_1 \lor x \in S_2} }} {{eqn | o = \land       | r = \paren {y \in T_1 \lor y \in T_2}       | c = {{Defof|Cartesian Product}} and {{Defof|Set Union}} }} {{eqn | ll= \leadstoandfrom       | o =        | r = \paren {\paren {x \in S_1 \lor x \in S_2} \land y \in T_1} }} {{eqn | o = \lor       | r = \paren {\paren {x \in S_1 \lor x \in S_2} \land y \in T_2}       | c = Rule of Distribution }} {{eqn | ll= \leadstoandfrom       | o =        | r = \paren {x \in S_1 \land y \in T_1} }} {{eqn | o = \lor       | r = \paren {x \in S_2 \land y \in T_1} }} {{eqn | o = \lor       | r = \paren {x \in S_1 \land y \in T_2} }} {{eqn | o = \lor       | r = \paren {x \in S_2 \land y \in T_2}       | c = Rule of Distribution }} {{eqn | ll= \leadstoandfrom       | o =        | r = \tuple {x, y} \in \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \cup \paren {S_1 \times T_2} \cup \paren {S_2 \times T_1}       | c = {{Defof|Cartesian Product}} and {{Defof|Set Union}} }} {{end-eqn}} {{qed}}
Let $\tuple {x, y} \in \paren {S_1 \times S_2} \setminus \paren {T_1 \times T_2}$. Then: {{begin-eqn}} {{eqn | l = \tuple {x, y} \in S_1 \times S_2       | o = \land       | r = \tuple {x, y} \notin T_1 \times T_2       | c = {{Defof|Set Difference}} }} {{eqn | ll= \leadstoandfrom       | l = x \in S_1 \land y \in S_2       | o = \land       | r = \neg \paren {x \in T_1 \land y \in T_2}       | c = {{Defof|Cartesian Product}} }} {{eqn | ll= \leadstoandfrom       | l = x \in S_1 \land y \in S_2       | o = \land       | r = \paren {x \notin T_1 \lor y \notin T_2}       | c = De Morgan's Laws: Disjunction of Negations }} {{eqn | ll= \leadstoandfrom       | l = \paren {x \in S_1 \land y \in S_2 \land x \notin T_1}       | o = \lor       | r = \paren {x \in S_1 \land y \in S_2 \land y \notin T_2}       | c = Rule of Distribution }} {{eqn | ll= \leadstoandfrom       | l = \paren {x \in S_1 \setminus T_1 \land y \in S_2}       | o = \lor       | r = \paren {x \in S_1 \land y \in S_2 \setminus T_2}       | c = {{Defof|Set Difference}} }} {{eqn | ll= \leadstoandfrom       | l = \paren {\tuple {x, y} \in \paren {S_1 \setminus T_1} \times S_2}       | o = \lor       | r = \paren {\tuple {x, y} \in S_1 \times \paren {S_2 \setminus T_2} }       | c = {{Defof|Cartesian Product}} }} {{eqn | ll= \leadstoandfrom       | l = \tuple {x, y}       | o = \in       | r = \paren {S_1 \times \paren {S_2 \setminus T_2} } \cup \paren {\paren {S_1 \setminus T_1} \times S_2}       | c = {{Defof|Set Union}} }} {{end-eqn}} The result follows from the definition of subset and set equality. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexive === {{:Diagonal Relation is Reflexive}} {{qed|lemma}} === Symmetric === {{:Diagonal Relation is Symmetric}} {{qed|lemma}} === Transitive === {{:Diagonal Relation is Transitive}} {{qed}}
Let us verify the conditions for an equivalence in turn. === Reflexivity === For $\RR$ to be reflexive means: :$\forall x \in S: \tuple {x, x} \in S \times S$ which is trivial by definition of the Cartesian product $S \times S$. {{qed|lemma}} === Symmetry === For $\RR$ to be symmetric means: :$\forall x, y \in S: \tuple {x, y} \in S \times S \land \tuple {y, x} \in S \times S$ Since we have by definition of Cartesian product that: :$\forall x, y \in S: \tuple {y, x} \in S \times S$ this follows by True Statement is implied by Every Statement. {{qed|lemma}} === Transitivity === For $\RR$ to be transitive means: :$\tuple {x, y} \in S \times S \land \tuple {y, z} \in S \times S \implies \tuple {x, z} \in S \times S$ By definition of Cartesian product, we have that: :$\forall x, z \in S: \tuple {x, z} \in S \times S$ hence by True Statement is implied by Every Statement, it follows that $\RR$ is transitive. {{qed|lemma}} Having verified all three conditions, we conclude $\RR$ is an equivalence. {{qed}}
This follows from set equality and Equality of Ordered Pairs. A '''relation''' on $S \times T$ is an ordered triple, thus: {{begin-eqn}} {{eqn | l = \RR_1       | r = \tuple {S_1, T_1, R_1}       | c = where $R_1 \subseteq S_1 \times T_1$ }} {{eqn | l = \RR_2       | r = \tuple {S_2, T_2, R_2}       | c = where $R_2 \subseteq S_2 \times T_2$ }} {{end-eqn}} From Equality of Ordered Tuples, $\RR_1 = \RR_1$ {{iff}}: {{begin-eqn}} {{eqn | l = S_1       | r = S_2 }} {{eqn | l = T_1       | r = T_2 }} {{eqn | l = R_1       | r = R_2 }} {{end-eqn}} Then we have that $R_1$ and $R_2$ are sets of ordered pairs. Thus by definition of set equality: :$R_1 = R_2 \iff \paren {\forall \tuple {x, y} \in \S_1 \times T_2: \tuple {x, y} \in R_1 \iff \tuple {x, y} \in R_2}$ Hence the result. {{Qed}}
We have the definitions: {{begin-eqn}} {{eqn | l = \map \RR s       | r = \set {t \in T: \tuple {s, t} \in \RR}       | c = {{Defof|Image of Element under Relation}} }} {{eqn | l = \RR \sqbrk {\set s}       | r = \set {t \in T: \exists s \in \set s: \tuple {s, t} \in \RR}       | c = {{Defof|Image of Subset under Relation}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = t         | o = \in       | r = \map \RR s }} {{eqn | ll= \leadsto       | l = \tuple {s, t}       | o = \in       | r = \RR }} {{eqn | ll= \leadsto       | lo= \exists s \in \set s:       | l = \tuple {s, t}       | o = \in       | r = \RR       | c = {{Defof|Singleton}} }} {{eqn | ll= \leadsto       | l = t       | o = \in       | r = \RR \sqbrk {\set s} }} {{eqn | ll= \leadsto       | l = \map \RR s       | o = \subseteq       | r = \RR \sqbrk {\set s}       | c = {{Defof|Subset}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = t       | o = \in       | r = \RR \sqbrk {\set s} }} {{eqn | ll= \leadsto       | lo= \exists s \in \set s:       | l = \tuple {s, t}       | o = \in       | r = \RR }} {{eqn | ll= \leadsto       | l = \tuple {s, t}       | o = \in       | r = \RR       | c = as $r \in \set s \implies r = s$ by Singleton Equality }} {{eqn | ll= \leadsto       | l = t       | o = \in       | r = \map \RR s }} {{eqn | ll= \leadsto       | l = \RR \sqbrk {\set s}       | o = \subseteq       | r = \map \RR s }} {{end-eqn}} Finally: :$\map \RR s = \RR \sqbrk {\set s}$ by definition of set equality. {{Qed}} Category:Relation Theory Category:Singletons 8ype9fcu1jses6a8x537szz5bvbbcnt
Suppose $\RR \sqbrk A \not \subseteq \RR \sqbrk B$. {{begin-eqn}} {{eqn | l = \RR \sqbrk A       | o = \nsubseteq       | r = \RR \sqbrk B }} {{eqn | ll= \leadsto       | l = \exists t \in \RR \sqbrk A: \exists \tuple {s, t} \in \RR: s       | o = \notin       | r = B       | c = {{Defof|Image of Subset under Relation}} }} {{eqn | ll= \leadsto       | l = \exists s \notin B: s       | o = \in       | r = A       | c = {{Defof|Ordered Pair}} }} {{eqn | ll= \leadsto       | l = A       | o = \nsubseteq       | r = B       | c = {{Defof|Subset}} }} {{end-eqn}} The result follows by the Rule of Transposition. {{qed}}
From Image of Singleton under Relation: :$\map \RR s = \RR \sqbrk {\set s}$ From Singleton of Element is Subset: :$s \in A \implies \set s \subseteq A$ The result follows from Image of Subset is Subset of Image. {{qed}} Category:Relation Theory 81ns2wlo73dzga5xv99qsw98k988h3w
{{begin-eqn}} {{eqn | l = A       | o = \subseteq       | r = \Dom \RR       | c =  }} {{eqn | ll= \leadsto       | l = \RR \sqbrk A       | o = \subseteq       | r = \Img \RR       | c = Image of Subset is Subset of Image }} {{eqn | o = \subseteq       | r = T       | c =  }} {{end-eqn}} {{qed}} Category:Relation Theory hy4x25yvdjd9rj1cqotvh3g8hhde5km
Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$. The domain of $\RR_1$ is $S_1$. The composite of $\RR_1$ and $\RR_2$ is defined as: :$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$ From this definition: :$\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$. Thus the domain of $\RR_2 \circ \RR_1$ is $S_1$. Thus: :$\Dom {\RR_2 \circ \RR_1} = S_1 = \Dom {\RR_1}$ {{qed}} Category:Composite Relations 174sj1pl3roals6ya0vhrygzissq3nb
Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$. The codomain of $\RR_2$ is $S_3$. The composite of $\RR_1$ and $\RR_2$ is defined as: :$\RR_2 \circ \RR_1 = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$ From this definition: :$\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$ Thus the codomain of $\RR_2 \circ \RR_1$ is $S_3$. Thus: :$\Cdm {\RR_2 \circ \RR_1} = S_3 = \Cdm {\RR_2}$ {{qed}} Category:Composite Relations 1ud451rt4knvb7jiqbis137ljeuaqx9
Let $f \subseteq S \times T$ be a mapping. Then: {{begin-eqn}} {{eqn | o =        | r = \forall x \in S: \exists y \in T: \tuple {x, y} \in f       | c = {{Defof|Mapping}} }} {{eqn | o = \implies       | r = \forall x \in S: x \in \Preimg f       | c = {{Defof|Preimage of Mapping}} }} {{eqn | o = \implies       | r = S \subseteq \Preimg f       | c = {{Defof|Subset}} }} {{end-eqn}} From Preimage of Relation is Subset of Domain, we have that $\Preimg f \subseteq S$. The result follows from the definition of set equality. {{qed}} Category:Preimages under Mappings 1u6shpayoujr0isjcfwbj0lesidhakp
The preimage of $\mathcal R$ is defined as: :$\Preimg {\mathcal R} = \set {s \in \Dom {\mathcal R}: \exists t \in \Rng {\mathcal R}: \tuple {s, t} \in \mathcal R}$ Hence: {{begin-eqn}} {{eqn | l = s       | o = \in       | r = \Preimg {\mathcal R}       | c =  }} {{eqn | ll= \leadsto       | l = s       | o = \in       | r = \Dom {\mathcal R}       | c = {{Defof|Preimage of Relation}} }} {{eqn | ll= \leadsto       | l = \Preimg {\mathcal R}       | o = \subseteq       | r = \Dom {\mathcal R}       | c = Subset of Set with Propositional Function }} {{end-eqn}} {{qed}} Category:Relation Theory k59tjbklrx0yccrj8xlj3fexm3y3uju
Suppose $A \subseteq S$. We have: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = A }} {{eqn | ll= \leadsto       | l = \set x       | o = \subseteq       | r = A       | c = Singleton of Element is Subset }} {{eqn | ll= \leadsto       | l = \RR \sqbrk {\set x}       | o = \subseteq       | r = \RR \sqbrk A       | c = Image of Subset under Relation is Subset of Image }} {{eqn | ll= \leadsto       | l = \RR^{-1} \sqbrk {\RR \sqbrk {\set x} }       | o = \subseteq       | r = \RR^{-1} \sqbrk {\RR \sqbrk A}       | c = Image of Subset under Relation is Subset of Image: Corollary 1 }} {{end-eqn}} Then we have that $\RR$ is a left-total relation. Thus: :$\exists t \in T: \tuple {x, t} \in \RR$ Hence by definition of inverse relation: :$\tuple {t, x} \in \RR^{-1}$ Hence: {{begin-eqn}} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \RR^{-1} \sqbrk {\RR \sqbrk A}       | c = {{Defof|Relation}} }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \paren {\RR^{-1} \circ \RR} \sqbrk A       | c = {{Defof|Composition of Relations}} }} {{end-eqn}} So by definition of subset: :$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$ {{qed}}
Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$ be relations. We assume that: :$\Dom {\RR_2} = \Cdm {\RR_1}$ where $\Dom \RR$ denotes domain and $\Cdm \RR$ denotes codomain of a relation $\RR$. This is necessary for $\RR_2 \circ \RR_1$ to exist. From the definition of an inverse relation, we have: :$\Dom {\RR_2} = \Cdm {\RR_2^{-1} }$ :$\Cdm {\RR_1} = \Dom {\RR_1^{-1} }$ So we confirm that $\RR_1^{-1} \circ \RR_2^{-1}$ is defined. {{begin-eqn}} {{eqn | l = \RR_2 \circ \RR_1       | r = \set {\tuple {x, z}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}       | c = {{Defof|Composition of Relations}} }} {{eqn | ll= \leadsto       | l = \paren {\RR_2 \circ \RR_1}^{-1}       | r = \set {\tuple {z, x}: \tuple {x, z} \in \RR_2 \circ \RR_1}       | c = {{Defof|Inverse Relation}} }} {{eqn | r = \set {\tuple {z, x}: x \in S_1, z \in S_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}       | c = {{Defof|Composition of Relations}} }} {{eqn | r = \set {\tuple {z, x}: z \in S_3, x \in S_1: \exists y \in S_2: \tuple {z, y} \in \RR_2^{-1} \land \tuple {y, x} \in \RR_1^{-1} }       | c = {{Defof|Inverse Relation}} }} {{eqn | r = \RR_1^{-1} \circ \RR_2^{-1}       | c = {{Defof|Composition of Relations}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = S_1 \cap S_2       | o = \subseteq       | r = S_1       | c = Intersection is Subset }} {{eqn | ll= \leadsto       | l = \RR \sqbrk {S_1 \cap S_2}       | o = \subseteq       | r = \RR \sqbrk {S_1}       | c = Image of Subset is Subset of Image }} {{end-eqn}} {{begin-eqn}} {{eqn | l = S_1 \cap S_2       | o = \subseteq       | r = S_2       | c = Intersection is Subset }} {{eqn | ll= \leadsto       | l = \RR \sqbrk {S_1 \cap S_2}       | o = \subseteq       | r = \RR \sqbrk {S_2}       | c = Image of Subset is Subset of Image }} {{end-eqn}} {{begin-eqn}} {{eqn | ll= \leadsto       | l = \RR \sqbrk {S_1 \cap S_2}       | o = \subseteq       | r = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}       | c = Intersection is Largest Subset }} {{end-eqn}} {{qed}}
We have that $\RR^{-1}$ is a relation The result follows from Image of Union under Relation. {{qed}}
This follows from Image of Intersection under Relation, and the fact that $\mathcal R^{-1}$ is itself a relation, and therefore obeys the same rules. {{qed}} Category:Relation Theory Category:Set Intersection bvvhm3d3k4ssf054yk53qc4wbzl6epq
We have that $\mathcal R^{-1}$ is itself a relation The result then follows from Image of Set Difference under Relation. {{qed}} Category:Preimages under Relations Category:Set Difference 09fl9rygc9mkeiqhl5fucr63f9pru6w
From the definition of restriction: :$\forall x \in X: \mathcal R \restriction_X \left({x}\right) = \mathcal R \left({x}\right)$ Thus: :$\forall x \in X: \exists t \in T: \left({x, t}\right) \in \mathcal R \restriction_X$ But $\left({x, t}\right)$ is also (by definition) in $\mathcal R$. It follows that: :$\mathcal R \restriction_X \subseteq \mathcal R$ {{qed}} Category:Restrictions kb9s90y6df7l76lo3j7hpdxrnt7sg0e
From the definition of null relation: :$\RR = \O$ === Antireflexivity === This follows directly from the definition: :$\RR = \O \implies \forall x \in S: \tuple {x, x} \notin \RR$ and so $\RR$ is antireflexive. {{qed|lemma}} === Symmetry === It follows vacuously that: :$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR$ and so $\RR$ is symmetric. {{qed|lemma}} === Transitivity === It follows vacuously that: :$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$ and so $\RR$ is transitive. {{qed}}
Let $\mathcal R$ be symmetric. {{begin-eqn}} {{eqn | l=\left({x, y}\right) \in \mathcal R       | o=\implies       | r=\left({y, x}\right) \in \mathcal R       | c=Definition of Symmetric }} {{eqn | l=\left({y, x}\right) \in \mathcal R       | o=\implies       | r=\left({x, y}\right) \in \mathcal R       | c=Definition of Symmetric }} {{eqn | ll=\implies       | l=(\left({x, y}\right) \in \mathcal R       | o=\iff       | r=\left({y, x}\right) \in \mathcal R)       | c=Definition of Biconditional }} {{end-eqn}} {{qed}} Category:Symmetric Relations 1ivm5w9dfywugeqhvcxevx91ds9sauu
=== Necessary Condition === Let $\RR$ be both symmetric and antisymmetric Suppose $\RR \not \subseteq \Delta_S$. Then: :$\exists \tuple {x, y} \in \RR: x \ne y$ But then as $\RR$ is symmetric: :$\tuple {y, x} \in \RR$ So we have: :$\tuple {x, y} \in \RR$ and: :$\tuple {y, x} \in \RR$ where $x \ne y$. Thus $\RR$ is not antisymmetric, contrary to hypothesis. So our assumption that $\RR \not \subseteq \Delta_S$ is false. That is: :$\RR \subseteq \Delta_S$ The result follows by definition of coreflexive. {{qed|lemma}} === Sufficient Condition === Let $\RR$ be coreflexive. Then: {{begin-eqn}} {{eqn | l = \tuple {x, y}       | o = \in       | r = \RR }} {{eqn | ll= \leadsto       | l = x       | r = y       | c = Definition of Coreflexive Relation }} {{eqn | ll= \leadsto       | l = \tuple {x, y}       | r = \tuple {y, x}       | c = Equality of Ordered Pairs }} {{eqn | ll= \leadsto       | l = \tuple {y, x}       | o = \in       | r = \RR }} {{end-eqn}} and so by definition $\RR$ is symmetric. Let $\tuple {x, y} \in \RR$ and $\tuple {y, x} \in \RR$. From the above, this can happen only when $x = y$. That is, by definition, $\RR$ is antisymmetric. So $\RR$ is both symmetric and antisymmetric. Hence the result. {{qed}}
Let $\RR \ne \O$. Then: :$\exists \tuple {x, y} \in \RR$ Now, either $\tuple {y, x} \in \RR$ or $\tuple {y, x} \notin \RR$. If $\tuple {y, x} \in \RR$, then $\RR$ is not asymmetric. If $\tuple {y, x} \notin \RR$, then $\RR$ is not symmetric. Therefore, if $\RR \ne \O$, $\RR$ can not be both symmetric and asymmetric. Let $\RR = \O$. Then: :$\forall x, y \in S: \tuple {x, y} \notin \RR \land \tuple {y, x} \notin \RR$ From Biconditional as Disjunction of Conjunctions: :$\tuple {p \land q} \lor \tuple {\neg p \land \neg q} \vdash p \iff q$ Thus: :$\forall x, y \in S: \tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$ So the condition for symmetry is fulfilled, and $\RR = \O$ is symmetric. As $\RR = \O$: :$\forall x, y \in S: \tuple {x, y} \notin \RR$ From False Statement implies Every Statement $\neg p \vdash p \implies q$, we can deduce: :$\forall x, y \in S: \tuple {x, y} \notin \RR$ therefore: :$\forall x, y \in S: \tuple {x, y} \in \RR \implies \forall x, y \in S: \tuple {y, x} \notin \RR$ This is the definition of asymmetric. Thus the only relation $\RR$ to be both symmetric and asymmetric is $\RR = \O$. {{qed}}
=== 1 iff 2 === See Inversion Mapping is Automorphism iff Group is Abelian. {{qed|lemma}} {{explain|There's a better link that proves this one, but IIRC it's difficult to find. I've been tidying up this recently, I'll see if I can find it in a bit.}} === 1 iff 3 === See Group Abelian iff Cross Cancellation Property. {{qed|lemma}} === 1 iff 4 === See Group Abelian iff Middle Cancellation Property. {{qed|lemma}} Hence all four statements are logically equivalent. {{qed}} Category:Abelian Groups lagjvdvlr6t1nx0th4h4fsxoubxkp92
From Relation both Symmetric and Asymmetric is Null, the empty set ''is'' both symmetric and asymmetric. This is why the assumption that $\mathcal R \ne \varnothing$. Let $\mathcal R$ be symmetric. Then from the definition of Asymmetric, $\mathcal R$ is not asymmetric. Also, from the definition of Non-Symmetric, it is not non-symmetric. Let $\mathcal R$ be asymmetric. Then from the definition of Symmetric, $\mathcal R$ is not symmetric. Also, from the definition of Non-Symmetric, it is not non-symmetric. If $\mathcal R$ is not symmetric, and $\mathcal R$ is not asymmetric, then from the definition of Non-Symmetric, it is non-symmetric. {{Qed}}
Let $\RR$ be asymmetric. Then from the definition of asymmetric: :$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$ Thus: :$\neg \exists \tuple {x, y} \in \RR: \tuple {y, x} \in \RR$ Thus: :$\set {\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR} = \O$ Thus: :$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$ is vacuously true. {{qed}}
Let $\RR$ be asymmetric. Then, by definition: :$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$ {{AimForCont}} $\tuple {x, x} \in \RR$. Then: {{begin-eqn}} {{eqn | l = \tuple {x, x} \in \RR       | o = \implies       | r = \tuple {x, x} \notin \RR       | c = {{Defof|Asymmetric Relation}} }} {{eqn | ll= \leadsto       | l = \tuple {x, x} \notin \RR       | o =       | c = Proof by Contradiction }} {{end-eqn}} Thus $\RR$ is antireflexive. {{qed}}
Let $\mathcal R \subseteq S \times S$ be antireflexive and transitive. That is: {{begin-eqn}} {{eqn | l=\forall x \in S:       | o=       | r=\left({x, x}\right) \notin \mathcal R       | c=Definition of Antireflexive }} {{eqn | l=\left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R       | o=\implies       | r=\left({x, z}\right) \in \mathcal R       | c=Definition of Transitive }} {{end-eqn}} We have that $\mathcal R$ is not null. Suppose $\mathcal R$ is not asymmetric. So, by definition, $\exists \left({x, y}\right) \in \mathcal R: \left({y, x}\right) \in \mathcal R$. Then from the transitivity of $\mathcal R$ that would mean $\left({x, x}\right) \in \mathcal R$. But that would contradict the antireflexivity of $\mathcal R$. Therefore $\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \notin \mathcal R$ and $\mathcal R$ has been shown to be asymmetric. {{qed}}