hcju/proofwikiqa huggingface dataset:
Suppose $\mathcal R \subseteq S \times S$ is not antireflexive. Then $\exists x \in S: \left({x, x}\right) \in \mathcal R$. (In the case of $\mathcal R$ being reflexive, the property holds for ''all'' $x \in S$.) Thus $\mathcal R$ is not antitransitive: {{begin-eqn}} {{eqn | o =       | r = \left({x, x}\right) \in \mathcal R \land \left({x, x}\right) \in \mathcal R       | c = }} {{eqn | o = \implies       | r = \left({x, x}\right) \in \mathcal R       | c = Rule of Idempotence }} {{end-eqn}} ... which means $\exists x \in S$ such that the condition for antitransitivity is broken. So $\mathcal R \subseteq S \times S$ has to be antireflexive. {{Qed}}
Let $S$ be a set on which $\RR$ is a relation which is symmetric, transitive and  serial. As $\RR$ is symmetric: :$x \mathrel \RR y \implies y \mathrel \RR x$ As $\RR$ is transitive: :$x \mathrel \RR y \land y \mathrel \RR x \implies x \mathrel \RR x$ As $\RR$ is serial: :$\forall x \in S: \exists y \in S: x \mathrel \RR y$ Let $x \in S$. Then {{begin-eqn}} {{eqn | lo= \exists y \in S:       | l = \tuple {x, y}       | o = \in       | r = \RR       | c = as $\RR$ is serial }} {{eqn | ll= \leadsto       | l = \tuple {y, x}       | o = \in       | r = \RR       | c = as $\RR$ is symmetric }} {{eqn | ll= \leadsto       | l = \tuple {x, x}       | o = \in       | r = \RR       | c = as $\RR$ is transitive }} {{end-eqn}} Thus: :$\forall x: x \mathrel \RR x$ and by definition $\RR$ is reflexive. It follows by definition that such a relation is an equivalence relation. {{qed}}
=== Reflexivity === {{:Inverse of Reflexive Relation is Reflexive}} === Antireflexivity === {{:Inverse of Antireflexive Relation is Antireflexive}} === Non-Reflexivity === {{:Inverse of Non-Reflexive Relation is Non-Reflexive}} === Symmetry === {{:Inverse of Symmetric Relation is Symmetric}} === Asymmetry === {{:Inverse of Asymmetric Relation is Asymmetric}} === Antisymmetry === {{:Inverse of Antisymmetric Relation is Antisymmetric}} === Non-Symmetry === {{:Inverse of Non-Symmetric Relation is Non-Symmetric}} === Transitivity === {{:Inverse of Transitive Relation is Transitive}} === Antitransitivity === {{:Inverse of Antitransitive Relation is Antitransitive}} === Non-Transitivity === {{:Inverse of Non-Transitive Relation is Non-Transitive}} Category:Inverse Relations 2ruzcjj4s8bs7p8muxq0jmun8d4duzo
=== Reflexivity === ==== Restriction of Reflexive Relation is Reflexive ==== {{:Restriction of Reflexive Relation is Reflexive}} ==== Restriction of Antireflexive Relation is Antireflexive ==== {{:Restriction of Antireflexive Relation is Antireflexive}} === Symmetry === ==== Restriction of Symmetric Relation is Symmetric ==== {{:Restriction of Symmetric Relation is Symmetric}} ==== Restriction of Asymmetric Relation is Asymmetric ==== {{:Restriction of Asymmetric Relation is Asymmetric}} ==== Restriction of Antisymmetric Relation is Antisymmetric ==== {{:Restriction of Antisymmetric Relation is Antisymmetric}} === Transitivity === ==== Restriction of Transitive Relation is Transitive ==== {{:Restriction of Transitive Relation is Transitive}} ==== Restriction of Antitransitive Relation is Antitransitive ==== {{:Restriction of Antitransitive Relation is Antitransitive}} === Connectedness === ==== Restriction of Connected Relation is Connected ==== {{:Restriction of Connected Relation is Connected}}
=== Definition 1 implies Definition 2 === Let $\RR$ be a relation. We use a Proof by Contraposition by showing that: :$\Delta_S \not \subseteq \RR \implies \exists x \in S: \tuple {x, x} \notin \RR$ Thus, suppose: :$\Delta_S \not \subseteq \RR$ Then by definition of Diagonal Relation: :$\exists \tuple {x, x} \in S \times S: \tuple {x, x} \notin \RR$ Thus: :$\exists x \in S: \tuple {x, x} \notin \RR$ From Rule of Transposition it follows that: :$\forall x \in S: \tuple {x, x} \in \RR \implies \Delta_S \subseteq \RR$ {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\RR$ be a relation which fulfils the condition: :$\Delta_S \subseteq \RR$ Then: {{begin-eqn}} {{eqn | l = \forall x \in S: \tuple {x, x}       | o = \in       | r = \Delta_S       | c = {{Defof|Diagonal Relation}} }} {{eqn | ll= \leadsto       | l = \forall x \in S: \tuple {x, x}       | o = \in       | r = \RR       | c = {{Defof|Subset}} }} {{end-eqn}} Thus $\RR$ is reflexive by Definition 1. {{Qed}}
=== Necessary Condition === Let $\mathcal R$ be an antireflexive relation. Let $\left({x, y}\right) \in \Delta_S \cap \mathcal R$. By definition: : $\left({x, y}\right) \in \Delta_S \implies x = y$ Likewise, by definition: : $\left({x, y}\right) \in \mathcal R \implies x \ne y$. Thus: : $\Delta_S \cap \mathcal R = \left\{{\left({x, y}\right): x = y \land x \ne y}\right\}$ and so: : $\Delta_S \cap \mathcal R = \varnothing$ === Sufficient Condition === Let $\Delta_S \cap \mathcal R = \varnothing$. Then by definition: : $\forall \left({x, y}\right) \in \mathcal R: \left({x, y}\right) \notin \Delta_S$ Thus: : $\not \exists \left({x, y}\right) \in \mathcal R: x = y$ Thus by definition, $\mathcal R$ is antireflexive. {{Qed}} Category:Reflexive Relations oylgje196y4qtnwbu8ovx5rjfvuzzyd
=== $(1)$ implies $(2)$ === Suppose $\RR \subseteq \RR^{-1}$. Then: {{begin-eqn}} {{eqn | l = \tuple {a, b}       | o = \in       | r = \RR^{-1}       | c = }} {{eqn | ll= \leadsto       | l = \tuple {b, a}       | o = \in       | r = \RR       | c = {{Defof|Inverse Relation}} }} {{eqn | ll= \leadsto       | l = \tuple {b, a}       | o = \in       | r = \RR^{-1}       | c = {{Defof|Subset}} (as $\RR \subseteq \RR^{-1}$) }} {{eqn | ll= \leadsto       | l = \tuple {a, b}       | o = \in       | r = \RR       | c = {{Defof|Inverse Relation}} }} {{end-eqn}} Hence $\RR^{-1} \subseteq \RR$. {{qed|lemma}} === $(2)$ implies $(1)$ === Since $(1)$ implies $(2)$ was already established above, interpreting it on $\RR^{-1}$ yields: :$\RR^{-1} \subseteq \paren {\RR^{-1} }^{-1}$ By Inverse of Inverse Relation, $\paren {\RR^{-1} }^{-1} = \RR$. Hence $(2)$ implies $(1)$. {{qed|lemma}} === $(3)$ {{iff}} $(1)$ and $(2)$ === This is precisely the definition of set equality. {{qed}} Category:Inverse Relations qrykpsv23jatvbaui9xdj6kmnxr2493
=== Definition 1 implies Definition 3 === Let $\RR$ be a relation which fulfils the condition: :$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR$ Then: {{begin-eqn}} {{eqn | o =        | r = \tuple {x, y} \in \RR       | c =  }} {{eqn | o = \leadsto       | r = \tuple {y, x} \in \RR       | c = by hypothesis }} {{eqn | o = \leadsto       | r = \tuple {x, y} \in \RR^{-1}       | c = {{Defof|Inverse Relation}} }} {{eqn | o = \leadsto       | r = \RR \subseteq \RR^{-1}       | c = {{Defof|Subset}} }} {{end-eqn}} Hence $\RR$ is symmetric by definition 3. {{qed|lemma}} === Definition 3 implies Definition 2 === Let $\RR$ be a relation which fulfils the condition: :$\RR \subseteq \RR^{-1}$ Then by Inverse Relation Equal iff Subset: :$\RR = \RR^{-1}$ Hence $\RR$ is symmetric by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\RR$ be a relation which fulfils the condition: :$\RR^{-1} = \RR$ Then: {{begin-eqn}} {{eqn | o =       | r = \tuple {x, y} \in \RR }} {{eqn | o = \leadsto       | r = \tuple {x, y} \in \RR^{-1}       | c = as $\RR^{-1} = \RR$ }} {{eqn | o = \leadsto       | r = \tuple {y, x} \in \RR       | c = {{Defof|Inverse Relation}} }} {{end-eqn}} Hence $\RR$ is symmetric by definition 1. {{Qed}}
Let $\mathcal R$ be symmetric. Then from Relation equals Inverse iff Symmetric: :$\mathcal R = \mathcal R^{-1}$ The result follows. {{qed}}
Note that this result holds for any $\mathcal R \subseteq S \times T$, and does not require that $\left({S, \mathcal R}\right)$ necessarily be a relational structure. {{begin-eqn}} {{eqn | o=       | r=\left({a, b}\right) \in \mathcal R^{-1} \circ \mathcal R       | c= }} {{eqn | o=\implies       | r=\exists c \in T: \left({a, c}\right) \in \mathcal R \land \left({c, b}\right) \in \mathcal R^{-1}       | c=Definition of Composite Relation }} {{eqn | o=\implies       | r=\exists c \in T: \left({b, c}\right) \in \mathcal R \land \left({c, a}\right) \in \mathcal R^{-1}       | c=Definition of Inverse Relation }} {{eqn | o=\implies       | r=\left({b, a}\right) \in \mathcal R^{-1} \circ \mathcal R       | c=Definition of Composite Relation }} {{end-eqn}} Thus $\left({a, b}\right) \in \mathcal R^{-1} \circ \mathcal R \implies \left({b, a}\right) \in \mathcal R^{-1} \circ \mathcal R$ and thus $\mathcal R^{-1} \circ \mathcal R$ is symmetric. As $\mathcal R = \left({\mathcal R^{-1}}\right)^{-1}$ from Inverse of Inverse Relation, it follows that $\mathcal R \circ \mathcal R^{-1} = \left({\mathcal R^{-1}}\right)^{-1} \circ \mathcal R^{-1}$ is likewise a symmetric relation. The domain of $\mathcal R^{-1} \circ \mathcal R$ is $S$ from Domain of Composite Relation, as is its codomain from Codomain of Composite Relation and the definition of Inverse Relation. Similarly, the codomain of $\mathcal R \circ \mathcal R^{-1}$ is $T$, as is its domain. This completes the proof. {{Qed}} Category:Symmetric Relations is2tjx47f8rfg6k18o9p64rn6vhq4nw
=== Definition 1 implies Definition 2 === Let $\RR$ be a relation which fulfils the condition: :$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$ Then: {{begin-eqn}} {{eqn | l = \tuple {x, z}       | o = \in       | r = \RR \circ \RR       | c =  }} {{eqn | ll= \leadsto       | l = \exists y \in \RR: \tuple {x, y} \in \RR       | o = \land       | r = \tuple {y, z} \in \RR       | c = Definition of Composite Relation }} {{eqn | ll= \leadsto       | l = \tuple {x, z}       | o = \in       | r = \RR       | c = $\RR$ is transitive }} {{eqn | ll= \leadsto       | l = \RR \circ \RR       | o = \subseteq       | r = \RR       | c = Definition of Subset }} {{end-eqn}} Thus $\RR$ is transitive by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\RR$ be a relation that fulfils the condition: :$\RR \circ \RR \subseteq \RR$ {{AimForCont}} $\RR$ does not fulfil the condition: :$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$ Then: {{begin-eqn}} {{eqn | l = \exists \paren {\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR}: \tuple {x, z}       | o = \notin       | r = \RR       | c = {{Defof|Non-Transitive Relation}} }} {{eqn | ll= \leadsto       | l = \exists \tuple {x, z} \in \RR \circ \RR: \tuple {x, z}       | o = \notin       | r = \RR       | c = {{Defof|Composition of Relations}} }} {{eqn | ll= \leadsto       | l = \RR \circ \RR       | o = \not \subseteq       | r = \RR       | c = {{Defof|Subset}} }} {{end-eqn}} This contradicts our statement that $\RR \circ \RR \subseteq \RR$. Hence by Proof by Contradiction $\RR$ does fulfils the condition: :$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$ Thus $\RR$ is transitive by definition 1. {{Qed}}
From the definition, a many-to-one relation $\RR \subseteq S \times T$ is such that: :$\RR \subseteq S \times T: \forall x \in \Dom \RR: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR \implies y_1 = y_2$ Also from the definition, a one-to-many relation $\RR \subseteq S \times T$ is such that: :$\RR \subseteq S \times T: \forall y \in \Img \RR: \tuple {x_1, y} \in \RR \land \tuple {x_2, y} \in \RR \implies x_1 = x_2$ The inverse of a relation $\RR$ is defined as: :$\RR^{-1} = \set {\tuple {t, s}: \tuple {s, t} \in \RR$ Also from the definition of inverse relation, we have that: :$\Dom \RR = \Rng {\RR^{-1} }$ :$\Rng \RR = \Dom {\RR^{-1} }$ So, let $\RR$ be a many-to-one relation. Putting the above all together, we have: :$\RR^{-1} \subseteq T \times S: \forall x \in \Img {\RR^{-1} }: \tuple {y_1, x} \in \RR^{-1} \land \tuple {y_2, x} \in \RR^{-1} \implies y_1 = y_2$ and it can be seen that $\RR^{-1}$ is one-to-many. Similarly, let $\RR$ be a one-to-many relation. This gives us: :$\RR^{-1} \subseteq T \times S: \forall y \in \Dom {\RR^{-1} }: \tuple {y, x_1} \in \RR^{-1} \land \tuple {y, x_2} \in \RR^{-1} \implies x_1 = x_2$ and it can be seen that $\RR^{-1}$ is many-to-one. {{qed}} Category:Inverse Relations g2tynfvzl6y05zu20qqtctz07oh9pna
As $\mathcal R$ is one-to-many, it follows from Inverse of Many-to-One Relation is One-to-Many that $\mathcal R^{-1}$ is many-to-one. Let $\left({x, z}\right) \in \mathcal R^{-1} \circ \mathcal R$. Then: : $\exists y \in S: \left({x, y}\right) \in \mathcal R, \left({y, z}\right) \in \mathcal R^{-1}$ As $\left({x, y}\right) \in \mathcal R$, from the definition of inverse relation: :$\left({y, x}\right) \in \mathcal R^{-1}$ As $\mathcal R^{-1}$ is many-to-one: : $\left({y, z}\right) = \left({y, x}\right)$ Thus if $\left({x, z}\right) \in \mathcal R^{-1} \circ \mathcal R$, it follows that: : $\left({x, z}\right) = \left({x, x}\right)$ Thus: : $\mathcal R^{-1} \circ \mathcal R \subseteq \Delta_X$ The result follows by definition of coreflexive relation. {{Qed}} Category:Reflexive Relations k8wchgaag457a6dv3zi7wjzu5032jhb
{{begin-eqn}} {{eqn | l = \forall x \in S: \tuple {x, x}       | o = \in       | r = \mathcal R       | c = {{Defof|Equivalence Relation}}: $\mathcal R$ is Reflexive }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \eqclass x {\mathcal R}       | c = {{Defof|Equivalence Class}} }} {{end-eqn}} {{qed}}
In the notation $\tuple {p, q}$-graph, $p$ is its order and $q$ its size. That is, $p$ is the number of vertices in $G$, and $q$ is the number of edges in $G$. Each edge is incident to exactly two vertices. The degree of each vertex is defined as the number of edges to which it is incident. So when we add up the degrees of all the vertices, we are counting all the edges of the graph twice. {{qed}}
=== Necessary Condition === First we prove that $\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$. Suppose: : $\tuple {x, y} \in \RR: x, y \in S$ Then: {{begin-eqn}} {{eqn | l = z       | o = \in       | r = \eqclass x \RR       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {x, z}       | o = \in       | r = \RR       | c = {{Defof|Equivalence Class}} }} {{eqn | ll= \leadsto       | l = \tuple {z, x}       | o = \in       | r = \RR       | c = {{Defof|Equivalence Relation}}: $\RR$ is symmetric }} {{eqn | ll= \leadsto       | l = \tuple {z, y}       | o = \in       | r = \RR       | c = {{Defof|Equivalence Relation}}: $\RR$ is transitive }} {{eqn | ll= \leadsto       | l = \tuple {y, z}       | o = \in       | r = \RR       | c = {{Defof|Equivalence Relation}}: $\RR$ is symmetric }} {{eqn | ll= \leadsto       | l = z       | o = \in       | r = \eqclass y \RR       | c = {{Defof|Equivalence Class}} }} {{end-eqn}} So: :$\eqclass x \RR \subseteq \eqclass y \RR$ Now: {{begin-eqn}} {{eqn | l = \tuple {x, y} \in \RR       | o = \implies       | r = \eqclass x \RR \subseteq \eqclass y \RR       | c = (see above) }} {{eqn | l = \tuple {x, y} \in \RR       | o = \implies       | r = \tuple {y, x} \in \RR       | c = {{Defof|Equivalence Relation}}: $\RR$ is symmetric }} {{eqn | o = \leadsto       | r = \eqclass y \RR \subseteq \eqclass x \RR       | c = from above }} {{eqn | o = \leadsto       | r = \eqclass y \RR = \eqclass x \RR       | c = {{Defof|Set Equality}} }} {{end-eqn}} ... so we have shown that: :$\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$. {{qed|lemma}} === Sufficient Condition === Next we prove that $\eqclass x \RR = \eqclass y \RR \implies \tuple {x, y} \in \RR$. By definition of set equality: :$\eqclass x \RR = \eqclass y \RR$ means: :$\paren {x \in \eqclass x \RR \iff x \in \eqclass y \RR}$ So by definition of equivalence class: :$\tuple {y, x} \in \RR$ Hence by definition of equivalence relation: $\RR$ is symmetric :$\tuple {x, y} \in \RR$ So we have shown that :$\eqclass x \RR = \eqclass y \RR \implies \tuple {x, y} \in \RR$ Thus, we have: {{begin-eqn}} {{eqn | l = \tuple {x, y} \in \RR       | o = \implies       | r = \eqclass x \RR = \eqclass y \RR       | c =  }} {{eqn | l = \eqclass x \RR = \eqclass y \RR       | o = \implies       | r = \tuple {x, y} \in \RR       | c =  }} {{end-eqn}} {{qed|lemma}} So by equivalence: :$\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y\RR$ {{qed}}
First we show that: :$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$ Suppose two $\RR$-classes are ''not'' disjoint: {{begin-eqn}} {{eqn | l = \eqclass x \RR \cap \eqclass y \RR \ne \O       | o = \leadsto       | r = \exists z: z \in \eqclass x \RR \cap \eqclass y \RR       | c = {{Defof|Empty Set}} }} {{eqn | o = \leadsto       | r = \exists z: z \in \eqclass x \RR \land z \in \eqclass y \RR       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadsto       | r = \exists z: \paren {\tuple {x, z} \in \RR} \land \paren {\tuple {y, z} \in \RR}       | c = {{Defof|Equivalence Class}} }} {{eqn | o = \leadsto       | r = \exists z: \paren {\tuple {x, z} \in \RR} \land \paren {\tuple {x, y} \in \RR}       | c = {{Defof|Symmetric Relation}}: $\RR$ is symmetric }} {{eqn | o = \leadsto       | r = \tuple {x, y} \in \RR       | c = {{Defof|Transitive Relation}}: $\RR$ is transitive }} {{end-eqn}} Thus we have shown that $\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \tuple {x, y} \in \RR$. Therefore, by the Rule of Transposition: :$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$ Now we show that: :$\eqclass x \RR \cap \eqclass y \RR = \O \implies \tuple {x, y} \notin \RR$ Suppose $\tuple {x, y} \in \RR$. {{begin-eqn}} {{eqn | o =        | r = y \in \eqclass y \RR       | c = {{Defof|Equivalence Class}} }} {{eqn | o =        | r = \tuple {x, y} \in \RR       | c = by hypothesis }} {{eqn | o = \leadsto       | r = y \in \eqclass x \RR       | c = {{Defof|Equivalence Class}} }} {{eqn | o = \leadsto       | r = y \in \eqclass x \RR \land y \in \eqclass y \RR       | c = Rule of Conjunction }} {{eqn | o = \leadsto       | r = y \in \eqclass x \RR \cap \eqclass y \RR       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadsto       | r = \eqclass x \RR \cap \eqclass y \RR \ne \O       | c = {{Defof|Empty Set}} }} {{end-eqn}} Thus we have shown that: :$\tuple {x, y} \in \RR \implies \eqclass x \RR \cap \eqclass y \RR \ne \O$ Therefore, by the Rule of Transposition: :$\eqclass x \RR \cap \eqclass y \RR = \O \implies \paren {x, y} \notin \RR$ Using the rule of Biconditional Introduction on these results: :$\eqclass x \RR \cap \eqclass y \RR = \O \iff \paren {x, y} \notin \RR$ and the proof is complete.
Suppose that for $x, y \in S$: :$\eqclass x {\RR} \cap \eqclass y \RR \ne \O$ Let: :$z \in \eqclass x \RR$ :$z \in \eqclass y \RR$ Then by definition of equivalence class: :$\tuple {x, z} \in \RR$ :$\tuple {y, z} \in \RR$ Let $c \in \eqclass x \RR$. That is: :$\tuple {x, c} \in \RR$ By definition of equivalence relation, $\RR$ is symmetric so: :$\tuple {z, x} \in \RR$ By definition of equivalence relation, $\RR$ is transitive so: :$\tuple {z, x} \in \RR \land \tuple {x, c} \in \RR \implies \tuple {z, c} \in \RR$ and :$\tuple {y, z} \in \RR \land \tuple {z, c} \in \RR \implies \tuple {y, c} \in \RR$ So we have $c \in \eqclass y \RR$. By definition of subset: :$\eqclass x \RR \subseteq \eqclass y \RR$ Similarly, let $c \in \eqclass y \RR$. That is: :$\tuple {y, c} \in \RR$ By definition of equivalence relation, $\RR$ is symmetric so: :$\tuple {z, y} \in \RR$ By definition of equivalence relation, $\RR$ is transitive so: :$\tuple {z, y} \in \RR \land \tuple {y, c} \in \RR \implies \tuple {z, c} \in \RR$ and :$\tuple {x, z} \in \RR \land \tuple {z, c} \in \RR \implies \tuple {x, c} \in \RR$ So we have $c \in \eqclass x \RR$. By definition of subset: :$\eqclass y \RR \subseteq \eqclass x \RR$ That is: :$\eqclass x \RR \subseteq \eqclass y \RR$ and :$\eqclass y \RR \subseteq \eqclass x \RR$ By definition of set equality: :$\eqclass x \RR = \eqclass y \RR$ Thus: :$\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \eqclass x \RR = \eqclass y \RR$ and the result follows. {{qed}}
To prove that $S / \RR$ is a partition of $S$, we have to prove: :$(1): \quad \displaystyle \bigcup {S / \RR} = S$ :$(2): \quad \eqclass x \RR \ne \eqclass y \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$ :$(3): \quad \forall \eqclass x \RR \in S / \RR: \eqclass x \RR \ne \O$ Taking each proposition in turn: === Union of Equivalence Classes is Whole Set === The set of $\RR$-classes constitutes the whole of $S$: {{:Union of Equivalence Classes is Whole Set}}{{qed|lemma}} === Equivalence Classes are Disjoint === {{:Equivalence Classes are Disjoint/Proof 1}}{{qed|lemma}} === Equivalence Class is not Empty === {{:Equivalence Class is not Empty}}{{qed|lemma}} Thus all conditions for $S / \RR$ to be a partition are fulfilled. {{qed}}
This follows directly from the Fundamental Theorem on Equivalence Relations: the set of $\mathcal R$-classes forms a partition of $S$. {{qed}}
: $(1)$ and $(2)$ are equivalent because, by Equivalence Class is Unique, $\eqclass x {\mathcal R}$ is the unique $\mathcal R$-class to which $x$ belongs, and $\eqclass y {\mathcal R}$ is the unique $\mathcal R$-class to which $y$ belongs. As these are unique for each, they must therefore be the same set. : $(2)$ is equivalent to $(3)$ by Equivalence Class holds Equivalent Elements. : $(3)$ is equivalent to $(4)$ by the definition of Equivalence Class. : $(3)$ is equivalent to $(5)$ through dint of the symmetry of $\mathcal R$ and the definition of Equivalence Class. : $(3)$ is equivalent to $(6)$ from Equivalence Classes are Disjoint. {{qed}}
Let $\mathbb S$ be a partition of $S$. From Relation Induced by Partition, we define the relation $\RR$ on $S$ by: :$\RR = \set {\tuple {x, y}: \paren {\exists T \in \mathbb S: x \in T \land y \in T} }$ === Test for Equivalence === We are to show that $\RR$ is an equivalence relation. Checking in turn each of the critera for equivalence: ==== Reflexive ==== $\RR$ is reflexive: {{begin-eqn}} {{eqn | o =       | r = \paren {x \in T \land x \in T}       | c = Rule of Idempotence }} {{eqn | o = \leadstoandfrom       | r = \tuple {x, x} \in \RR       | c = Definition of $\RR$ }} {{end-eqn}} {{qed|lemma}} ==== Symmetric ==== $\RR$ is symmetric: {{begin-eqn}} {{eqn | o =        | r = \tuple {x, y} \in \RR       | c =  }} {{eqn | o = \leadsto       | r = x \in T \land y \in T       | c = Definition of $\RR$ }} {{eqn | o = \leadsto       | r = y \in T \land x \in T       | c = Rule of Commutation }} {{eqn | o = \leadsto       | r = \tuple {y, x} \in \RR       | c = Definition of $\RR$ }} {{end-eqn}} {{qed|lemma}} ==== Transitive ==== $\RR$ is transitive: {{begin-eqn}} {{eqn | l = x \in T \land y \in T       | o = \leadstoandfrom       | r = \tuple {x, y} \in \RR       | c = Definition of $\RR$ }} {{eqn | l = y \in T \land z \in T       | o = \leadstoandfrom       | r = \tuple {y, z} \in \RR       | c = Definition of $\RR$ }} {{eqn | l = \paren {x \in T \land y \in T} \land \paren {y \in T \land z \in T}       | o = \leadstoandfrom       | r = \tuple {x, z} \in \RR       | c = Definition of $\RR$ }} {{end-eqn}} {{qed|lemma}} So $\RR$ is reflexive, symmetric and transitive. Hence, by definition, $\RR$ is an equivalence relation. {{qed|lemma}} === Test for Uniqueness === Now by definition of a partition, we have that: :$\mathbb S$ partitions $S \implies \forall x \in S: \exists T \in \mathbb S: x \in T$ Also: {{begin-eqn}} {{eqn | l = x, y \in T       | o = \leadsto       | r = \tuple {x, y} \in \RR       | c = Definition of $\RR$ }} {{eqn | l = x \in T_1, y \in T_2, T_1 \ne T_2       | o = \leadsto       | r = \tuple{x, y} \notin \RR       | c = {{Defof|Set Partition}} }} {{end-eqn}} Thus $\mathbb S$ is the set of $\RR$-classes constructed above, and no other relation can be constructed in this way. {{qed}}
Let $\RR$ be an equivalence relation on $S$. From the Fundamental Theorem on Equivalence Relations, we have that the equivalence classes of $\RR$ form a partition. {{qed|lemma}} Let $S$ be partitioned into subsets by a relation $\RR$. From Relation Induced by Partition is Equivalence, $\RR$ is an equivalence relation. {{qed}}