=== 1 implies 2 === {{begin-eqn}} {{eqn | l = \tuple {x, y}       | o = \in       | r = \map \TT \equiv }} {{eqn | ll = \leadstoandfrom       | l = x       | o = \equiv       | r = y       | c = {{Defof|Graph of Relation}} }} {{eqn | ll = \leadsto       | l = x       | o = \sim       | r = y       | c = from $1$ }} {{eqn | ll = \leadstoandfrom       | l = \tuple {x, y}       | o = \in       | r = \map \TT \sim       | c = {{Defof|Graph of Relation}} }} {{end-eqn}} By definition of subset, $\map \TT \equiv \subseteq \map \TT \sim$. {{qed|lemma}} === 2 implies 1 === {{begin-eqn}} {{eqn | l = x       | o = \equiv       | r = y       | c = {{Defof|Equivalence Class}} }} {{eqn | ll = \leadstoandfrom       | l = \tuple {x, y}       | o = \in       | r = \map \TT \equiv       | c = {{Defof|Graph of Relation}} }} {{eqn | o = \subseteq       | r = \map \TT \sim       | c = from $2$ }} {{eqn | ll = \leadstoandfrom       | l = x       | o = \sim       | r = y       | c = {{Defof|Graph of Relation}} }} {{end-eqn}} {{qed|lemma}} === 1 implies 3 === Let $x \in X$. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \eqclass x \equiv }} {{eqn | ll = \leadstoandfrom       | l = y       | o = \equiv       | r = x       | c = {{Defof|Equivalence Class}} }} {{eqn | ll = \leadsto       | l = y       | o = \sim       | r = x       | c = from $1$ }} {{eqn | ll = \leadstoandfrom       | l = y       | o = \in       | r = \eqclass x \sim       | c = {{Defof|Equivalence Class}} }} {{end-eqn}} By definition of subset, $\eqclass x \equiv \subseteq \eqclass x \sim$. {{qed|lemma}} === 3 implies 1 === For this proof and the next, we use this result implied by $3$: Let $y \in X$. $3$ implies: :$\exists z \in X: \eqclass y \equiv \subseteq \eqclass z \sim$ By Element in its own Equivalence Class: :$y \in \eqclass y \equiv \subseteq \eqclass z \sim$ :$y \in \eqclass y \sim$ By contrapositive of Equivalence Classes are Disjoint: :$\eqclass y \sim = \eqclass z \sim$ Hence $\eqclass y \equiv \subseteq \eqclass y \sim$. Thus: {{begin-eqn}} {{eqn | l = x       | o = \equiv       | r = y }} {{eqn | ll = \leadstoandfrom       | l = x       | o = \in       | r = \eqclass y \equiv       | c = {{Defof|Equivalence Class}} }} {{eqn | o = \subseteq       | r = \eqclass y \sim }} {{eqn | ll = \leadstoandfrom       | l = x       | o = \sim       | r = y       | c = {{Defof|Equivalence Class}} }} {{end-eqn}} {{qed|lemma}} === 3 implies 4 === Let $x \in X$. {{begin-eqn}} {{eqn | l = \bigcup_{y \mathop \in \eqclass x \sim} \eqclass y \equiv       | o = \subseteq       | r = \bigcup_{y \mathop \in \eqclass x \sim} \eqclass y \sim       | c = Set Union Preserves Subsets }} {{eqn | r = \bigcup_{\substack {y \mathop \in X \\ y \mathop \sim x} } \eqclass y \sim       | c = {{Defof|Equivalence Class}} }} {{eqn | r = \bigcup_{\substack {y \mathop \in X \\ y \mathop \sim x} } \eqclass x \sim       | c = Equivalence Class holds Equivalent Elements }} {{eqn | r = \eqclass x \sim       | c =  }} {{end-eqn}} so $\eqclass x \sim$ is a union of equivalence classes under $\equiv$. Hence $\eqclass x \sim$ is saturated under $\equiv$. {{qed|lemma}} === 4 implies 3 === Let $x \in X$. Then $\eqclass x \sim$ is saturated under $\equiv$: :$\displaystyle \exists S \subseteq X: \bigcup_{y \mathop \in S} \eqclass y \equiv = \eqclass x \sim$ By Element in its own Equivalence Class: :$x \in \eqclass x \sim$ :$x \in \eqclass x \equiv$ Hence by definition of Union of Family: :$\exists z \in S: x \in \eqclass z \equiv$ By contrapositive of Equivalence Classes are Disjoint: {{begin-eqn}} {{eqn | l = \eqclass x \equiv       | r = \eqclass z \equiv }} {{eqn | o = \subseteq       | r = \bigcup_{y \mathop \in S} \eqclass y \equiv       | c = Set is Subset of Union }} {{eqn | r = \eqclass x \sim }} {{end-eqn}} {{qed}}