Theorem. Show that the definition of the ordered pair $(x,y) := \{ \{ x \}, \{ x, y \} \}$ obeys the property (3.5) $$(x,y) = (x',y') \leftrightarrow x' = x \wedge y' = y.$$
Proof.

Two sets are equal iff every element of one set is element of the other set and vice versa: $$\label{eqn_idSets} \{ \{ x \}, \{ x, y \} \} = \{ \{ x' \}, \{ x', y' \} \} \leftrightarrow (\forall z) \: z \in \{ \{ x \}, \{ x, y \} \} \leftrightarrow z \in \{ \{ x' \}, \{ x', y' \} \}.$$ Since $\{ \{ x \}, \{ x, y \} \}$ is, according to the pair set axiom (3.3), a set, we also have $$z \in \{ \{ x \}, \{ x, y \} \} \leftrightarrow z = \{ x \} \vee z = \{ x, y \}.$$ Inserting this identity into (\ref{eqn_idSets}) yields $$(\forall z) \: z = \{ x \} \vee z = \{ x, y \} \leftrightarrow z = \{ x' \} \vee z = \{ x', y' \},$$ or, equivalently, $$\label{eqn_expand} (\forall z) \: (z = \{ x \} \vee z = \{ x, y \} \rightarrow z = \{ x' \} \vee z = \{ x', y' \}) \wedge (z = \{ x \} \vee z = \{ x,y \} \leftarrow z = \{ x' \} \vee z = \{ x',y' \}).$$ If $x = y$, (\ref{eqn_idSets}) requires that $$(\forall z) \: z \in \{ \{ x \} \} \leftrightarrow z \in \{ \{ x' \}, \{ x', y' \} \}.$$ In this case the equivalence holds iff $x' = y'$ such that $x = x'$ and $y = y'$.

If $x \neq y$, (\ref{eqn_expand}) yields two implications for the 1st clause of the conjunction: \begin{align}\label{eqn_impxneqy1} z = \{ x \} & \rightarrow z = \{ x' \} \vee z = \{ x', y' \}, \\ \label{eqn_impxneqy2} z = \{ x,y \} & \rightarrow z = \{ x' \} \vee z = \{ x', y' \}. \\ \end{align} Since for two sets to be equal they need to have the same number of elements, ($\ref{eqn_impxneqy1}$) and ($\ref{eqn_impxneqy2}$) reduce to \begin{align} \label{eqn_impxneqy3} z = \{ x \} & \rightarrow z = \{ x' \}, \\ \label{eqn_impxneqy4} z = \{ x,y \} & \rightarrow z = \{ x', y' \}. \\ \end{align} From ($\ref{eqn_impxneqy3}$) we have that $x = x'$ while ($\ref{eqn_impxneqy4}$) requires that $$\{ x,y \} = \{ x', y' \} \stackrel{(\ref{eqn_impxneqy3})}{\longrightarrow} \{ x,y \} = \{ x, y' \} \rightarrow y = y'.$$ Summarizing the above yields $$\{ \{ x \}, \{ x, y \} \} = \{ \{ x' \}, \{ x', y' \} \} \leftrightarrow (x = x' \wedge y = y').$$ $\Box$

2015 FEB 07 (v1.0)
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