Theorem. The set $\{ x, \{ x, y \} \}$ obeys the property $$ \underbrace{\{ x, \{ x, y \} \} = \{ x', \{ x', y' \} \}}_{(i)} \leftrightarrow (x = x') \wedge (y = y'). $$
Proof.

Two sets are equal iff every element of one set is in the other set and vice versa. The proposition (i) is therefore equivalent to $$ (\forall z) \: \underbrace{z \in \{ x, \{ x, y \} \} \leftrightarrow z \in \{ x', \{ x', y' \} \}}_{(ii)}. $$ The axiom of pair set requires that $$ z \in \{ x, \{ x, y \} \} \leftrightarrow z = x \vee z = \{ x, y \}. $$ Now suppose $z = x \wedge z = \{ x, y \} $. According to the (axiomatic) symmetric and transitive property of the equation relation we get $ x = \{ x, y \}. $ But this requires that there is an element in $x$ which is $x$. This contradicts the axiom of regularity: $$ \neg ( x = \{ x, y \} \rightarrow x \in x ) \leftrightarrow ( x \notin x \rightarrow x \neq \{ x, y \} ). $$ From this it follows that either $z = x$ or $z = \{x, y\}$, but not both. A listing of all required implications is given as follows (it's a conjunction since (ii) has to hold for all z): $$ (z = x \rightarrow z = x' \vee z = \{ x', y' \} ) \wedge (z = \{ x, y \} \rightarrow z = x' \vee z = \{ x', y' \}) \wedge (z = x' \rightarrow z = x \vee z = \{ x, y \} ) \wedge (z = \{ x', y' \} \rightarrow z = x \vee z = \{ x, y \}). $$ Assuming $(z = x \rightarrow z = \{ x', y' \}) \wedge (z = \{ x', y' \} \rightarrow z = \{ x, y \})$, the transitive property of implication has it that $z = x \rightarrow z = \{ x, y \}$ which is equivalent to saying that $x = \{ x, y \}$. Which is false due to the axiom of regularity (see above).

Going through all the possible permutations we find that the only valid implications are $$ (z = x \rightarrow z = x') \wedge (z = \{ x, y \} \rightarrow z = \{ x', y' \}). $$ Or, equivalently, $$ (x = x') \wedge (y = y'). $$

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