Prove that if $x\not=y,$ then $\{a\}\not=\{x,y\}$ for any sets $a$ only using three given axioms

Question

Prove that if $x \not = y$ , then $\{ a \} \not = \{ x , y \}$ for any set $a$.

It is a trivial problem, but I wanted to make my life more difficult and prove it using only three axioms if possible :

1. The Axiom of Equality $\forall x \forall y (x = y \rightarrow \forall z (x \in z\leftrightarrow y \in z))$
2. The Axiom of Extensionality $\forall x \forall y (x = y \leftrightarrow \forall u (u \in x \leftrightarrow u \in y))$
3. The Axiom of Pairing $\forall a \forall b \exists z (\forall x (x \in z \leftrightarrow [x = a \vee x = b]))$

Here is what I have so far :

Suppose $\{ a \} = \{x , y\}$. Let $u \in \{ a \}$. The set $\{ a \}$ was constructed by applying the pairing axiom to the sets $a$ and $a$ , and so, from the part of the axiom that says $(x \in z \leftrightarrow [x = a \vee x = b])$ , we have $u = a$. The extensionality axiom implies $u \in \{x , y \}$. Similarly, we get $u = x \vee u = y$. So from the transitivity that follows from the extensionality axiom, we have $a = x \vee a = y$. What next ?

Of course none of that would be a problem if we could prove that it is always the case that $x \in \{ x , y \}$ , but I could not find any rigorous definition of membership $( \in )$ , so I have to rely only on those three axioms.

Answer 1

You only need the axiom of equality. Take $x \ne y$.

Suppose that $\{a\}= \{x,y\}$. Then we have $u \in \{x,y\}$ if and only if $u \in \{a\}$

Since $x\in \{x,y\}$ then $x \in \{a\}$, so $x=a$.

Since $y\in \{x,y\}$ then $y \in \{a\}$, so $y=a$.

This means $x=y$, which contradicts the hypothesis.

Answer 2

It is a Confusing Issue (at least for me !) , where the many $a$ , $x$ , $y$ variables are re-used in the theorem & axioms.

Here is my attempt , not sure whether it is valid or is workable or can be tweaked to work out.

Consider the Case $\{ x, y \} = \{a \}$ with the Axiom of Extensionality.
It will imply that $x=a$ , when $x \in \{ x, y \} \iff x \in \{a \}$
It will also imply that $y=a$ , when $y \in \{ x, y \} \iff y \in \{a \}$
Then we get $x=a=y$ , then $x=y$

Contradiction.

Hence $\{ x, y \} \not = \{a \}$