I am aware that the question is a duplicate. However, I would like to clear up a more general principle regarding instantiation through this proof which I have outlined at the end.
Suppose there are arbitrary elements $x \in A$ and $y \in B$
$(x,y) \in A \times B \implies (x,y) \in B \times A \implies x \in A \land y \in B$
Since the choice of $x$ and $y$ was arbitrary, $\forall x \in A$ $\forall y \in B(x \in B \land y \in A)$
Since the above proposition is true for all $y \in B$ and $B$ is not empty, it must be true for some $y_0 \in B$
Therefore, $\forall x \in A(x \in B \land y_0 \in A)$
$\implies \forall x \in A(x \in B)$
$\implies A \subseteq B$
Using a similar argument, it can be shown that $B \subseteq A$
Thus, $A = B$
I know that $y=y_0$ can be used in the beginning, but I wanted to find a proof where the instantiation was at the end. My main doubt regarding the proof is the part where I have shown that $\forall x \in A(x \in B)$ is true for some random $y_0$.
Are there any improvements I could make to this proof?
Best Answer
Let $x\in A$ and $y\in B$. Then $(x,y)\in A\times B=B\times A$, and, hence $x\in B$ and $y\in A$. That is $A\subseteq B$, and $B\subseteq A$ (as the same time).