Deduction of $\forall y \exists x (x=y)$

Question

Give a deduction of $\forall y \exists x (x=y)$

My thinking is I can prove $\exists x(x=y)$ then use generalization theorem， which is equivalent to prove $\neg \forall x \neg (x=y)$ but I got stuck proving this... thanks in advance

Sorry that I did not specify what deduction system that I want to use here, it is the deduction sequence where ai is either Axiom or original hypothesis or obtained from MP

Answer 1

Hint:

$$\matrix{ y=y \\ \hline \exists x.x=y \\ \hline \forall y.\exists x.x=y } $$

Answer 2

For any arbitrary value $v$, because we always witnesses $v=v$ by the Law of Identity, therefore ...

Since this is so for any arbitrary value , then we deduce ...

$\blacksquare$.

Answer 3

I'll assume Enderton's axiom system, with Deduction Th and Generalization Th.

1) $\vdash \forall x \lnot(x=y) \to \lnot(y=y)$ --- Ax.2 : $∀x α → α(t/x)$

2) $\vdash (y=y) \to \lnot \forall x \lnot(x=y)$ --- from 1) by Ax.1: Taut and modus ponens

3) $\vdash \forall x(x=x)$ --- Ax.5 : generalization of $x=x$

4) $\vdash \forall x(x=x) \to (y=y)$ --- Ax.2

5) $\vdash y=y$ --- from 4) and 3) by mp

6) $\vdash \exists x (x=y)$ --- from 5) and 2) by mp with abbreviation : $\exists$ for $\lnot \forall \lnot$

$\vdash \forall y \exists x(x=y)$ --- from 6) by Gen Th.