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How to prove in a formal system (like HFOL) the following formula $\forall x\exists y(x = y)$?

Obviously, we would like to use the axiom $x = x$ and then use a rule for introducing $\exists$, but how do we get $x = y$?

user91015
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    Please articulate what you mean by HFOL. You don't mention the domains of $x, y$. This is crucial, because if $x \in \mathbb R$, and $y\in \mathbb N$, then the statement is not true. But if $x, y$ are both in the same domain, then yes, $\forall x\exists y(x=y)$. – amWhy Mar 04 '20 at 22:03
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    It is a basic proof system for first-order logic with equality. – user91015 Mar 04 '20 at 22:09
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    From $\forall x(x=x)$ use universal instantiation $a=a$. Then existential generalization $\exists y(a=y)$. Then universal generalization $\forall x\exists y(x=y)$. –  Mar 04 '20 at 22:09
  • @flan but by the existential generalization, shouldn't we replace all the occurrences of $a$ with $y$? – user91015 Mar 04 '20 at 22:23
  • See my answer posted just now at https://math.stackexchange.com/questions/2033513/deduction-of-forall-y-exists-x-x-y – Dan Christensen Mar 04 '20 at 23:22
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    @user91015 Nope. You do not need to replace all occurrences of $a$. Should $a=a$, there will exist some $y$ where $a=y$. – Graham Kemp Mar 05 '20 at 02:32

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Here's a possible formal proof in a Fitch-style natural deduction system:

\begin{align*} & 1. ~ a = a \mathrm{~(a.k.a.~}(a = y) [ y / a ] \mathrm{)} & \qquad & {=}I \\ & 2. ~ \exists y, a = y \mathrm{~(a.k.a.~}(\exists y, x = y) [ x / a ] \mathrm{)} & \qquad & {\exists}I, 1 \\ & 3. ~ \forall x, \exists y, x = y & \qquad & {\forall}I, 2 \end{align*}

Daniel Schepler
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