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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

struggling for math
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  • You might want to make explicit the domain of $x, y$, or, for your purposes that $x,y$ share the same domain. – amWhy Nov 27 '16 at 22:43
  • For this type of question, it is vital to specify the precise formal system that you want to use for the deduction. Otherwise, we could just say that the statement given above is an axiom of our system (the statement is logically valid, after all). Different systems can be quite distinctive about how the achieve this kind of deduction - and in even more basic respects, such as having linear or tree-shaped deductions. Also, not all systems even have $\exists$ as a basic symbol, so the real goal might be to deduce $\forall y \lnot \forall x \lnot (x = y)$. – Carl Mummert Nov 27 '16 at 22:48
  • @TheGreatDuck，that was a typo. I just fixed it. Sorry about the confusion – struggling for math Nov 28 '16 at 02:48

4 Answers4

3

Hint:

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

Berci
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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.

Mauro ALLEGRANZA
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It works even if you cannot assume a non-empty universe by using a proof by contradiction. Start by assuming to the contrary...

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Dan Christensen
  • 14,529
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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 ...

$\blacksquare$.

Graham Kemp
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