Here's an argument that's quite clearly valid, but which I'm having trouble proving in Natural Deduction:

$\exists x~\exists y~\lnot x=y \vdash\forall x~\exists y~\lnot x=y$

The informal reasoning: the premise is that there are at least two distinct elements; this implies that for every element, there is at least one element which is distinct from it.

I presume the proof will end (and I'm using what I believe is called a proof tree):

$$\exists y~\lnot a=y\over\forall x~\exists y~\lnot x=y$$

with the top sentence following from an application of negation elimination or from $\lnot a=b$ (or some constant). But after hours of trying I haven't been able to complete a proof. Can someone advise me?

  • 2
    $\begingroup$ Are you using intuitionistic logic? If so then this is not provable. Try proving it assuming also the law of excluded middle $\forall P. P \vee \neg P$ $\endgroup$
    – Max New
    Jan 3, 2018 at 17:48
  • 3
    $\begingroup$ Or more specifically, decidable equality should do $\forall x. \forall y. (x = y) \vee \neg (x = y)$ $\endgroup$
    – Max New
    Jan 3, 2018 at 17:56

2 Answers 2


Here's a proof in Fitch ... you can probably convert it to a proof in your system:

enter image description here


Proof sketch

1) $∃x ∃y \ ¬(x=y)$ --- premise

2) $¬∀x ∃y \ ¬(x=y)$ --- assumed [a]

3) $¬(a=b)$ --- from 1) by two $∃$-elim

4) $∃x ∀y \ (x=y)$ --- from 2) by equivalences: to be derived

5) $∀y \ (c=y)$ --- from 4) by $∃$-elim

6) $c=a$ --- from 5)

7) $c=b$ --- from 5)

8) $a=b$ --- from 6) and 7) by transitivity of equality

9) $\bot$ --- from 2) and 8)

Now we can close all the $∃$-elim's, because there are no free variables in 9). Thus:

$∀x ∃y \ ¬ (x=y)$ --- from 2) by Double Negation [that is not intuitionistically valid; see also @Max New's comment], discharging [a].


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