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How does one prove $\vdash (\exists x)(A\rightarrow(\forall x)A)$?

I can think of using duality: $\vdash(\exists x)(A\wedge B) \equiv A\wedge (\exists x) B$

However, I'm not too sure how to formally go about it. Thanks for the help :)

Thomas Edenbauer
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  • Is $A$ a property of $x$ or is it independent? If it is independent, then it is essentially $A\to A$. – Javi Jul 20 '19 at 21:13
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    How you formally go about it is a function of what formal system you are using, so we would need to know more about that to answer. (Also I don't think that axiom is or should be called duality. Also it is only valid if $x$ is not free in $A$.) Anyhow, this may help for the intuition and for searching the site for duplicates https://en.wikipedia.org/wiki/Drinker_paradox – spaceisdarkgreen Jul 20 '19 at 21:23

1 Answers1

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The negation is
$\forall x (A \wedge \neg \forall x A)$
iff
$\forall x A \wedge \forall x \neg \forall x A)$
iff
$\forall x A \wedge \neg \forall x A)$
which is clearly false.

William Elliot
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  • It's not that clearly false as both this and the original statement rely on the assumption of a non-empty domain. This assumption isn't always made that explicit and, of course, it need not be made, though it usually is in traditional accounts. – Derek Elkins left SE Jul 21 '19 at 08:16
  • Proof from the axioms, not that satisfying stuff of valid models.. @DerekElkins – William Elliot Jul 21 '19 at 11:37
  • There are clearly things in the rules/axioms that imply that the domain is non-empty in traditional accounts. This can be explicit, e.g. $\exists x.\top$, but more often it is implicit, e.g. in the rules for "null quantification" which state $A$ entails $\exists x.A$ when $x$ is not free in $A$. If we set $A$ to $\top$, which certainly doesn't contain $x$ free, we get $\top$ entails $\exists x.\top$. A more careful accounting of the rules will not have this entailment. You will need these "bad" cases of null quantification (or something equivalent) to prove this statement. – Derek Elkins left SE Jul 21 '19 at 15:14
  • @DerekElkins. See edit. An axiomatic proof without that superfluous domain thing. The problem is about |-, not about |=. – William Elliot Jul 21 '19 at 22:36
  • The second step uses null quantification as I said would be necessary. If you write the rules of inference carefully, that step isn't one of the rules of inference. That we usually do allow this rule of inference corresponds to the semantic fact that we usually require domains to be non-empty. (We have soundness and completeness.) If you allow empty domains, then the rule is not sound. If the domain is empty, then $\forall x.P(x)$ for any $P(x)$ is vacuously true. Choose $A=\bot$ in the original formula and you immediately get $\exists x.\top$. – Derek Elkins left SE Jul 22 '19 at 18:03