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∀x(¬Qx → Px)

∃x¬Px

∴ ∃xQx

For the above argument, is it possible to construct a counter-model to show that the argument is invalid? If not, what is the reasoning that it is valid.

For example, through trial and error, creating a Domain {1,2} where the extension of Q is an empty set, {}, and P has an extension of {1}, is this a countermodel that shows that the argument is invalid? All premises have to be true with the conclusion false.

D.Ronald
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2 Answers2

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No, the reasoning is valid (for non-empty domains), and in your model for the element $2$ you have that $Q$ does not hold (so $\lnot Q$ does hold) and $P$ does not hold, so $\lnot Q(2) \to P(2)$ is false, so it's not a model for $\forall x: (\lnot Qx \to Px)$.

Henno Brandsma
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    Note: the second, existential condition implies that the domain is not empty. – Berci Sep 25 '19 at 10:34
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    In short: $\neg Px, \neg Qx\to Px\vDash Qx$ is a valid argument, so if $\exists x~\neg Px$ and $\forall x~(\neg Qx\to Px)$ are both satisfied then $\exists x~Qx$ will be satisfied too. – Graham Kemp Sep 25 '19 at 14:18
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If one suspects that the argument is invalid one might first try using a tree proof. If it is invalid then one of the branches will not close and one could use that branch to construct a countermodel.

In this case, the argument is valid and so one could construct a tree proof as this generator shows:

enter image description here

One could also show validity by using natural deduction with this Fitch-style proof checker:

enter image description here

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

Tree Proof Generator. https://www.umsu.de/trees/

Frank Hubeny
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