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I am claiming that $$\forall xP(x) \vee \neg \forall yP(y)$$ is a tautology, since if we denote $\varphi = \forall xP(x)$ and we know that $(\varphi \vee \neg\varphi)$ is a tautology in propositional logic and is equivalent to our given formula then the given formula is a tautology.

However, my text says that it is not a tautology.

I can't find a way to show that this formula is indeed not a tautology.

ryang
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Yarin
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    Every propositional-logic tautology (which you've correctly argued this to be) is automatically a first-order validity (first-order tautology). P.S. This question is not a duplicate of the above suggested link. – ryang Jul 02 '23 at 15:18
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    @ryang Note that the two disjuncts are not actually negations of each other here - the variable change, while trivial to us, does prevent this from being a (substitution instance of a) propositional tautology. – Noah Schweber Jul 02 '23 at 15:22
  • @NoahSchweber Yes of course; thanks for the correction! This given formula isn't a propositional-logic tautology but is certainly an FOL validity (FOL tautology). – ryang Jul 02 '23 at 15:37

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$$\forall xP(x) \vee \neg \forall yP(y)$$

The given formula has truth-functional form $$A∨¬B,$$ so is not a (propositional-logic) tautology. Because it is equivalent to $$\forall xP(x) \vee \neg \forall xP(x),$$ it certainly is a first-order validity. You might call it a "first-order tautology".

While some FOL texts consider "validity" and "tautology" synonymous, your textbook is using a strict definition of "tautology" referring specifically to propositional-logic tautologies.

ryang
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  • Just to make sure that I understood, because the given formula is equivalent to $$\forall xP(x) \vee \neg \forall xP(x)$$ we can say that it is equivalent to propositional-logic tautology and therefore is FOL tautology ? – Yarin Jul 02 '23 at 15:55
  • I didn’t actually explain why it’s a validity. Your indirect reasoning isn’t wrong. – ryang Jul 03 '23 at 00:20