Is the following expression a tautology?

Question

$\forall x\,(P(x)\rightarrow Q(x))\rightarrow (\exists y\,P(y)\rightarrow\exists z\,Q(z))$

I believe the sentece is a tautoloogy. Can someone confirm?

Answer 1

$\forall x\,(P(x)\rightarrow Q(x))\rightarrow (\exists y\,P(y)\rightarrow\exists z\,P(z))$ is trivially a logical truth, since $C = (\exists y\,P(y)\rightarrow\exists z\,P(z))$ is a trivial logical truth, and if $C$ is a logical truth, so is $A \to C$ for any $A$.

You probably meant, however, to ask whether $\forall x\,(P(x)\rightarrow Q(x))\rightarrow (\exists y\,P(y)\rightarrow\exists z\,Q(z))$ is a logical truth. But yes, this is too. If any $P$ (should there be one) is a $Q$, then if there is a $P$ then there is a $Q$.

Answer 2

Original Post:

As stated, the right-side (the consequent) of the implication is always true, because $\exists yP(y) \equiv \exists z P(z)$.

In an implication: $P\rightarrow Q$, whenever $Q$ is true, so is $P\rightarrow Q$ true.

Hence, since the right-side of your implication is always true, so is the entire implication. That means it is a tautology.

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Updated post

Likewise, if you intended to write $$\forall x(P(x)\rightarrow Q(x)) \rightarrow (\exists y P(y) \rightarrow \exists z Q(z))$$ it too is a tautology, but for a different reason than the above. If the antecedent is false, the whole implication is necessarily true. And if the antecedent is true, then the consequent is true, regardless of whether such an $x$ exists, because if there is any $x$ such that $P(x) \rightarrow Q(x)$, then the consequent necessarily follows. (And if no such $x$ exists, well, then the consequent is true because its antecedent $P(y)$ is false, which means the implication on the right-hand side is necessarily true.)

Answer 3

It's a tautology. The other answers here explain why very well, but just for an extra perspective, let me show you that it's the case with a tableau. You start with the negation of the formula then apply a series of contradiction-hunting rules to get

,

which is closed (i.e., its branches end in contradictions), implying that the original formula is a tautology.