for all $x$, $P(x)$ implies there exists $x$, $P(x)$?

Question

$$\begin{align*} \big[\forall x\,P(x)\to\exists x\,P(x)\big]&\iff\left[\big(\neg\forall x\,P(x)\big)\lor\exists x\,P(x)\right]\\ &\iff\exists x\,\neg P(x)\lor\exists x\,P(x)\;, \end{align*}$$

so is

$$\forall x\,P(x)\to\exists x\,P(x)$$

true?

Answer 1

Yes, both (1) and (2) are valid first-order sentences:

1. $$\vDash ∀xP(x)→∃xP(x)$$
2. $$\vDash ∃x¬P(x)∨∃xP(x)$$

which means that they hold for every model $M$ of the concerned language.

What you are probably concerned about is the ontological comitment underlying both sentences, namely, the existential presupposition of objects that we made when asserting them. (For instance (2) says that, for every property $P$, there exists a guy that either have it or not.)

But note that this existential presupposition of first order logic is nothing but a technical consequence of the constraint that the universe of $M$ must be non-empty. Is there a reason why it is so? Well, because it is useful. Particularly, note that without this constraint we would have to change the inference rules of natural deduction, in order to guarantee soundness.

Alternatively, you can also look for free logics, a non-classical variant of first order logic that may allow for a domain to be empty (and, also, may allow for terms to not denote any object).