Vacuous statements and explosion

Question

So my understanding of vacuous statements is as follows: For any statement $P$, the statement $(\forall x \in \emptyset)(P(x))$. This can be argued as follows: Assume for contradiction $\neg [(\forall x \in \emptyset)(P(x))]$. Then by DeMorgan, $(\exists x \in \emptyset)(\neg P(x))$, i.e. if a "for all" statement fails, there must be a witness to the failure. But $\not \exists x \in \emptyset$, so we cannot have a witness, a contradiction. Thus $(\forall x \in \emptyset)(P(x))$.

But this seems as if it would entail explosion, as $[(\forall x \in \emptyset)(\neg P(x))] \Rightarrow \neg [(\forall x \in \emptyset)(P(x))]$. So it seems as if I've shown the contradiction $[(\forall x \in \emptyset)(P(x))] \land \neg [(\forall x \in \emptyset)(P(x))]$. Why does this not imply explosion?

Answer 1

You assert that $$[(\forall x \in \emptyset)(\neg P(x))] \Rightarrow \neg [(\forall x \in \emptyset)(P(X))]$$ without providing any justification for it. It's simply false.

Answer 2

Okay, let us break the chain of your logic down.   It is just two steps.   You have:

$$\begin{align*} \forall x\in D:P(x) ~ \implies & ~ \exists x\in D:P(x) \tag 1 \\[1ex] \exists x\in D:P(x) ~ \iff & ~ \neg \forall x\in D:\neg P(x) \tag 2 \\[1ex] ~ \\[0ex]\hline \\[1ex] \therefore \forall x\in\varnothing:P(x) \implies & ~ \neg \forall x\in \varnothing: \neg P(x) \tag {C!} \end{align*}$$

Now, step $(2)$ is from the Dual Negation equivalence.   That is okay.

The issue lies in that the weakening in step $(1)$ is only sound if the domain $D$ is not empty^†.   It is not justifiable if $D$ is in fact empty.

⟨ † : That caveat is too often omitted when it is implicit that we are dealing with a non-empty domain.   It is something to watch for when it is possible that the domain may be empty. ⟩

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"Anything in the empty set has property P" does not actually imply that "There's something in the empty set with property P".