How to semantically show $\forall x\neg \phi \vDash \neg\exists x \phi.$

Question

How to semantically show

$$\forall x\neg \phi \vDash \neg\exists x \phi.$$

I know the natural-deduction proof of this is not difficult, but how can I semantically prove this?

Answer 1

Consider a structure $\mathfrak A$ with domain $A$ such that $\mathfrak A \vDash \forall x \lnot \phi$ (in other words, a model for the formula, i.e. an interpretation satisfying the formula).

And assume that $\mathfrak A \nvDash \lnot \exists x \phi$.

This means that $\mathfrak A \vDash \exists x \phi$.

But we have, according to the previous relation, that: for some $d \in A$ (i.e. for some object in the domain of the interpretation) $\phi[d/x]$ holds (said otehrwise: $\mathfrak A, l \vDash \phi$, where the "environment" function $l: \text {var} \to A$ is such that $l(x)=d$; see Huth & Ryan, Logic in Computer Science, page 127-28) .

But we have that $\mathfrak A \vDash \forall x \lnot \phi$, that means that $\lnot \phi[a/x]$ holds for every $a \in A$ (i.e. $\mathfrak A, l(a/x) \vDash \lnot \phi$ holds, for all $a \in A$).

Contradiction.

Answer 2

Suppose that $\mathcal{M}\models\forall x\neg\phi(x)$, this means that for every $x\in\mathcal{M}$ (In the universe of $\mathcal{M}$) satisfies $\neg\phi(x)$. So, there is not $x\in\mathcal{M}$ such that satisfies $\phi(x)$, i.e $\mathcal{M}\models\neg\exists\phi(x).$