How to interpret $\exists x (\forall x \Phi (x))$?

Question

It's clear to me what the interpretation is when we have something like:

$$\exists x (\forall y \Phi(x, y))$$

or even how to interpret the formula when x or y are not variables in the expression $\Phi$, but when these are used in a seemingly contradictory fashion, what is the meaning of these expressions:

$$\exists x (\forall x \Phi (x))$$ $$\forall x (\exists x \Phi (x))$$

Are these even wffs?

Answer 1

One usually takes these to be well-formed formulas.

Let us take, for example, $\exists x\forall x \Phi(x)$. When we interpret this sentence, we examine $\forall x \Phi(x)$ for all free occurrences of $x$ in $\forall x\Phi(x)$. There are no such free occurrences, so $\exists x\forall x\Phi(x)$ is true in a structure $M$ precisely if $\forall x\Phi(x)$ is true in $M$.

More informally, the $\exists x$ in front has no effect. For that reason, one would never (except for the purposes of this question!) actually use the sentence $\exists x\forall x\Phi(x)$.

Answer 2

The "reason way" is (so to say) a "formal" one.

See Stephen Cole Kleene, Mathematical Logic (1967), page 127 : if $x$ is not free in $A$, then

$\vdash \forall x A \equiv A$ and $\vdash \exists x A \equiv A$.

We will prove the first one :

$\forall x A \vdash A$ --- by $\forall$-elim

$\vdash \forall x A \rightarrow A$ --- by $\rightarrow$-intro --- (A)

$A \vdash A$

$A \vdash \forall x A$ --- by $\forall$-intro, $x$ not free in $A$

$\vdash A \rightarrow \forall x A$ --- by $\rightarrow$-intro --- (B)

$\vdash \forall x A \equiv A$ --- from (A) and (B) by $\equiv$-intro.

The same for $\vdash \exists x A \equiv A$.

Thus, if we apply them to $A := \forall x B$, because $x$ is not free in $A$, we have that :

$\exists x \forall x B \equiv \forall x B$.

Now, forgetting the "uselessness" of it, its "interpretation" is exactly the same as that of $\forall x B$.