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

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?

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These expressions don't look well-formed to me. Have you run across them somewhere? – Grumpy Parsnip Apr 2 '14 at 20:04
If one is able to conclude $\exists x\Phi(x)$ from earlier inference for some formula $\Phi$, then by Generalization, you should be able to derive the second formula. However, it looks rather odd to me, and I have not seen its usage in any texts. – Bill Province Apr 2 '14 at 20:19
@Grumpy In Metamath, this type of expression is quite well-defined, although people tend not to use it because it gives them a headache to think about. The interpretation is that since you can change bound variables on the inner quantifier, $\forall x\,\Phi(x)\leftrightarrow\forall y\,\Phi(y)$ implies $\exists x\forall x\,\Phi(x)\leftrightarrow\exists x\forall y\,\Phi(y)$. – Mario Carneiro Apr 2 '14 at 20:26
@MarioCarneiro: thanks for the clarification. – Grumpy Parsnip Apr 2 '14 at 20:59

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

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Thanks, Andre. Does the same reasoning also apply to the second formula? Would we conclude that the two formulas assert different things? – Bill Province Apr 2 '14 at 20:22
The argument for the second formula is the same. In $\forall x\exists x\Phi(x)$, the universal quantifier has no effect. The first sentence is logically equivalent to $\forall x\Phi(x)$, and the second is logically equivalent to $\exists x\Phi(x)$. Please note that in each case it is the front quantifier that makes no difference. – André Nicolas Apr 2 '14 at 20:29
Is this interpretation consistent with the one I just posted as a comment (rename bound variables with preference to inner bindings until you have something sensible)? – Mario Carneiro Apr 2 '14 at 20:30
@AndréNicolas If the domain of discourse is empty, then $\exists x\forall x\,x=x$ is false, but $\forall x\,x=x$ is true. – Mario Carneiro Apr 2 '14 at 20:33
@MarioCarneiro: In traditional Model Theory, the underlying set $|M|$ of an $L$-structure $M$ is non-empty. – André Nicolas Apr 2 '14 at 20:36

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 $\forall x B$.

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