Is $\vDash \exists x ( Q x \to \forall x Qx)$ a valid sentence?

Question

Is $\vDash \exists x ( Q x \to \forall x Qx)$ a valid sentence?

$Q$ is a unitary relation.

I suppose that $\vDash Q x \to \forall x Qx$ , which is equivalent to $\vDash Q x \to \forall y Qy$ is invalid, since there is a structure $\mathfrak{A}$ with the universe $|\mathfrak{A}|=\{a,b\}$ plus one relation $Q = \{a\}$ and a function $s$ which sends the variable $x$ to $a$. But I got confused henceforth this point.

I'm inclined to reason that, since $x$ is bounded, the part $\forall x$ is redundant, the sentence should be valid.

Answer 1

Whether it's even well-formed depends on the low-level details of how you define syntax.

But even if it is well-formed in the syntax you use, using a variable $x$ as a dummy variable in a context where $x$ already has meaning is usually a bad idea.

That said, typically in syntax that allows such a thing, a variable acquires the innermost meaning. Therefore

$$\exists x ( Q x \to \forall x Qx)$$

is the same expression as

$$\exists x ( Q x \to \forall y Qy)$$

and is a different expression than

$$\exists x ( Q x \to \forall y Qx)$$

Answer 2

Yes, it is.

The "flaw" in your purported counterexample to the validity of:

(A) --- $∃x(Qx \rightarrow ∀xQx)$

is that you are "reading" it as : $∃xQx \rightarrow ∀xQx$, which is not [this is implied by the structure $\mathcal A$ with universe $|\mathcal A| = \{ a,b \}$ such that $Q^{\mathcal A} = \{ a \}$].

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We prove the vaildity of (A) in two ways.

Formal proof

We have that :

$\vdash (∀xβ \rightarrow α) \leftrightarrow ∃x(β \rightarrow α)$, if $x$ does not occur free in α [see this post for the proof].

If we apply it to (A), we get :

(B) ---$∀xQx \rightarrow ∀xQx$

which is clearly valid.

---

Semantic proof

Consider an interpretation $\mathcal A$ whatever and consider the conditional : $Qx \rightarrow ∀xQx$ : if $Qx$ holds for all $x$, then the conditional is true; thus, also $∃x(Qx \rightarrow ∀xQx)$ is.

If there is an object $a \in |\mathcal A|$ such that $a \notin Q^{\mathcal A}$ , then $\forall xQx$ is false.

Now, what is the condition for $\mathcal A \vDash \exists x \varphi[s]$ , i.e. such that $\mathcal A$ satisfy $\varphi$ with the assignment $s : Var \mapsto |\mathcal A|$ ?

It is :

for some $a \in |\mathcal A|$, we have $\mathcal A \vDash \varphi[s(x|a)]$.

Thus, if we consider an assignment $s$, we have that $\mathcal A \vDash (Qx \rightarrow ∀xQx)[s(x|a)]$ [because $\forall xQx$ is false precisely because $Q$ does not hold for $a$, and so we have : $False \rightarrow False$, which is $True$ ].

But, having shown that :

for some $a \in |\mathcal A|$, we have $\mathcal A \vDash (Qx \rightarrow ∀xQx)[s(x|a)]$

we may conclude with :

$\mathcal A \vDash \exists x (Qx \rightarrow ∀xQx)[s]$.

But this holds for $s$ and $\mathcal A$ whatever; thus :

$\vDash \exists x (Qx \rightarrow ∀xQx)$.

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Comment

A further "informal" argument is the following.

By tautological equivalence, we may rewrite : $\exists x (Qx \rightarrow ∀xQx)$ as :

$\exists x (\lnot Qx \lor ∀xQx)$.

$\exists$ "works" as a "generalized" disjunction; thus, we can "distribute" it over $\lor$, obtaining the equivalent formula :

$\exists x \lnot Qx \lor ∀xQx$,

due to the fact that $\exists x \forall xQx$ has the same meaning of $\forall xQx$, $x$ being already bound.

But $\exists x \lnot Qx \lor ∀xQx$ is clearly valid; it means that either all objects are $Q$ or there is some object which is not, and this must hold in all universes.