Universal closure of a formula

I am confused about the following: I read yesterday that for a formula $\phi(x_1,\ldots,x_n)$ in a first order language $\mathcal{L}$ and an $\mathcal{L}$-structure $\mathcal{A}$, $\mathcal{A} \models \phi(x_1,\ldots,x_n)$ iff $\mathcal{A} \models \forall x_1 \ldots \forall x_n \phi(x_1,\ldots,x_n)$. This seems perfectly fine to me; if I understand it right, it's like saying something like $x^2 \geq 0$ is true iff $\forall x (x^2 \geq 0)$ is true.

Now consider a predicate $Q(x)$; then according to the above $\models Q(x)$ iff $\models \forall x Q(x)$. Isn't this equivalent to $\models Q(x) \leftrightarrow \forall x Q(x)$? However, $\not \models Q(x) \rightarrow \forall x Q(x)$; take for example $\mathcal{A} = (A, Q^{\mathcal{A}})$, where $A=\{a,b\}$, $Q^{\mathcal{A}}=\{a\}$ and $w: \mathrm{Var} \rightarrow A$, $w(x) = a$. What am I doing wrong?

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I believe you are mixing two different conventions:

• Some authors include a variable assignment along with each model when defining the satisfaction relation, so that a model consists of a structure $\mathcal{A}$ and a function $w$ assigning some element of $|\mathcal{A}|$ to each variable. It is possible to have two such functions $w,w'$ such that $\mathcal{A},w \models \phi$ but $\mathcal{A},w' \not \models \phi$. Using only this definition, it is impossible to write $\mathcal{A} \models \phi$ when $\phi$ has free variables, as the definition requires $w$ to be fixed first. Enderton's book uses essentially this approach, writing $\models_\mathcal{A}\, \phi[w]$.

• Other authors go on to define $\mathcal{A} \models \phi$, when $\phi$ has free variables, to mean $\mathcal{A},w\models \phi$ for every variable assignment function $w$ from $\mathcal{A}$. Letting $\phi^u$ be the universal closure of $\phi$, these authors would indeed say that $\mathcal{A} \models \phi$ if and only if $\mathcal{A} \models \phi^u$, because this is an immediate consequence of their definitions. This approach is used by Kleene's 1967 book and I believe it is also common in universal algebra.

One additional point of confusion is that the authors who take the first approach continue by defining $\models \phi$'' to mean that $\mathcal{A},w\models \phi$ for every structure $\mathcal{A}$ in the language and every variable assignment $w$ from $\mathcal{A}$. So authors who take the first approach can still prove that $\models \phi$ if and only if $\models \phi^u$ even if $\phi$ has free variables. This is exercise 6 on page 99 of Enderton's book, for example.

In the example in the second paragraph in the question, the trouble is that under the second convention I described, $\mathcal{A} \not \models Q(x)$, although it is true that $\mathcal{A},w \models Q(x)$. In fact neither $Q(x)$ nor $(\forall x)Q(x)$ is a logically valid formula.

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Thank you for your answer. I am following the second convention. So, under this convention, is it wrong to use $\models\phi$ (meaning that $\mathcal{A} \models \phi$ for any structure $\mathcal{A}$)? My question basically was how can $\models Q(x)$ iff $\models \forall x Q(x)$ when $\not \models Q(x) \rightarrow \forall x Q(x)$? (for any $Q$) Doesn't $\models Q(x)$ iff $\models \forall x Q(x)$ imply that $\models Q(x) \rightarrow \forall x Q(x)$? – Andrew Jan 10 '12 at 22:51
No, it doesn't. "$\models Q(x)$ implies $\models (\forall x) Q(x)$" says that if $Q(x)$ holds in absolutely every interpretation, then $(\forall x)Q(x)$ holds in absolutely every interpretation. That's a true claim. But "$\models Q(x) \to (\forall x)Q(x)$" claims that in any particular interpretation where $Q(x)$ holds, $(\forall x)Q(x)$ must also hold, even if $Q(x)$ may not hold in other interpretations. That's a false claim. – Carl Mummert Jan 10 '12 at 23:10
Dear Carl: Thank you very much for having pointed out the stupid mistake contained in my (now deleted) answer to this question. That was very kind of you! – Pierre-Yves Gaillard Jan 12 '12 at 23:19