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I know the definitions for satisfaction of a sentence in a model but if there are free variables in a formula does that mean that it isn't true or false? (as determined by the model)Thanks

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up vote 1 down vote accepted

The answer depends on the fine print of the semantic story.

One common line makes it come out that a wff with free variables counts as true-in-$\mathcal{M}$ just when its universal closure comes out true.

But alternatively you can have it that a wff with free variables gets no true-value with respect to $\mathcal{M}\ $ -- it only gets a value with respect to $\mathcal{M} + \mathcal{I}$ where $\mathcal{I}$ is an assignment function mapping variables to elements of $\mathcal{M}$'s domain (in effect giving the variables an interpretation as temporary names).

In some ways, the second story is more natural: the first conforms with informal mathematical practice of treating sentences with free variables as implicitly universally quantified. But you can go either way.

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There are three ways of resolving this. One way is to say that a formula $\phi (\vec{x})$ is satisfied if and only if the sentence $\forall \vec{x} . \phi (\vec{x})$ is satisfied. Another is to insist that it makes no sense to ask about the truth value of $\phi (\vec{x})$ without an assignment of values to the free variables $\vec{x}$. The third is to expand the notion of truth value somewhat and say that the valuation of a formula $\phi (\vec{x})$ with $n$ free variables $\vec{x}$ in a model $A$ is the subset $$[\vec{x} . \phi (\vec{x})]_A = \{ \vec{a} \in A^n : A \vDash \phi[\vec{a}] \}$$ and note that in the case $n = 0$, there are only two truth values, as usual.

The third interpretation has the advantage of making it easier to bridge syntax and semantics: for example, $T \vdash \phi (\vec{x}) \to \psi (\vec{x})$ if and only if $[\vec{x}. \phi (\vec{x}))]_A \subseteq [\vec{x}. \psi (\vec{x})]_A$ for all models $A$ of the theory $T$. Note that this is not true if we reduce the valuations of formulae to $\{ \bot, \top \}$! For example, take $T$ to be the theory of rings, $\phi (x)$ to be the formula "$x \ne 0$", and $\psi (x)$ to be the formula "$x$ is invertible". Then, $(\forall x. \phi (x)) \to (\forall x. \psi(x))$ is certainly true in all rings because $\forall x. \phi (x)$ is false; but $\forall x. (\phi (x) \to \psi (x))$ is true only in a field!

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