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I'm learning predicate logic and I'm reading some lecture notes from my professor that say: From theorem XXX (that is, "In a model M, if two assignments σ and σ' are such that σ(x) = σ'(x) for every variable x free in a formula α, then σ satisfies α iff σ' satisfies α") it turns out that, in any model, either a closed formula is satisfied by all assignments or it is satisfied by none. For suppose that two assignments σ and σ' are such that σ satisfies a formula α while σ' does not. Then from theorem XXX we get by contraposition that α contains some free variable x such that σ(x) ≠ σ'(x). But this entails that α is not closed. So, if α is closed, there are no two such assignments.

My question concerns the definition of a closed formula and more specifically the part of the text I put in bold: couldn't we simply say that a closed formula contains no free variable at all (i.e., remove the "such that σ(x) ≠ σ'(x)" part)? Isn't this precisely the definition of a closed formula (i.e., no free variables)? Should we really specify that a closed formula doesn't contain any free variable such that σ(x) ≠ σ'(x)?

Your help would be greatly appreciated.

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    *Closed* formula is exactly (as you say) a formula with **no** free vars... but the issue is to prove the consequence of Th.XXX. – 2017-01-23
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    If σ and σ' are such that σ satisfies a formula α while σ' does not, this means that it is **not true** that σ satisfies α iff σ' satisfies α. Thus, contraposing Th.XXX, we get : it is **not true** that : for every variable x *free* in α, σ(x) = σ'(x). But "**not** for every" is equiv to "exists (at least one) **not**". i.e. there is a var x *free* in α such that **not** σ(x) = σ'(x). – 2017-01-23
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    The idea is quite simple: the *satisfaction* of a formula α by an assignment σ (i.e. $\mathfrak M, σ \vDash α$) depends on "what" σ assigns to the free vars in α. If there are no free vars (i.e. α is *closed*), all assignments act in the same way. – 2017-01-23

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A formula is indeed closed if it contains no free variables.

The source of confusion lies elsewhere. Let us say that two assignments $\sigma$ and $\sigma'$ agree for a formula $\alpha$ if for every $x \in \mathrm{fv}(\alpha)$ we have that $\sigma(x)=\sigma'(x)$.

Now theorem XXX says that if assignments $\sigma$ and $\sigma'$ agree for a formula $\alpha$, then $\sigma \models \alpha \iff \sigma' \models \alpha$.

We now want to prove that for every closed $\alpha$ we have that either there is no $\sigma$ such that $\sigma \models \alpha$ or every $\sigma$ has that $\sigma \models \alpha$.

So let $\alpha$ be an arbitrary formula and suppose to that there is a $\sigma$ and a $\sigma'$ such that $\sigma \models \alpha$ but $\sigma' \not\models \alpha$. Theorem XXX now tells us that $\sigma$ and $\sigma'$ do not agree for $\alpha$. But then, by definition of agreement, this means that there must be an $x \in \mathrm{fv}(\alpha)$ where $\sigma(x)\neq \sigma'(x)$. This then means that $\alpha$ contains at least one free variable and therefore cannot be closed.