Consider this proposition in first-order logic:

For any interpretation $I$, any closed formula $\phi$ and any two valuations $\rho$, $\sigma$. $I\rho \models \phi \iff I\sigma \models \phi$

This is incredibly obvious yet i don't see how to prove it, the obvious induction doesn't really work since a formula is only closed when you have all the quantifiers.


You need a preliminary result, like in Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001), page 86 :

THEOREM 22A Assume that $s_1$ and $s_2$ are functions from $V$ [the set of variables] into $|\mathfrak A|$ [the domain of the structure] which agree at all variables (if any) that occur free in the wff $\varphi$.


$\mathfrak A \vDash \varphi[s_1]$ iff $\mathfrak A \vDash \varphi[s_2]$.

The proof is by induction on the complexity of $\varphi$.

Then [page 87] :

COROLLARY 22B For a sentence $\sigma$, either

(a) $\mathfrak A$ satisfies $\sigma$ with every function $s$ from $V$ intto $|\mathfrak A|$, or

(b) $\mathfrak A$ does not satisfy $\sigma$ with any such function.

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  • $\begingroup$ Interesting. I did think of proving 22A. But my intuition somehow told me it didn't make sense. $\endgroup$ – Bumphe Dec 2 '14 at 19:25

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