Deducible implies satisfiable (first order logic)

Question

Let $\sigma$ be an $L$-sentence and $\Sigma$ an $L$-theory. Then, if $\Sigma$ deduces $\sigma$, it also satisfies it. I've tried proving this by induction on the formal proof for $\sigma$ from $\Sigma$ but it is fairly confusing, is there any easier way to prove this? Thanks

Answer 1

No, induction on proof complexity is the only way to go - remember that deducibility itself is defined recursively, so it's hard to see how to analyze it without using induction.

The key point to note is that the proof rules are truth preserving. For instance, the last step in a minimal counterexample (= a proof $\pi$ whose conclusion is not entailed semantically, and has the least length among such) cannot be $\wedge$-introduction: if $\Sigma\models a$ and $\Sigma\models b$ then $\Sigma\models a\wedge b$ (just check the $\wedge$ clause in the definition of $\models$), so if $p=a\wedge b$ is $\Sigma$-provable but not $\Sigma$-entailed then either $a$ or $b$ would also have the same property via a shorter proof (indeed, a proper subproof of $\pi$).

You'll continue like this, and show that for every proof rule, that rule can't be the last step in a minimal counterexample. That shows that indeed $\Sigma\vdash p\implies\Sigma\models p$.