This is an easy consequence (Doets calls it Compactness Theorem (version 2)) in Kees Doets' Basic Model theory:

Let $\Sigma$ a set of sentences and $\phi$ a sentence.

If $\Sigma \models \phi$, then for some finite $\Delta \subset\Sigma$, $\Delta \models \phi$.

The proof by contradiction reads:

"If $\phi$ does not logically follow from some $\Delta \subset \Sigma$, then the set $\Sigma \cup \{\neg \phi\}$ is finitely satisfiable and therefore, by compactness, satisfiable."

How do we know that $\Sigma \cup \{\neg \phi\}$ is finitely satisfiable?

From the fact that $\phi$ does not logically follow from some $\Delta$, we know that $\neg \phi$ follows from every $\Delta \subset \Sigma$. But how does this imply that there's a model for every finite subset of the set of sentences?


First, a quick comment. Your statement "From the fact that $\varphi$ does not logically follow from some $\Delta$, we know that $\neg\varphi$ follows from every $\Delta\subseteq \Sigma$." is incorrect: when $A$ is a set of sentences, "$A\not\models b$" is different from "$A\models\neg b$." We could have $b$ be independent of $A$ - that is, $A\not\models b$ and $A\not\models\neg b$.

What is true is that since $\varphi$ isn't entailed by any finite $\Delta\subseteq\Sigma$, $\neg\varphi$ must be compatible with every finite $\Delta\subseteq\Sigma$ in the sense that, for every $\Delta\subseteq\Sigma$, the set $\Delta\cup\{\neg\varphi\}$ will be satisfiable. (It's arguably more natural to say "consistent with" instead of "compatible with," but consistency is a syntactic property which isn't needed in this purely semantic question, so I want to avoid using language which might bring in confusion.)

This is in fact the crux of the problem!

Saying "$\Sigma\cup\{\neg\varphi\}$ is finitely satisfiable" just means "$\Delta\cup\{\neg\varphi\}$ is satisfiable for every finite $\Delta\subseteq\Sigma$." But "$\Delta\cup\{\neg\varphi\}$ is satisfiable" just means "there is some model of $\Delta\cup\{\neg\varphi\}$," which is to say $$\Delta\not\models\varphi$$ ("$\Delta\models\varphi$" means exactly "every model of $\Delta$ satisfies $\varphi$," which can't be true if there is some model of $\Delta\cup\{\neg\varphi\}$).

All that is to say that the statement "$\Sigma\cup\{\neg\varphi\}$ is finitely satisfiable" is equivalent to "For every finite $\Delta\subseteq\Sigma$, we have $\Delta\not\models\varphi$." But this latter statement is precisely our hypothesis "$\varphi$ does not logically follow from some [finite] $\Delta\subseteq\Sigma$!"

Note that the phrase "logically follow from" could confuse things here: it's meant in the semantic sense ($\models$), not the syntactic sense ($\vdash$).


My intuition about this does not (directly) use contradiction. Rather, it follows from the soundness and completeness of first-order logic:

If $\Sigma\vDash\phi$, then there is a proof of $\Sigma\vdash\phi$ (completeness), and since that proof is a finite object it mentions at most finitely many of the axioms of $\Sigma$. Therefore the same proof proves $\Delta\vdash\phi$ for some finite $\Delta\subseteq\Sigma$, and thus $\Delta\vDash\phi$ (soundness).

  • $\begingroup$ To be fair, bringing the completeness theorem into a purely semantic problem is a bit overkill given that compactness can be proved without it. $\endgroup$ – Noah Schweber Dec 8 '18 at 20:28
  • $\begingroup$ @NoahSchweber I thought this is a purely syntactic problem, and bringing in models would make it semantic, now? $\endgroup$ – Mike Dec 8 '18 at 20:31
  • $\begingroup$ @Craig You have things backwards. "$\models$" is a purely semantic notion ($\Sigma\models\varphi$ iff every model of $\Sigma$ satisfies $\varphi$); "$\vdash$" is syntactic ($\Sigma\vdash\varphi$ iff there is a proof of $\varphi$ from $\Sigma$). Since the problem involves only "$\models$," it is (a priori) purely semantic. $\endgroup$ – Noah Schweber Dec 8 '18 at 20:32
  • $\begingroup$ @NoahSchweber Ohhh, I thought that $\models$ was interpreted as logical deduction when used in relation to sets of formulas. That should clear everything up, thanks. $\endgroup$ – Mike Dec 8 '18 at 20:36
  • $\begingroup$ @NoahSchweber: Well, this is the only way I could think about proving compactness either. But I'll admit to a strong bias in favor of proof theory ... $\endgroup$ – Henning Makholm Dec 8 '18 at 20:51

Every $\Delta$ finite have $\Delta\vDash\lnot\varphi$, so every model of $\Delta$ is a model of $\Delta\cup\{\lnot\varphi\}$, and there is a model for every $\Delta$, namely, the model of $\Sigma$

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    $\begingroup$ So does Doets assume that $\Sigma$ has a model? $\endgroup$ – Mike Dec 8 '18 at 20:32
  • $\begingroup$ If there is no model to $\Sigma$, there is finite deduction sequence that proves contradiction, take $\Delta$ be all the elements of $\Sigma$ that appear in that sequence and you got a finite subset of $\Sigma$ that proves $\varphi$ $\endgroup$ – Holo Dec 8 '18 at 20:35
  • $\begingroup$ @Holo "If there is no model to $\Sigma$, there is finite deduction sequence that proves contradiction" This uses the completeness theorem, which is overkill here. $\endgroup$ – Noah Schweber Dec 8 '18 at 20:38
  • $\begingroup$ @NoahSchweber that is true.. but this is the easiest way to look at this(at least the easiest from the ways I know) $\endgroup$ – Holo Dec 8 '18 at 20:50
  • $\begingroup$ @Holo I disagree - $\vdash$ isn't needed at all, this is just a direct application of the basic definition of $\models$. $\endgroup$ – Noah Schweber Dec 8 '18 at 20:51

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