# Proving a simple consequence of the Compactness Theorem

I am self-learning logic, and trying to prove the following exercise using the Compactness Theorem:

Suppose $$T$$ is a theory for language $$L$$, and $$\sigma$$ is a sentence of $$L$$ such that $$T \models \sigma$$ ($$\sigma$$ is a consequence of $$T$$). Prove that there is some finite model $$T' \subseteq T$$ such that $$T' \models \sigma$$.

I get the sense that this should be easy to prove, but I am not coming up with a satisfying argument.

The argument I want to make is: $$T \models \sigma$$ iff $$T \cup \{\neg \sigma\}$$ is unsatisfiable, implying no $$T' \cup \{\neg \sigma\}$$ is satisfiable, where $$T'\subseteq T$$ is finite. From here I want to then derive the conclusion but am not sure how.

Alternatively, recalling that since $$T \models \sigma$$ means, by definition, for model $$U$$, if we have $$U \models T$$, then $$U \models \sigma$$. Would it then be wrong to argue that, if $$T \models \sigma$$, then $$T \cup \{\sigma\}$$ is satisfiable, and hence every finite subset of $$T \cup \{\sigma\}$$ is satisfiable by the Compactness Theorem, and hence we have a finite $$T'\subseteq T$$ such that $$T' \models \sigma$$?

"finite model" is incorrect. Are you sure the exercise doesn't say "finite subtheory" or "finite subset"?

There are a number of basic logical errors in your two arguments (correctly using quantifiers, negating statements, etc). I'll give you feedback on what you wrote, which hopefully will lead directly to a solution.

$$T \models \sigma$$ iff $$T \cup \{\neg \sigma\}$$ is unsatisfiable, implying no $$T' \cup \{\neg \sigma\}$$ is satisfiable, where $$T'\subseteq T$$ is finite.

No, $$T\cup \{\lnot \sigma\}$$ unsatisfiable does not imply $$T' \cup \{\neg \sigma\}$$ unsatisfiable for all finite $$T'$$. For example, suppose $$T = \{\sigma\}$$, $$\lnot \sigma$$ is satisfiable, and $$T' = \varnothing\subseteq T$$. Then $$T\cup \{\lnot \sigma\}$$ is unsatisfiable, but $$T'\cup \{\lnot \sigma\}$$ is satisfiable.

What $$T\cup \{\lnot\sigma\}$$ unsatisfiable does imply (by the compactness theorem!) is that there is some finite $$T'\subseteq T$$ such that $$T'\cup \{\lnot\sigma\}$$ is unsatisfiable.

if $$T \models \sigma$$, then $$T \cup \{\sigma\}$$ is satisfiable, and hence every finite subset of $$T \cup \{\sigma\}$$ is satisfiable by the Compactness Theorem, and hence we have a finite $$T'\subseteq T$$ such that $$T' \models \sigma$$?

(1) $$T\models \sigma$$ does not imply $$T\cup \{\sigma\}$$ is satisfiable, unless you already know that $$T$$ is satisfiable. $$T\models \sigma$$ means that every model of $$T$$ satisfies $$\sigma$$, but how do you know there are any models of $$T$$ in the first place?

(2) The implication from "$$T\cup \{\sigma\}$$ is satisfiable" to "every finite subset of $$T \cup \{\sigma\}$$ is satisfiable" is obvious and does not require the compactness theorem.

(3) $$T'\cup \{\sigma\}$$ is satisfiable does not imply $$T'\models \sigma$$. $$T'\cup \{\sigma\}$$ is satisfiable if there is some model of $$T'$$ which satisfies $$\sigma$$. $$T'\models \sigma$$ if every model of $$T'$$ satisfies $$\sigma$$.

• Thanks! Yes I misread the problem it was asking if there is a "finite sub theory" $T'$, not a finite model. In your fourth paragraph, is why is it the case that $T\cup \{\neg \sigma\}$ unsatisfiable implies there is some finite $T' \subseteq T$ s.t. $T' \cup \{\sigma\}$ is unsatisfiable? Specifically why $T'\cup \{\sigma\}$ unsatisfiable and not $T'\cup \{\neg\sigma\}$? Commented Jul 18 at 15:16
• @user918212 That was a typo. I'll fix it now. Commented Jul 18 at 15:22
• Thanks! I am embarrassed that I need to be lead on like this, but given that we know there is some finite $T' \subseteq T$ such that $T' \cup \{\neg \sigma\}$ is unsatisfiable, how do we conclude the existence of a finite $T' \models \sigma$? I get the sense there is some basic detail I am still missing here. Commented Jul 18 at 15:32
• @user918212 Write down the definitions of "$T'\models \sigma$" and of "$T'\cup\{\neg\sigma\}$ is unsatisfiable." Then think about them until it becomes clear that they say exactly the same thing. (It might be helpful to think what counterxamples would have to look like.) Commented Jul 18 at 16:09
• @AndreasBlass I see now, thank you! Commented Jul 18 at 16:40