# Compactness Theorem (logic) application.

I met a question which says: if we have $$A, B$$, both are nonempty theories such that every model of $$A$$ satisfies at least one formula (sentence if they are in the first-order language, the question indicates that there is no difference whether it's propositional or first-order) in $$B$$. Then it asks me to show there are some formulas/sentences, $$a_1....a_k$$ from $$A$$ and $$b_1...b_l$$ from $$B$$ such that $$(a_1...\land...a_k)\to(b_1...\lor...b_l)$$ is a tautology.

The question specifically says I should use the compactness theorem, but I don't see how to apply the theorem here. Any hints?

• Hint: Using the condition "every model of A satisfies at least one sentence in B", try to come up with a set of sentences that is inconsistent. Then apply compactness theorem to that set. Dec 7, 2020 at 17:36
• @halrankard2, so you mean, we need to first separate if B is satisfiable or not? If B is not satisfiable, then we just pick some b and not b so that it will be a tautology. But if B is satisfible, how to use the condition to construct such a set? Dec 7, 2020 at 18:12
• Analyzing whether or not B is satisfiable isn't fully leveraging the main connection between A and B. You know that every model of A is a model of some sentence in B. Therefore there is a certain collection of sentences (built using both A and B as ingredients) that is inconsistent. Dec 7, 2020 at 18:29
• Well it's hard to make an example that doesn't give away the answer. But suppose B consisted of just one sentence $\phi$. Then your assumption is that every model of A is a model of $\phi$. So $A\cup\{\neg\phi\}$ is an inconsistent set. Dec 7, 2020 at 22:21
• @qwerty You don't have that $A$ and $B$ are finite ...
– Ned
Dec 7, 2020 at 23:46

## 1 Answer

Hint: Notice that the hypothesis says exactly that $$\{\varphi,\neg \psi\mid \varphi\in A,\psi\in B\}$$ is inconsistent. By compactness, there is a finite inconsistent subset.