Compactness Theorem for Propositional Logic

Question

Here is the compactness theorem:

If every finite subset of $\Phi$ is satisfiable, then $\Phi$ is satisfiable.

Is the contrapositive the following?

If $\Phi$ is unsatisfiable (tautologically independent), then every finite subset of $\Phi$ is unsatisfiable.

Does unsatisfiability imply tautological independence?

Answer 1

The comment from @spaceisdarkgreen is correct, but maybe a little more explanation would help. The contrapositive of "if A then B" is "if it's not the case that B then it's not the case that A. In your situation, B is "$\Phi$ is satisfiable" so "it's not the case that B" is "$\Phi$ is unsatisfiable"; that part of your question was correct. The problem concerns the other half, where "it's not the case that A" is "it's not the case that every finite subset of $\Phi$ is satisfiable." Where you went wrong is that to deny that every finite subset is satisfiable is not the same as to assert that every finite subset is unsatisfiable; rather it is to assert that some finite subset is unsatisfiable. (For an easy analogy: To deny that all Americans are right-handed is not to assert that we're all left-handed, only that some of us are left-handed.)

To summarize, the error in your proposed contrapositive does not involve the compactness theorem directly nor the notion of contrapositive but rather the negation of universal assertions.

Answer 2

The comment that "The contrapositive is that if $\Phi$ is unsatisfiable then some finite subset of $\Phi$ is unsatisfiable," is correct of course.

A formula $\varphi$ is tautologically independent of a set $\Phi$ if $\varphi$ is not a tautological consequence of $\Phi$. That is, you can't prove $\varphi$ from $\Phi$ as a set of axioms. If $\Phi$ is unsatisfiable, then it is inconsistent. This means you can prove a contradiction starting with $\Phi$. If you can prove a contradiction, you can prove anything, including $\varphi$. This means that no formula would be tautologically independent of $\Phi$.

Answer 3

$$\text{If every finite subset of $\Phi$ is satisfiable, then $\Phi$ is satisfiable.}$$

The immediate contrapositive is: $$\text{If $\Phi$ is unsatisfiable, then not every finite subset of $\Phi$ is satisfiable.}$$

Looking at the consequent: the negation of "every Widget is" is "some Widget is not".   Refer to quantifier duality.$$\text{If $\Phi$ is unsatisfiable, then some finite subset of $\Phi$ is unsatisfiable.}$$