Can there be a formula $\psi$ such that $(\psi \rightarrow (¬\psi))$ is a theorem of L?

Question

Can there be a formula $\psi$ such that $(\psi \rightarrow (¬\psi))$ is a theorem of L?

I would like to check my answer. I thought:

No it cannot be a theorem of L. Because to be a theorem of L it must be a tautology but if we constructed the truth table we would see that it's not a tautology.

Answer 1

Yes, it can be theorem, let $\psi=(p\land (\neg p))$. The truth table for this will verify that $(\psi \rightarrow (¬\psi))$ is true and completeness will ensure it is a theorem.

The problem with your argument is that you want to attribute truth values to the formula $\psi$, but you should be attributing them to the propositional letters that occur in $(\psi \rightarrow (¬\psi))$. This is how valuations are defined.

Answer 2

A minor variation on Git Gud's theme: we have ⊥ ⊢ φ, and in particular ⊥ ⊢ $\neg \psi$, so if $\psi$ is false that would do it.