Consistent theory based on L and not(A->A) is a theorem

Question

I am working on this problem in which I have a theory $T$ based on language $\mathcal{L}$ and the only information we have is that T is consistent and $\vdash \lnot(A \rightarrow A)$. Given this information, how can I know if this theory is sound, complete and/or decidable?

My only guess is that I can say that $T$ is sound because since $T$ is consistent and we can derive $\lnot(A \rightarrow A)$ from axioms and inference rules, $\lnot(A \rightarrow A)$ is a theorem and because of that we can assume that $T$ is sound (because the premises and conclusions are true).

Am I correct? What else can I say about this theory?

Thank you!

Answer 1

I'll assume that we're working in the context of classical propositional logic.

The first thing to point out is that $\lnot(A \rightarrow A)$ leads to a contradiction: ($A \land \lnot A$). Since propositional logic itself is consistent, it is impossible to derive a contradiction, so $\vdash \lnot(A \rightarrow A)$ is impossible.

So perhaps your intention was $T \vdash \lnot(A \rightarrow A)$. This can only happen if $T$ is itself inconsistent.

Given that $T$ is inconsistent, $T$ is vacuously complete (since an inconsistent theory can prove any formula) and decidable (for the same reason). As for soundness, it's a property of the logical system, not any particular theory like $T$. Propositional logic is sound.