# $T\vdash\alpha$ iff there is a finite set $\{\alpha_1,...,\alpha_n\} \subseteq T$ such that $(\alpha_1\land ...\land \alpha_n) \to \alpha$ is valid.

Prove that $$T\vdash\alpha$$ iff there is a finite set $$\{\alpha_1,...,\alpha_n\}$$ of proper axioms of $$T$$ such that $$(\alpha_1\land ...\land \alpha_n) \to \alpha$$ is valid. Here, $$T$$ is a first order theory.

I have come up with the following proof, and it would be great if someone could help verify its correctness.

Proof:
If $$T\vdash\alpha$$, then by the compactness of $$\vdash$$, there are $$\{\alpha_1,...,\alpha_n\}\subseteq T$$, such that $$\{\alpha_1,...,\alpha_n\}\vdash \alpha$$. So, $$\vdash (\alpha_1 \land...\land \alpha_n) \to \alpha$$ follows from deduction theorem (and observing that $$\vdash A\to(B\to C) \leftrightarrow (A\land B)\to C$$, which can be generalized to $$n$$ terms).

If $$\vdash (\alpha_1 \land...\land \alpha_n) \to \alpha$$ holds, then by using deduction theorem (the other way round), we can arrive at $$\{\alpha_1,...,\alpha_n\}\vdash \alpha$$, and hence $$T\vdash \alpha$$ (dilution).

Does this seem fine? Please point out even the slightest of gaps and errors. Thank you!

• Looks fine to me. Side note, I do believe that there is a generalized version of the deduction theorem asserts the first step "for all $n$." I'll see if I can find it. Commented Dec 10, 2020 at 14:15
• Okay, that would be great! Commented Dec 10, 2020 at 14:16
• Yes, this is correct. Two minor notes. (1) The bit "which can be generalized to $n$ terms" can be made precise by induction. (2) Your use of compactness is correct, but in other (more popular) equivalent formulations compactness you might need to prove that you can use it in this form. So depending on what is the exact statement in your notes / book that might need a bit more work. Commented Dec 10, 2020 at 14:22
• Thank you @MarkKamsma Commented Dec 15, 2020 at 3:55