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!
