Having trouble with the forward direction of this proof. I assume that the decision problem for implication is solvable, so that for any set of sentences $T$, I can arrive at a yes or no answer to whether or not for any sentence $A$, $T \models A$.
I'm having trouble thinking of a way to prove this. I need to think of a sentence $A$ that can help me decide whether $T$ is valid or not. I've tried letting $A$ be a sentence that is true in every interpretation but that leads me nowhere since even if we knew $T \models A$, there could still be a member of $T$ that is false. Letting $A$ be a sentence that is false would be useful for showing that the decision problem for satisfiability is solvable, but this is not the case.
Any help that would lead me to being able to prove this would be greatly appreciated! Thanks in advance.
You want to show that if you can solve the problem of that takes a (finite?) set of sentences $T$ and a sentence $A$ and returns whether or not $T\vDash A$, then you can solve the problem that takes a sentence $A$ and returns whether or not $\vDash A$?
You can either solve the second problem using the first by setting $T = \emptyset$, or, if that is not allowed for some reason, let $T = \{p \lor \neg p\}$ (or any set of tautologies).
The meaty direction of the proof is the other way, using problem two to solve problem one.