Let A and B represent arbitrary formulas.
Also let 1 ≡ True and 0 ≡ False
Prove that A ⊨ B iff ⊨A → B
For my proof, I break down the biconditional into two conditionals and prove each conditional.
(A ⊨ B)→(⊨A → B)
(⊨A → B)→(A ⊨ B)
Assume (A ⊨ B)
Thus, by the definition of semantic consequence, there is no interpretation, f, of A and B such that f(B) = 0 while f(A) = 1
⊨A is a tautology, so it is always true by definition
Also, by definition, a material conditional is false iff the antecedent is true while the consequent is false
Since B is never false when A is true, we can infer ⊨A → B
When ⊨A → B, B must be true since ⊨A is a tautology
Thus, because B is always true, we can infer A⊨B
Therefore, A ⊨ B iff ⊨A → B ■
The italicized lines are the ones I'm not so sure about. For the first one, I would assume that A is not the same as ⊨A, so would it really be proper to make the inference I did?
The second italicized lines makes me a little anxious since it wasn't established (at least I don't think so) that A is always true, so there could be the possibility that A is false while B is true.
Thanks for your help in verifying my proof.