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.

  • This looks lke a parsing error to me: ⊨ A → B means ⊨ (A → B), not (⊨ A) → B. – Uwe Jan 21 '20 at 11:15

The assumption "A ⊨ B" does not mean that A is a tautology.

You have assumed f(A)=1, that means that A is true for an interpretation f.

The same for the second case : "When ⊨A → B, B must be true since ⊨A is a tautology" is wrong.

A → B is a tautology: this means that, whenever A is true also B is, and this is enough to conclude with A⊨B.

  • Doesn't the notation, ⊨A, mean that A is a consequence of the empty set, so ⊨A would be a tautology? – N. Bar Jan 15 '20 at 19:07

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