Question

Show that A1, ,An-taut Biff는 B.

Show that A1, ..., An |=taut B iff |=taut A1 ->A2 ->...->An -> B.

Answer 1

Proof.

It is an easy semantic exercise to see that the hypothesis yields (indeed we have done so in class) that

|=taut A1 → A2 . . . → An → B

   |-  A1 → A2 . . . → An → B

hence (by Hypothesis Strengthening)

  A1, . . . , An |- A1 → . . . → An → B

Applying modus ponens n times to (1) we get

  A1, . . . , An |- B

The above corollary is very convenient. It says that any (correct) schema A1, . . . , An |=taut B leads to a derived rule of inference, A1, . . . , An |- B.

In particular, combining with the “transitivity of `” Metatheorem known from class and text, we get