Theorem ibib 650

Description: Implication in terms of implication and biconditional.

Assertion

Ref	Expression
ibib	|- ((ph -> ps) <-> (ph -> (ph <-> ps)))

Proof of Theorem ibib

Step	Hyp	Ref	Expression
1		pm3.4 358	. . . . 5 |- ((ph /\ ps) -> (ph -> ps))
2		simpl 346	. . . . . 6 |- ((ph /\ ps) -> ph)
3	2	a1d 15	. . . . 5 |- ((ph /\ ps) -> (ps -> ph))
4	1, 3	impbid 574	. . . 4 |- ((ph /\ ps) -> (ph <-> ps))
5	4	ex 402	. . 3 |- (ph -> (ps -> (ph <-> ps)))
6		bi1 165	. . . 4 |- ((ph <-> ps) -> (ph -> ps))
7	6	com12 14	. . 3 |- (ph -> ((ph <-> ps) -> ps))
8	5, 7	impbid 574	. 2 |- (ph -> (ps <-> (ph <-> ps)))
9	8	pm5.74i 644	1 |- ((ph -> ps) <-> (ph -> (ph <-> ps)))

Syntax hints:   -> wi 3   <-> wb 163   /\ wa 240

This theorem is referenced by:  ibibr 651  ibd 654  pm5.501 655  zneo 7412

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7

This theorem depends on definitions:  df-bi 164  df-an 242

Copyright terms: Public domain