Theorem ibib 589

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		pm3.4 331	. . . . 5 |- ((ph /\ ps) -> (ph -> ps))
2		pm3.26 319	. . . . . 6 |- ((ph /\ ps) -> ph)
3	2	a1d 12	. . . . 5 |- ((ph /\ ps) -> (ps -> ph))
4	1, 3	impbid 515	. . . 4 |- ((ph /\ ps) -> (ph <-> ps))
5	4	ex 373	. . 3 |- (ph -> (ps -> (ph <-> ps)))
6		bi1 148	. . . 4 |- ((ph <-> ps) -> (ph -> ps))
7	6	com12 11	. . 3 |- (ph -> ((ph <-> ps) -> ps))
8	5, 7	impbid 515	. 2 |- (ph -> (ps <-> (ph <-> ps)))
9	8	pm5.74i 583	1 |- ((ph -> ps) <-> (ph -> (ph <-> ps)))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223

This theorem is referenced by:  ibibr 590  ibd 593  pm5.501 594  zneo 6157  zneoOLD 6158

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 147  df-an 225

Copyright terms: Public domain