Theorem bibif 681

Description: Transfer negation via an equivalence.

Assertion

Ref	Expression
bibif	|- (-. ps -> ((ph <-> ps) <-> -. ph))

Proof of Theorem bibif

Step	Hyp	Ref	Expression
1		bi1 148	. . . 4 |- ((ph <-> ps) -> (ph -> ps))
2	1	con3d 95	. . 3 |- ((ph <-> ps) -> (-. ps -> -. ph))
3	2	com12 11	. 2 |- (-. ps -> ((ph <-> ps) -> -. ph))
4		pm5.21 677	. . 3 |- ((-. ph /\ -. ps) -> (ph <-> ps))
5	4	expcom 374	. 2 |- (-. ps -> (-. ph -> (ph <-> ps)))
6	3, 5	impbid 516	1 |- (-. ps -> ((ph <-> ps) <-> -. ph))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146

This theorem is referenced by:  ntreq0 7708  top2ind 10548

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