Theorem baibd 875

Description: Move conjunction outside of biconditional. (Contributed by Mario Carneiro, 11-Sep-2015.)

Hypothesis

Ref	Expression
baibd.1	|- (ph -> (ps <-> (ch /\ th)))

Assertion

Ref	Expression
baibd	|- ((ph /\ ch) -> (ps <-> th))

Proof of Theorem baibd

Step	Hyp	Ref	Expression
1		baibd.1	. 2 |- (ph -> (ps <-> (ch /\ th)))
2		ibar 490	. . 3 |- (ch -> (th <-> (ch /\ th)))
3	2	bicomd 192	. 2 |- (ch -> ((ch /\ th) <-> th))
4	1, 3	sylan9bb 680	1 |- ((ph /\ ch) -> (ps <-> th))

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 177  df-an 360

This theorem is referenced by: (None)