Theorem bianabs 619

Description: Absorb a hypothesis into the second member of a biconditional. (Contributed by FL, 15-Feb-2007.)

Hypothesis

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

Assertion

Ref	Expression
bianabs	|- ( ph -> ( ps <-> ch ) )

Proof of Theorem bianabs

Step	Hyp	Ref	Expression
1		bianabs.1	. 2 |- ( ph -> ( ps <-> ( ph /\ ch ) ) )
2		ibar 301	. 2 |- ( ph -> ( ch <-> ( ph /\ ch ) ) )
3	1, 2	bitr4d 191	1 |- ( ph -> ( ps <-> ch ) )

This proof depends on syntax axioms:   -> wi 4   /\ wa 104   <-> wb 105

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This proof depends on definitions:  df-bi 117

This theorem is used by:  ceqsrexv  2956  opelopab2a  4407  ov  6208  ovg  6228  ltresr  8206