Theorem 3anibar 1106

Description: Remove a hypothesis from the second member of a biimplication. (Contributed by FL, 22-Jul-2008.)

Hypothesis

Ref	Expression
3anibar.1	|- ( ( ph /\ ps /\ ch ) -> ( th <-> ( ch /\ ta ) ) )

Assertion

Ref	Expression
3anibar	|- ( ( ph /\ ps /\ ch ) -> ( th <-> ta ) )

Proof of Theorem 3anibar

Step	Hyp	Ref	Expression
1		3anibar.1	. 2 |- ( ( ph /\ ps /\ ch ) -> ( th <-> ( ch /\ ta ) ) )
2		simp3 940	. . 3 |- ( ( ph /\ ps /\ ch ) -> ch )
3	2	biantrurd 299	. 2 |- ( ( ph /\ ps /\ ch ) -> ( ta <-> ( ch /\ ta ) ) )
4	1, 3	bitr4d 189	1 |- ( ( ph /\ ps /\ ch ) -> ( th <-> ta ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103   /\ w3a 919

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106

This theorem depends on definitions:  df-bi 115  df-3an 921

This theorem is referenced by:  frecsuclem3  6013  shftfibg  9708