Theorem 3anibar 1155

Description: Remove a hypothesis from the second member of a biconditional. (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 989	. . 3 |- ( ( ph /\ ps /\ ch ) -> ch )
3	2	biantrurd 303	. 2 |- ( ( ph /\ ps /\ ch ) -> ( ta <-> ( ch /\ ta ) ) )
4	1, 3	bitr4d 190	1 |- ( ( ph /\ ps /\ ch ) -> ( th <-> ta ) )

Syntax hints:   -> wi 4   /\ wa 103   <-> wb 104   /\ w3a 968

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116  df-3an 970

This theorem is referenced by:  frecsuclem  6374  shftfibg  10762  neiint  12795