Theorem 3anibar 1196

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 1030	. . 3 |- ( ( ph /\ ps /\ ch ) -> ch )
3	2	biantrurd 305	. 2 |- ( ( ph /\ ps /\ ch ) -> ( ta <-> ( ch /\ ta ) ) )
4	1, 3	bitr4d 191	1 |- ( ( ph /\ ps /\ ch ) -> ( th <-> ta ) )

This proof depends on syntax axioms:   -> wi 4   /\ wa 104   <-> wb 105   /\ w3a 1009

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  df-3an 1011

This theorem is used by:  frecsuclem  6677  shftfibg  11585  neiint  15246