Theorem 3anibar 1123

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 957	. . 3 |- ((ph /\ ps /\ ch) -> ch)
3	2	biantrurd 494	. 2 |- ((ph /\ ps /\ ch) -> (ta <-> (ch /\ ta)))
4	1, 3	bitr4d 247	1 |- ((ph /\ ps /\ ch) -> (th <-> ta))

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358   /\ w3a 934

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  df-3an 936

This theorem is referenced by: (None)