Theorem ompfl3 10422

Description: Remove a hypothesis from the second member of a biimplication.

Hypothesis

Ref	Expression
ompfl3.1	|- ((ph /\ ps /\ ch) -> (th <-> (ch /\ ta)))

Assertion

Ref	Expression
ompfl3	|- ((ph /\ ps /\ ch) -> (th <-> ta))

Proof of Theorem ompfl3

Step	Hyp	Ref	Expression
1		ompfl3.1	. 2 |- ((ph /\ ps /\ ch) -> (th <-> (ch /\ ta)))
2		ibar 645	. . 3 |- (ch -> (ta <-> (ch /\ ta)))
3	2	3ad2ant3 804	. 2 |- ((ph /\ ps /\ ch) -> (ta <-> (ch /\ ta)))
4	1, 3	bitr4d 533	1 |- ((ph /\ ps /\ ch) -> (th <-> ta))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223   /\ w3a 777

This theorem is referenced by:  plimfilOLD 10580

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7

This theorem depends on definitions:  df-bi 147  df-an 225  df-3an 779

Copyright terms: Public domain