Theorem bianabs 653

Description: Absorb a hypothesis into the second member of a biconditional. (Contributed by FL, 15-Feb-2007.)

Hypothesis

Ref	Expression
bianabs.1	|- (ph -> (ps <-> (ph /\ ch)))

Assertion

Ref	Expression
bianabs	|- (ph -> (ps <-> ch))

Proof of Theorem bianabs

Step	Hyp	Ref	Expression
1		bianabs.1	. 2 |- (ph -> (ps <-> (ph /\ ch)))
2		ibar 643	. 2 |- (ph -> (ch <-> (ph /\ ch)))
3	1, 2	bitr4d 531	1 |- (ph -> (ps <-> ch))

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

This theorem is referenced by:  ceqsrexv 1889  oprabval 4023  brecop 4306  ltprord 5134  clm2 7078  isph 8481  hlim2 9060  cmbrt 9527  cvbrt 10209  mdbrt 10221  dmdbrt 10226  hmph 10524

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

Copyright terms: Public domain