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Theorem 3anibar 1156
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
StepHypRef Expression
1 3anibar.1 . 2  |-  ( (
ph  /\  ps  /\  ch )  ->  ( th  <->  ( ch  /\ 
ta ) ) )
2 simp3 990 . . 3  |-  ( (
ph  /\  ps  /\  ch )  ->  ch )
32biantrurd 508 . 2  |-  ( (
ph  /\  ps  /\  ch )  ->  ( ta  <->  ( ch  /\ 
ta ) ) )
41, 3bitr4d 256 1  |-  ( (
ph  /\  ps  /\  ch )  ->  ( th  <->  ta )
)
Colors of variables: wff setvar class
Syntax hints:    -> wi 4    <-> wb 184    /\ wa 369    /\ w3a 965
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 185  df-an 371  df-3an 967
This theorem is referenced by:  neiint  18833  islinindfiss  31094  cpmatel  31177
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