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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
StepHypRef Expression
1 3anibar.1 . 2  |-  ( (
ph  /\  ps  /\  ch )  ->  ( th  <->  ( ch  /\ 
ta ) ) )
2 simp3 957 . . 3  |-  ( (
ph  /\  ps  /\  ch )  ->  ch )
32biantrurd 494 . 2  |-  ( (
ph  /\  ps  /\  ch )  ->  ( ta  <->  ( ch  /\ 
ta ) ) )
41, 3bitr4d 247 1  |-  ( (
ph  /\  ps  /\  ch )  ->  ( th  <->  ta )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 176    /\ wa 358    /\ w3a 934
This theorem is referenced by:  neiint  16841
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8
This theorem depends on definitions:  df-bi 177  df-an 360  df-3an 936
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