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Theorem bicomddOLD 26809
Description: Commute two sides of a biconditional in a deduction. (Contributed by Rodolfo Medina, 19-Oct-2010.) (Proof modification is discouraged.) (New usage is discouraged.)
Hypothesis
Ref Expression
bicomdd.1  |-  ( ph  ->  ( ps  ->  ( ch 
<->  th ) ) )
Assertion
Ref Expression
bicomddOLD  |-  ( ph  ->  ( ps  ->  ( th 
<->  ch ) ) )

Proof of Theorem bicomddOLD
StepHypRef Expression
1 bicomdd.1 . . . 4  |-  ( ph  ->  ( ps  ->  ( ch 
<->  th ) ) )
21imp 418 . . 3  |-  ( (
ph  /\  ps )  ->  ( ch  <->  th )
)
32bicomd 192 . 2  |-  ( (
ph  /\  ps )  ->  ( th  <->  ch )
)
43ex 423 1  |-  ( ph  ->  ( ps  ->  ( th 
<->  ch ) ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 176    /\ wa 358
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
  Copyright terms: Public domain W3C validator