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Theorem condandc 876
Description: Proof by contradiction. This only holds for decidable propositions, as it is part of the family of theorems which assume  -.  ps, derive a contradiction, and therefore conclude  ps. By contrast, assuming  ps, deriving a contradiction, and therefore concluding  -.  ps, as in pm2.65 654, is valid for all propositions. (Contributed by Jim Kingdon, 13-May-2018.)
Hypotheses
Ref Expression
condandc.1  |-  ( (
ph  /\  -.  ps )  ->  ch )
condandc.2  |-  ( (
ph  /\  -.  ps )  ->  -.  ch )
Assertion
Ref Expression
condandc  |-  (DECID  ps  ->  (
ph  ->  ps ) )

Proof of Theorem condandc
StepHypRef Expression
1 condandc.1 . . 3  |-  ( (
ph  /\  -.  ps )  ->  ch )
2 condandc.2 . . 3  |-  ( (
ph  /\  -.  ps )  ->  -.  ch )
31, 2pm2.65da 656 . 2  |-  ( ph  ->  -.  -.  ps )
4 notnotrdc 838 . 2  |-  (DECID  ps  ->  ( -.  -.  ps  ->  ps ) )
53, 4syl5 32 1  |-  (DECID  ps  ->  (
ph  ->  ps ) )
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4    /\ wa 103  DECID wdc 829
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 609  ax-in2 610  ax-io 704
This theorem depends on definitions:  df-bi 116  df-dc 830
This theorem is referenced by: (None)
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