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Theorem condisd 24943
Description: Proof by contradiction combined with a disjunction. (Contributed by FL, 20-Apr-2011.)
Hypotheses
Ref Expression
condisd.1  |-  ( (
ph  /\  ps )  ->  ch )
condisd.2  |-  ( (
ph  /\  -.  ps )  ->  th )
Assertion
Ref Expression
condisd  |-  ( ph  ->  ( ch  \/  th ) )

Proof of Theorem condisd
StepHypRef Expression
1 exmid 404 . 2  |-  ( ps  \/  -.  ps )
2 andi 837 . . 3  |-  ( (
ph  /\  ( ps  \/  -.  ps ) )  <-> 
( ( ph  /\  ps )  \/  ( ph  /\  -.  ps )
) )
3 condisd.1 . . . 4  |-  ( (
ph  /\  ps )  ->  ch )
4 condisd.2 . . . 4  |-  ( (
ph  /\  -.  ps )  ->  th )
53, 4orim12i 502 . . 3  |-  ( ( ( ph  /\  ps )  \/  ( ph  /\ 
-.  ps ) )  -> 
( ch  \/  th ) )
62, 5sylbi 187 . 2  |-  ( (
ph  /\  ( ps  \/  -.  ps ) )  ->  ( ch  \/  th ) )
71, 6mpan2 652 1  |-  ( ph  ->  ( ch  \/  th ) )
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4    \/ wo 357    /\ 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-or 359  df-an 360
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