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Theorem ifpor 1429
Description: The conditional operator implies the disjunction of its possible outputs. Dual statement of anifp 1428. (Contributed by BJ, 1-Oct-2019.)
Assertion
Ref Expression
ifpor  |-  (if- (
ph ,  ps ,  ch )  ->  ( ps  \/  ch ) )

Proof of Theorem ifpor
StepHypRef Expression
1 df-ifp 1421 . 2  |-  (if- (
ph ,  ps ,  ch )  <->  ( ( ph  /\ 
ps )  \/  ( -.  ph  /\  ch )
) )
2 simpr 462 . . 3  |-  ( (
ph  /\  ps )  ->  ps )
3 simpr 462 . . 3  |-  ( ( -.  ph  /\  ch )  ->  ch )
42, 3orim12i 518 . 2  |-  ( ( ( ph  /\  ps )  \/  ( -.  ph 
/\  ch ) )  -> 
( ps  \/  ch ) )
51, 4sylbi 198 1  |-  (if- (
ph ,  ps ,  ch )  ->  ( ps  \/  ch ) )
Colors of variables: wff setvar class
Syntax hints:   -. wn 3    -> wi 4    \/ wo 369    /\ wa 370  if-wif 1420
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 188  df-or 371  df-an 372  df-ifp 1421
This theorem is referenced by: (None)
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