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Theorem ifordc 3573
Description: Rewrite a disjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifordc  |-  (DECID  ph  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B )
) )

Proof of Theorem ifordc
StepHypRef Expression
1 exmiddc 836 . 2  |-  (DECID  ph  ->  (
ph  \/  -.  ph )
)
2 iftrue 3539 . . . . 5  |-  ( (
ph  \/  ps )  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  A )
32orcs 735 . . . 4  |-  ( ph  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  A )
4 iftrue 3539 . . . 4  |-  ( ph  ->  if ( ph ,  A ,  if ( ps ,  A ,  B ) )  =  A )
53, 4eqtr4d 2213 . . 3  |-  ( ph  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B )
) )
6 iffalse 3542 . . . 4  |-  ( -. 
ph  ->  if ( ph ,  A ,  if ( ps ,  A ,  B ) )  =  if ( ps ,  A ,  B )
)
7 biorf 744 . . . . 5  |-  ( -. 
ph  ->  ( ps  <->  ( ph  \/  ps ) ) )
87ifbid 3555 . . . 4  |-  ( -. 
ph  ->  if ( ps ,  A ,  B
)  =  if ( ( ph  \/  ps ) ,  A ,  B ) )
96, 8eqtr2d 2211 . . 3  |-  ( -. 
ph  ->  if ( (
ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B ) ) )
105, 9jaoi 716 . 2  |-  ( (
ph  \/  -.  ph )  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B )
) )
111, 10syl 14 1  |-  (DECID  ph  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B )
) )
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4    \/ wo 708  DECID wdc 834    = wceq 1353   ifcif 3534
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-11 1506  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-dc 835  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-if 3535
This theorem is referenced by:  nninfwlpoimlemg  7172
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