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Theorem ifor 3771
Description: Rewrite a disjunction in an if statement as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifor  |-  if ( ( ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B )
)

Proof of Theorem ifor
StepHypRef Expression
1 iftrue 3737 . . . 4  |-  ( (
ph  \/  ps )  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  A )
21orcs 384 . . 3  |-  ( ph  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  A )
3 iftrue 3737 . . 3  |-  ( ph  ->  if ( ph ,  A ,  if ( ps ,  A ,  B ) )  =  A )
42, 3eqtr4d 2470 . 2  |-  ( ph  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B )
) )
5 iffalse 3738 . . 3  |-  ( -. 
ph  ->  if ( ph ,  A ,  if ( ps ,  A ,  B ) )  =  if ( ps ,  A ,  B )
)
6 biorf 395 . . . 4  |-  ( -. 
ph  ->  ( ps  <->  ( ph  \/  ps ) ) )
76ifbid 3749 . . 3  |-  ( -. 
ph  ->  if ( ps ,  A ,  B
)  =  if ( ( ph  \/  ps ) ,  A ,  B ) )
85, 7eqtr2d 2468 . 2  |-  ( -. 
ph  ->  if ( (
ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B ) ) )
94, 8pm2.61i 158 1  |-  if ( ( ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B )
)
Colors of variables: wff set class
Syntax hints:   -. wn 3    \/ wo 358    = wceq 1652   ifcif 3731
This theorem is referenced by:  cantnflem1d  7636  cantnflem1  7637
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1555  ax-5 1566  ax-17 1626  ax-9 1666  ax-8 1687  ax-6 1744  ax-7 1749  ax-11 1761  ax-12 1950  ax-ext 2416
This theorem depends on definitions:  df-bi 178  df-or 360  df-an 361  df-tru 1328  df-ex 1551  df-nf 1554  df-sb 1659  df-clab 2422  df-cleq 2428  df-clel 2431  df-if 3732
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