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Theorem ifor 3747
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 3713 . . . 4  |-  ( (
ph  \/  ps )  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  A )
21orcs 384 . . 3  |-  ( ph  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  A )
3 iftrue 3713 . . 3  |-  ( ph  ->  if ( ph ,  A ,  if ( ps ,  A ,  B ) )  =  A )
42, 3eqtr4d 2447 . 2  |-  ( ph  ->  if ( ( ph  \/  ps ) ,  A ,  B )  =  if ( ph ,  A ,  if ( ps ,  A ,  B )
) )
5 iffalse 3714 . . 3  |-  ( -. 
ph  ->  if ( ph ,  A ,  if ( ps ,  A ,  B ) )  =  if ( ps ,  A ,  B )
)
6 biorf 395 . . . 4  |-  ( -. 
ph  ->  ( ps  <->  ( ph  \/  ps ) ) )
76ifbid 3725 . . 3  |-  ( -. 
ph  ->  if ( ps ,  A ,  B
)  =  if ( ( ph  \/  ps ) ,  A ,  B ) )
85, 7eqtr2d 2445 . 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 1649   ifcif 3707
This theorem is referenced by:  cantnflem1d  7608  cantnflem1  7609
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1552  ax-5 1563  ax-17 1623  ax-9 1662  ax-8 1683  ax-6 1740  ax-7 1745  ax-11 1757  ax-12 1946  ax-ext 2393
This theorem depends on definitions:  df-bi 178  df-or 360  df-an 361  df-tru 1325  df-ex 1548  df-nf 1551  df-sb 1656  df-clab 2399  df-cleq 2405  df-clel 2408  df-if 3708
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