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Theorem ifandc 3478
Description: Rewrite a conjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifandc  |-  (DECID  ph  ->  if ( ( ph  /\  ps ) ,  A ,  B )  =  if ( ph ,  if ( ps ,  A ,  B ) ,  B
) )

Proof of Theorem ifandc
StepHypRef Expression
1 df-dc 805 . 2  |-  (DECID  ph  <->  ( ph  \/  -.  ph ) )
2 iftrue 3449 . . . 4  |-  ( ph  ->  if ( ph ,  if ( ps ,  A ,  B ) ,  B
)  =  if ( ps ,  A ,  B ) )
3 ibar 299 . . . . 5  |-  ( ph  ->  ( ps  <->  ( ph  /\ 
ps ) ) )
43ifbid 3463 . . . 4  |-  ( ph  ->  if ( ps ,  A ,  B )  =  if ( ( ph  /\ 
ps ) ,  A ,  B ) )
52, 4eqtr2d 2151 . . 3  |-  ( ph  ->  if ( ( ph  /\ 
ps ) ,  A ,  B )  =  if ( ph ,  if ( ps ,  A ,  B ) ,  B
) )
6 simpl 108 . . . . . 6  |-  ( (
ph  /\  ps )  ->  ph )
76con3i 606 . . . . 5  |-  ( -. 
ph  ->  -.  ( ph  /\ 
ps ) )
87iffalsed 3454 . . . 4  |-  ( -. 
ph  ->  if ( (
ph  /\  ps ) ,  A ,  B )  =  B )
9 iffalse 3452 . . . 4  |-  ( -. 
ph  ->  if ( ph ,  if ( ps ,  A ,  B ) ,  B )  =  B )
108, 9eqtr4d 2153 . . 3  |-  ( -. 
ph  ->  if ( (
ph  /\  ps ) ,  A ,  B )  =  if ( ph ,  if ( ps ,  A ,  B ) ,  B ) )
115, 10jaoi 690 . 2  |-  ( (
ph  \/  -.  ph )  ->  if ( ( ph  /\ 
ps ) ,  A ,  B )  =  if ( ph ,  if ( ps ,  A ,  B ) ,  B
) )
121, 11sylbi 120 1  |-  (DECID  ph  ->  if ( ( ph  /\  ps ) ,  A ,  B )  =  if ( ph ,  if ( ps ,  A ,  B ) ,  B
) )
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4    /\ wa 103    \/ wo 682  DECID wdc 804    = wceq 1316   ifcif 3444
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 588  ax-in2 589  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-11 1469  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-dc 805  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-if 3445
This theorem is referenced by:  isumss  11115
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