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Theorem ifor 3969
 Description: Rewrite a disjunction in an if statement as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)
Assertion
Ref Expression
ifor

Proof of Theorem ifor
StepHypRef Expression
1 iftrue 3928 . . . 4
21orcs 394 . . 3
3 iftrue 3928 . . 3
42, 3eqtr4d 2485 . 2
5 iffalse 3931 . . 3
6 biorf 405 . . . 4
76ifbid 3944 . . 3
85, 7eqtr2d 2483 . 2
94, 8pm2.61i 164 1
 Colors of variables: wff setvar class Syntax hints:   wn 3   wo 368   wceq 1381  cif 3922 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1603  ax-4 1616  ax-5 1689  ax-6 1732  ax-7 1774  ax-10 1821  ax-11 1826  ax-12 1838  ax-13 1983  ax-ext 2419 This theorem depends on definitions:  df-bi 185  df-or 370  df-an 371  df-tru 1384  df-ex 1598  df-nf 1602  df-sb 1725  df-clab 2427  df-cleq 2433  df-clel 2436  df-if 3923 This theorem is referenced by:  cantnflem1d  8105  cantnflem1  8106  cantnflem1dOLD  8128  cantnflem1OLD  8129
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