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Theorem ifor 3988
 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 3947 . . . 4
21orcs 394 . . 3
3 iftrue 3947 . . 3
42, 3eqtr4d 2501 . 2
5 iffalse 3950 . . 3
6 biorf 405 . . . 4
76ifbid 3963 . . 3
85, 7eqtr2d 2499 . 2
94, 8pm2.61i 164 1
 Colors of variables: wff setvar class Syntax hints:  -.wn 3  \/wo 368  =wceq 1395  ifcif 3941 This theorem is referenced by:  cantnflem1d  8128  cantnflem1  8129  cantnflem1dOLD  8151  cantnflem1OLD  8152 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1618  ax-4 1631  ax-5 1704  ax-6 1747  ax-7 1790  ax-10 1837  ax-11 1842  ax-12 1854  ax-13 1999  ax-ext 2435 This theorem depends on definitions:  df-bi 185  df-or 370  df-an 371  df-tru 1398  df-ex 1613  df-nf 1617  df-sb 1740  df-clab 2443  df-cleq 2449  df-clel 2452  df-if 3942
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