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

Step	Hyp	Ref	Expression
1		iftrue 3947	. . . 4
2	1	orcs 394	. . 3
3		iftrue 3947	. . 3
4	2, 3	eqtr4d 2501	. 2
5		iffalse 3950	. . 3
6		biorf 405	. . . 4
7	6	ifbid 3963	. . 3
8	5, 7	eqtr2d 2499	. 2
9	4, 8	pm2.61i 164	1

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