Theorem ifor 3703

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

Step	Hyp	Ref	Expression
1		iftrue 3669	. . . 4 |- ((ph \/ ps) -> if((ph \/ ps), A, B) = A)
2	1	orcs 383	. . 3 |- (ph -> if((ph \/ ps), A, B) = A)
3		iftrue 3669	. . 3 |- (ph -> if(ph, A, if(ps, A, B)) = A)
4	2, 3	eqtr4d 2388	. 2 |- (ph -> if((ph \/ ps), A, B) = if(ph, A, if(ps, A, B)))
5		iffalse 3670	. . 3 |- (-. ph -> if(ph, A, if(ps, A, B)) = if(ps, A, B))
6		biorf 394	. . . 4 |- (-. ph -> (ps <-> (ph \/ ps)))
7	6	ifbid 3681	. . 3 |- (-. ph -> if(ps, A, B) = if((ph \/ ps), A, B))
8	5, 7	eqtr2d 2386	. 2 |- (-. ph -> if((ph \/ ps), A, B) = if(ph, A, if(ps, A, B)))
9	4, 8	pm2.61i 156	1 |- if((ph \/ ps), A, B) = if(ph, A, if(ps, A, B))

Syntax hints:  -. wn 3   \/ wo 357   = wceq 1642  ifcif 3663

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-if 3664

This theorem is referenced by: (None)