Theorem ifor 4135

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 4092	. . . 4 |- ( ( ph \/ ps ) -> if ( ( ph \/ ps ) , A , B ) = A )
2	1	orcs 409	. . 3 |- ( ph -> if ( ( ph \/ ps ) , A , B ) = A )
3		iftrue 4092	. . 3 |- ( ph -> if ( ph , A , if ( ps , A , B ) ) = A )
4	2, 3	eqtr4d 2659	. 2 |- ( ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )
5		iffalse 4095	. . 3 |- ( -. ph -> if ( ph , A , if ( ps , A , B ) ) = if ( ps , A , B ) )
6		biorf 420	. . . 4 |- ( -. ph -> ( ps <-> ( ph \/ ps ) ) )
7	6	ifbid 4108	. . 3 |- ( -. ph -> if ( ps , A , B ) = if ( ( ph \/ ps ) , A , B ) )
8	5, 7	eqtr2d 2657	. 2 |- ( -. ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )
9	4, 8	pm2.61i 176	1 |- if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) )

Syntax hints:   -. wn 3   \/ wo 383   = wceq 1483   ifcif 4086

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-if 4087

This theorem is referenced by:  cantnflem1d  8585  cantnflem1  8586