Theorem ifordc 3601

Description: Rewrite a disjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.)

Assertion

Ref	Expression
ifordc	|- (DECID ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )

Proof of Theorem ifordc

Step	Hyp	Ref	Expression
1		exmiddc 837	. 2 |- (DECID ph -> ( ph \/ -. ph ) )
2		iftrue 3567	. . . . 5 |- ( ( ph \/ ps ) -> if ( ( ph \/ ps ) , A , B ) = A )
3	2	orcs 736	. . . 4 |- ( ph -> if ( ( ph \/ ps ) , A , B ) = A )
4		iftrue 3567	. . . 4 |- ( ph -> if ( ph , A , if ( ps , A , B ) ) = A )
5	3, 4	eqtr4d 2232	. . 3 |- ( ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )
6		iffalse 3570	. . . 4 |- ( -. ph -> if ( ph , A , if ( ps , A , B ) ) = if ( ps , A , B ) )
7		biorf 745	. . . . 5 |- ( -. ph -> ( ps <-> ( ph \/ ps ) ) )
8	7	ifbid 3583	. . . 4 |- ( -. ph -> if ( ps , A , B ) = if ( ( ph \/ ps ) , A , B ) )
9	6, 8	eqtr2d 2230	. . 3 |- ( -. ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )
10	5, 9	jaoi 717	. 2 |- ( ( ph \/ -. ph ) -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )
11	1, 10	syl 14	1 |- (DECID ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )

Syntax hints:   -. wn 3   -> wi 4   \/ wo 709  DECID wdc 835   = wceq 1364   ifcif 3562

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-11 1520  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-dc 836  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-if 3563

This theorem is referenced by:  nninfwlpoimlemg  7250