Theorem ifordc 3575

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 836	. 2 |- (DECID ph -> ( ph \/ -. ph ) )
2		iftrue 3541	. . . . 5 |- ( ( ph \/ ps ) -> if ( ( ph \/ ps ) , A , B ) = A )
3	2	orcs 735	. . . 4 |- ( ph -> if ( ( ph \/ ps ) , A , B ) = A )
4		iftrue 3541	. . . 4 |- ( ph -> if ( ph , A , if ( ps , A , B ) ) = A )
5	3, 4	eqtr4d 2213	. . 3 |- ( ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )
6		iffalse 3544	. . . 4 |- ( -. ph -> if ( ph , A , if ( ps , A , B ) ) = if ( ps , A , B ) )
7		biorf 744	. . . . 5 |- ( -. ph -> ( ps <-> ( ph \/ ps ) ) )
8	7	ifbid 3557	. . . 4 |- ( -. ph -> if ( ps , A , B ) = if ( ( ph \/ ps ) , A , B ) )
9	6, 8	eqtr2d 2211	. . 3 |- ( -. ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )
10	5, 9	jaoi 716	. 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 708  DECID wdc 834   = wceq 1353   ifcif 3536

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 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-11 1506  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-dc 835  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-if 3537

This theorem is referenced by:  nninfwlpoimlemg  7175