Theorem ifordc 3564

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 831	. 2 |- (DECID ph -> ( ph \/ -. ph ) )
2		iftrue 3531	. . . . 5 |- ( ( ph \/ ps ) -> if ( ( ph \/ ps ) , A , B ) = A )
3	2	orcs 730	. . . 4 |- ( ph -> if ( ( ph \/ ps ) , A , B ) = A )
4		iftrue 3531	. . . 4 |- ( ph -> if ( ph , A , if ( ps , A , B ) ) = A )
5	3, 4	eqtr4d 2206	. . 3 |- ( ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )
6		iffalse 3534	. . . 4 |- ( -. ph -> if ( ph , A , if ( ps , A , B ) ) = if ( ps , A , B ) )
7		biorf 739	. . . . 5 |- ( -. ph -> ( ps <-> ( ph \/ ps ) ) )
8	7	ifbid 3547	. . . 4 |- ( -. ph -> if ( ps , A , B ) = if ( ( ph \/ ps ) , A , B ) )
9	6, 8	eqtr2d 2204	. . 3 |- ( -. ph -> if ( ( ph \/ ps ) , A , B ) = if ( ph , A , if ( ps , A , B ) ) )
10	5, 9	jaoi 711	. 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 703  DECID wdc 829   = wceq 1348   ifcif 3526

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-11 1499  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-dc 830  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-if 3527

This theorem is referenced by:  nninfwlpoimlemg  7151