Theorem ifandc 3563

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

Assertion

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

Proof of Theorem ifandc

Step	Hyp	Ref	Expression
1		df-dc 830	. 2 |- (DECID ph <-> ( ph \/ -. ph ) )
2		iftrue 3531	. . . 4 |- ( ph -> if ( ph , if ( ps , A , B ) , B ) = if ( ps , A , B ) )
3		ibar 299	. . . . 5 |- ( ph -> ( ps <-> ( ph /\ ps ) ) )
4	3	ifbid 3547	. . . 4 |- ( ph -> if ( ps , A , B ) = if ( ( ph /\ ps ) , A , B ) )
5	2, 4	eqtr2d 2204	. . 3 |- ( ph -> if ( ( ph /\ ps ) , A , B ) = if ( ph , if ( ps , A , B ) , B ) )
6		simpl 108	. . . . . 6 |- ( ( ph /\ ps ) -> ph )
7	6	con3i 627	. . . . 5 |- ( -. ph -> -. ( ph /\ ps ) )
8	7	iffalsed 3536	. . . 4 |- ( -. ph -> if ( ( ph /\ ps ) , A , B ) = B )
9		iffalse 3534	. . . 4 |- ( -. ph -> if ( ph , if ( ps , A , B ) , B ) = B )
10	8, 9	eqtr4d 2206	. . 3 |- ( -. ph -> if ( ( ph /\ ps ) , A , B ) = if ( ph , if ( ps , A , B ) , B ) )
11	5, 10	jaoi 711	. 2 |- ( ( ph \/ -. ph ) -> if ( ( ph /\ ps ) , A , B ) = if ( ph , if ( ps , A , B ) , B ) )
12	1, 11	sylbi 120	1 |- (DECID ph -> if ( ( ph /\ ps ) , A , B ) = if ( ph , if ( ps , A , B ) , B ) )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 103   \/ 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:  isumss  11354