Theorem ifandc 3508

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 820	. 2 |- (DECID ph <-> ( ph \/ -. ph ) )
2		iftrue 3479	. . . 4 |- ( ph -> if ( ph , if ( ps , A , B ) , B ) = if ( ps , A , B ) )
3		ibar 299	. . . . 5 |- ( ph -> ( ps <-> ( ph /\ ps ) ) )
4	3	ifbid 3493	. . . 4 |- ( ph -> if ( ps , A , B ) = if ( ( ph /\ ps ) , A , B ) )
5	2, 4	eqtr2d 2173	. . 3 |- ( ph -> if ( ( ph /\ ps ) , A , B ) = if ( ph , if ( ps , A , B ) , B ) )
6		simpl 108	. . . . . 6 |- ( ( ph /\ ps ) -> ph )
7	6	con3i 621	. . . . 5 |- ( -. ph -> -. ( ph /\ ps ) )
8	7	iffalsed 3484	. . . 4 |- ( -. ph -> if ( ( ph /\ ps ) , A , B ) = B )
9		iffalse 3482	. . . 4 |- ( -. ph -> if ( ph , if ( ps , A , B ) , B ) = B )
10	8, 9	eqtr4d 2175	. . 3 |- ( -. ph -> if ( ( ph /\ ps ) , A , B ) = if ( ph , if ( ps , A , B ) , B ) )
11	5, 10	jaoi 705	. 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 697  DECID wdc 819   = wceq 1331   ifcif 3474

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 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-11 1484  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-dc 820  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-if 3475

This theorem is referenced by:  isumss  11160