Theorem ifsbdc 3615

Description: Distribute a function over an if-clause. (Contributed by Jim Kingdon, 1-Jan-2022.)

Hypotheses

Ref	Expression
ifsbdc.1	|- ( if ( ph , A , B ) = A -> C = D )
ifsbdc.2	|- ( if ( ph , A , B ) = B -> C = E )

Assertion

Ref	Expression
ifsbdc	|- (DECID ph -> C = if ( ph , D , E ) )

Proof of Theorem ifsbdc

Step	Hyp	Ref	Expression
1		exmiddc 841	. 2 |- (DECID ph -> ( ph \/ -. ph ) )
2		iftrue 3607	. . . . 5 |- ( ph -> if ( ph , A , B ) = A )
3		ifsbdc.1	. . . . 5 |- ( if ( ph , A , B ) = A -> C = D )
4	2, 3	syl 14	. . . 4 |- ( ph -> C = D )
5		iftrue 3607	. . . 4 |- ( ph -> if ( ph , D , E ) = D )
6	4, 5	eqtr4d 2265	. . 3 |- ( ph -> C = if ( ph , D , E ) )
7		iffalse 3610	. . . . 5 |- ( -. ph -> if ( ph , A , B ) = B )
8		ifsbdc.2	. . . . 5 |- ( if ( ph , A , B ) = B -> C = E )
9	7, 8	syl 14	. . . 4 |- ( -. ph -> C = E )
10		iffalse 3610	. . . 4 |- ( -. ph -> if ( ph , D , E ) = E )
11	9, 10	eqtr4d 2265	. . 3 |- ( -. ph -> C = if ( ph , D , E ) )
12	6, 11	jaoi 721	. 2 |- ( ( ph \/ -. ph ) -> C = if ( ph , D , E ) )
13	1, 12	syl 14	1 |- (DECID ph -> C = if ( ph , D , E ) )

Syntax hints:   -. wn 3   -> wi 4   \/ wo 713  DECID wdc 839   = wceq 1395   ifcif 3602

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-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-11 1552  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-dc 840  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-if 3603

This theorem is referenced by:  fvifdc  5649  lgsneg  15703  lgsdilem  15706