Theorem ifsbdc 3363

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 777	. 2 |- (DECID ph -> ( ph \/ -. ph ) )
2		iftrue 3356	. . . . 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 3356	. . . 4 |- ( ph -> if ( ph , D , E ) = D )
6	4, 5	eqtr4d 2116	. . 3 |- ( ph -> C = if ( ph , D , E ) )
7		iffalse 3359	. . . . 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 3359	. . . 4 |- ( -. ph -> if ( ph , D , E ) = E )
11	9, 10	eqtr4d 2116	. . 3 |- ( -. ph -> C = if ( ph , D , E ) )
12	6, 11	jaoi 668	. 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 661  DECID wdc 775   = wceq 1284   ifcif 3351

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-dc 776  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-if 3352

This theorem is referenced by:  fvifdc  5217