Theorem ifeq3da 29365

Description: Given an expression C containing if ( ps , E , F ), substitute (hypotheses .1 and .2) and evaluate (hypotheses .3 and .4) it for both cases at the same time. (Contributed by Thierry Arnoux, 13-Dec-2021.)

Hypotheses

Ref	Expression
ifeq3da.1	|- ( if ( ps , E , F ) = E -> C = G )
ifeq3da.2	|- ( if ( ps , E , F ) = F -> C = H )
ifeq3da.3	|- ( ph -> G = A )
ifeq3da.4	|- ( ph -> H = B )

Assertion

Ref	Expression
ifeq3da	|- ( ph -> if ( ps , A , B ) = C )

Proof of Theorem ifeq3da

Step	Hyp	Ref	Expression
1		iftrue 4092	. . . . 5 |- ( ps -> if ( ps , E , F ) = E )
2		ifeq3da.1	. . . . 5 |- ( if ( ps , E , F ) = E -> C = G )
3	1, 2	syl 17	. . . 4 |- ( ps -> C = G )
4	3	adantl 482	. . 3 |- ( ( ph /\ ps ) -> C = G )
5		ifeq3da.3	. . . 4 |- ( ph -> G = A )
6	5	adantr 481	. . 3 |- ( ( ph /\ ps ) -> G = A )
7	4, 6	eqtr2d 2657	. 2 |- ( ( ph /\ ps ) -> A = C )
8		iffalse 4095	. . . . 5 |- ( -. ps -> if ( ps , E , F ) = F )
9		ifeq3da.2	. . . . 5 |- ( if ( ps , E , F ) = F -> C = H )
10	8, 9	syl 17	. . . 4 |- ( -. ps -> C = H )
11	10	adantl 482	. . 3 |- ( ( ph /\ -. ps ) -> C = H )
12		ifeq3da.4	. . . 4 |- ( ph -> H = B )
13	12	adantr 481	. . 3 |- ( ( ph /\ -. ps ) -> H = B )
14	11, 13	eqtr2d 2657	. 2 |- ( ( ph /\ -. ps ) -> B = C )
15	7, 14	ifeqda 4121	1 |- ( ph -> if ( ps , A , B ) = C )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 384   = wceq 1483   ifcif 4086

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-if 4087

This theorem is referenced by:  circlemeth  30718