Theorem ifeq3da 30293

Description: Given an expression 𝐶 containing if(𝜓, 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(𝜓, 𝐸, 𝐹) = 𝐸 → 𝐶 = 𝐺)
ifeq3da.2	⊢ (if(𝜓, 𝐸, 𝐹) = 𝐹 → 𝐶 = 𝐻)
ifeq3da.3	⊢ (𝜑 → 𝐺 = 𝐴)
ifeq3da.4	⊢ (𝜑 → 𝐻 = 𝐵)

Assertion

Ref	Expression
ifeq3da	⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = 𝐶)

Proof of Theorem ifeq3da

Step	Hyp	Ref	Expression
1		iftrue 4471	. . . . 5 ⊢ (𝜓 → if(𝜓, 𝐸, 𝐹) = 𝐸)
2		ifeq3da.1	. . . . 5 ⊢ (if(𝜓, 𝐸, 𝐹) = 𝐸 → 𝐶 = 𝐺)
3	1, 2	syl 17	. . . 4 ⊢ (𝜓 → 𝐶 = 𝐺)
4	3	adantl 484	. . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝐶 = 𝐺)
5		ifeq3da.3	. . . 4 ⊢ (𝜑 → 𝐺 = 𝐴)
6	5	adantr 483	. . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝐺 = 𝐴)
7	4, 6	eqtr2d 2855	. 2 ⊢ ((𝜑 ∧ 𝜓) → 𝐴 = 𝐶)
8		iffalse 4474	. . . . 5 ⊢ (¬ 𝜓 → if(𝜓, 𝐸, 𝐹) = 𝐹)
9		ifeq3da.2	. . . . 5 ⊢ (if(𝜓, 𝐸, 𝐹) = 𝐹 → 𝐶 = 𝐻)
10	8, 9	syl 17	. . . 4 ⊢ (¬ 𝜓 → 𝐶 = 𝐻)
11	10	adantl 484	. . 3 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝐶 = 𝐻)
12		ifeq3da.4	. . . 4 ⊢ (𝜑 → 𝐻 = 𝐵)
13	12	adantr 483	. . 3 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝐻 = 𝐵)
14	11, 13	eqtr2d 2855	. 2 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝐵 = 𝐶)
15	7, 14	ifeqda 4500	1 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = 𝐶)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 398   = wceq 1531  ifcif 4465

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1905  ax-6 1964  ax-7 2009  ax-8 2110  ax-9 2118  ax-ext 2791

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1775  df-sb 2064  df-clab 2798  df-cleq 2812  df-clel 2891  df-if 4466

This theorem is referenced by:  circlemeth  31904