Theorem elimifd 30559

Description: Elimination of a conditional operator contained in a wff 𝜒. (Contributed by Thierry Arnoux, 25-Jan-2017.)

Hypotheses

Ref	Expression
elimifd.1	⊢ (𝜑 → (if(𝜓, 𝐴, 𝐵) = 𝐴 → (𝜒 ↔ 𝜃)))
elimifd.2	⊢ (𝜑 → (if(𝜓, 𝐴, 𝐵) = 𝐵 → (𝜒 ↔ 𝜏)))

Assertion

Ref	Expression
elimifd	⊢ (𝜑 → (𝜒 ↔ ((𝜓 ∧ 𝜃) ∨ (¬ 𝜓 ∧ 𝜏))))

Proof of Theorem elimifd

Step	Hyp	Ref	Expression
1		exmid 895	. . . 4 ⊢ (𝜓 ∨ ¬ 𝜓)
2	1	biantrur 534	. . 3 ⊢ (𝜒 ↔ ((𝜓 ∨ ¬ 𝜓) ∧ 𝜒))
3	2	a1i 11	. 2 ⊢ (𝜑 → (𝜒 ↔ ((𝜓 ∨ ¬ 𝜓) ∧ 𝜒)))
4		andir 1009	. . 3 ⊢ (((𝜓 ∨ ¬ 𝜓) ∧ 𝜒) ↔ ((𝜓 ∧ 𝜒) ∨ (¬ 𝜓 ∧ 𝜒)))
5	4	a1i 11	. 2 ⊢ (𝜑 → (((𝜓 ∨ ¬ 𝜓) ∧ 𝜒) ↔ ((𝜓 ∧ 𝜒) ∨ (¬ 𝜓 ∧ 𝜒))))
6		iftrue 4431	. . . . 5 ⊢ (𝜓 → if(𝜓, 𝐴, 𝐵) = 𝐴)
7		elimifd.1	. . . . 5 ⊢ (𝜑 → (if(𝜓, 𝐴, 𝐵) = 𝐴 → (𝜒 ↔ 𝜃)))
8	6, 7	syl5 34	. . . 4 ⊢ (𝜑 → (𝜓 → (𝜒 ↔ 𝜃)))
9	8	pm5.32d 580	. . 3 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ (𝜓 ∧ 𝜃)))
10		iffalse 4434	. . . . 5 ⊢ (¬ 𝜓 → if(𝜓, 𝐴, 𝐵) = 𝐵)
11		elimifd.2	. . . . 5 ⊢ (𝜑 → (if(𝜓, 𝐴, 𝐵) = 𝐵 → (𝜒 ↔ 𝜏)))
12	10, 11	syl5 34	. . . 4 ⊢ (𝜑 → (¬ 𝜓 → (𝜒 ↔ 𝜏)))
13	12	pm5.32d 580	. . 3 ⊢ (𝜑 → ((¬ 𝜓 ∧ 𝜒) ↔ (¬ 𝜓 ∧ 𝜏)))
14	9, 13	orbi12d 919	. 2 ⊢ (𝜑 → (((𝜓 ∧ 𝜒) ∨ (¬ 𝜓 ∧ 𝜒)) ↔ ((𝜓 ∧ 𝜃) ∨ (¬ 𝜓 ∧ 𝜏))))
15	3, 5, 14	3bitrd 308	1 ⊢ (𝜑 → (𝜒 ↔ ((𝜓 ∧ 𝜃) ∨ (¬ 𝜓 ∧ 𝜏))))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399   ∨ wo 847   = wceq 1543  ifcif 4425

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-ex 1788  df-sb 2073  df-clab 2715  df-cleq 2728  df-clel 2809  df-if 4426

This theorem is referenced by:  elim2if  30560