Theorem elimifd 28746

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 430	. . . 4 ⊢ (𝜓 ∨ ¬ 𝜓)
2	1	biantrur 526	. . 3 ⊢ (𝜒 ↔ ((𝜓 ∨ ¬ 𝜓) ∧ 𝜒))
3	2	a1i 11	. 2 ⊢ (𝜑 → (𝜒 ↔ ((𝜓 ∨ ¬ 𝜓) ∧ 𝜒)))
4		andir 908	. . 3 ⊢ (((𝜓 ∨ ¬ 𝜓) ∧ 𝜒) ↔ ((𝜓 ∧ 𝜒) ∨ (¬ 𝜓 ∧ 𝜒)))
5	4	a1i 11	. 2 ⊢ (𝜑 → (((𝜓 ∨ ¬ 𝜓) ∧ 𝜒) ↔ ((𝜓 ∧ 𝜒) ∨ (¬ 𝜓 ∧ 𝜒))))
6		iftrue 4042	. . . . 5 ⊢ (𝜓 → if(𝜓, 𝐴, 𝐵) = 𝐴)
7		elimifd.1	. . . . 5 ⊢ (𝜑 → (if(𝜓, 𝐴, 𝐵) = 𝐴 → (𝜒 ↔ 𝜃)))
8	6, 7	syl5 33	. . . 4 ⊢ (𝜑 → (𝜓 → (𝜒 ↔ 𝜃)))
9	8	pm5.32d 669	. . 3 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ (𝜓 ∧ 𝜃)))
10		iffalse 4045	. . . . 5 ⊢ (¬ 𝜓 → if(𝜓, 𝐴, 𝐵) = 𝐵)
11		elimifd.2	. . . . 5 ⊢ (𝜑 → (if(𝜓, 𝐴, 𝐵) = 𝐵 → (𝜒 ↔ 𝜏)))
12	10, 11	syl5 33	. . . 4 ⊢ (𝜑 → (¬ 𝜓 → (𝜒 ↔ 𝜏)))
13	12	pm5.32d 669	. . 3 ⊢ (𝜑 → ((¬ 𝜓 ∧ 𝜒) ↔ (¬ 𝜓 ∧ 𝜏)))
14	9, 13	orbi12d 742	. 2 ⊢ (𝜑 → (((𝜓 ∧ 𝜒) ∨ (¬ 𝜓 ∧ 𝜒)) ↔ ((𝜓 ∧ 𝜃) ∨ (¬ 𝜓 ∧ 𝜏))))
15	3, 5, 14	3bitrd 293	1 ⊢ (𝜑 → (𝜒 ↔ ((𝜓 ∧ 𝜃) ∨ (¬ 𝜓 ∧ 𝜏))))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195   ∨ wo 382   ∧ wa 383   = wceq 1475  ifcif 4036

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-if 4037

This theorem is referenced by:  elim2if  28747