Theorem elimifd 29362

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 431	. . . 4 ⊢ (𝜓 ∨ ¬ 𝜓)
2	1	biantrur 527	. . 3 ⊢ (𝜒 ↔ ((𝜓 ∨ ¬ 𝜓) ∧ 𝜒))
3	2	a1i 11	. 2 ⊢ (𝜑 → (𝜒 ↔ ((𝜓 ∨ ¬ 𝜓) ∧ 𝜒)))
4		andir 912	. . 3 ⊢ (((𝜓 ∨ ¬ 𝜓) ∧ 𝜒) ↔ ((𝜓 ∧ 𝜒) ∨ (¬ 𝜓 ∧ 𝜒)))
5	4	a1i 11	. 2 ⊢ (𝜑 → (((𝜓 ∨ ¬ 𝜓) ∧ 𝜒) ↔ ((𝜓 ∧ 𝜒) ∨ (¬ 𝜓 ∧ 𝜒))))
6		iftrue 4092	. . . . 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 671	. . 3 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ (𝜓 ∧ 𝜃)))
10		iffalse 4095	. . . . 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 671	. . 3 ⊢ (𝜑 → ((¬ 𝜓 ∧ 𝜒) ↔ (¬ 𝜓 ∧ 𝜏)))
14	9, 13	orbi12d 746	. 2 ⊢ (𝜑 → (((𝜓 ∧ 𝜒) ∨ (¬ 𝜓 ∧ 𝜒)) ↔ ((𝜓 ∧ 𝜃) ∨ (¬ 𝜓 ∧ 𝜏))))
15	3, 5, 14	3bitrd 294	1 ⊢ (𝜑 → (𝜒 ↔ ((𝜓 ∧ 𝜃) ∨ (¬ 𝜓 ∧ 𝜏))))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 196   ∨ wo 383   ∧ 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:  elim2if  29363