Theorem elimif 4519

Description: Elimination of a conditional operator contained in a wff 𝜓. (Contributed by NM, 15-Feb-2005.) (Proof shortened by NM, 25-Apr-2019.)

Hypotheses

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

Assertion

Ref	Expression
elimif	⊢ (𝜓 ↔ ((𝜑 ∧ 𝜒) ∨ (¬ 𝜑 ∧ 𝜃)))

Proof of Theorem elimif

Step	Hyp	Ref	Expression
1		iftrue 4487	. . 3 ⊢ (𝜑 → if(𝜑, 𝐴, 𝐵) = 𝐴)
2		elimif.1	. . 3 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐴 → (𝜓 ↔ 𝜒))
3	1, 2	syl 18	. 2 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
4		iffalse 4490	. . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, 𝐵) = 𝐵)
5		elimif.2	. . 3 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐵 → (𝜓 ↔ 𝜃))
6	4, 5	syl 18	. 2 ⊢ (¬ 𝜑 → (𝜓 ↔ 𝜃))
7	3, 6	cases 1058	1 ⊢ (𝜓 ↔ ((𝜑 ∧ 𝜒) ∨ (¬ 𝜑 ∧ 𝜃)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1569  ifcif 4481

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-if 4482

This theorem is used by:  eqif  4523  elif  4525  ifel  4526  ftc1anclem5  38709  clsk1indlem2  45136