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Theorem elimif 4523
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
StepHypRef Expression
1 iftrue 4491 . . 3 (𝜑 → if(𝜑, 𝐴, 𝐵) = 𝐴)
2 elimif.1 . . 3 (if(𝜑, 𝐴, 𝐵) = 𝐴 → (𝜓 ↔ 𝜒))
31, 2syl 18 . 2 (𝜑 → (𝜓 ↔ 𝜒))
4 iffalse 4494 . . 3 (¬ 𝜑 → if(𝜑, 𝐴, 𝐵) = 𝐵)
5 elimif.2 . . 3 (if(𝜑, 𝐴, 𝐵) = 𝐵 → (𝜓 ↔ 𝜃))
64, 5syl 18 . 2 (¬ 𝜑 → (𝜓 ↔ 𝜃))
73, 6cases 1058 1 (𝜓 ↔ ((𝜑 ∧ 𝜒) ∨ (¬ 𝜑 ∧ 𝜃)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570  ifcif 4485
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-if 4486
This theorem is used by:  eqif  4527  elif  4529  ifel  4530  ftc1anclem5  38820  clsk1indlem2  45245
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