Theorem elif 4078

Description: Membership in a conditional operator. (Contributed by NM, 14-Feb-2005.)

Assertion

Ref	Expression
elif	⊢ (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 ∈ 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 ∈ 𝐶)))

Proof of Theorem elif

Step	Hyp	Ref	Expression
1		eleq2 2677	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ 𝐴 ∈ 𝐵))
2		eleq2 2677	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ 𝐴 ∈ 𝐶))
3	1, 2	elimif 4072	1 ⊢ (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 ∈ 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 ∈ 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 195   ∨ wo 382   ∧ wa 383   ∈ wcel 1977  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:  clsk1indlem3  37361