Theorem elif 4429

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 2873	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ 𝐴 ∈ 𝐵))
2		eleq2 2873	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ 𝐴 ∈ 𝐶))
3	1, 2	elimif 4423	1 ⊢ (𝐴 ∈ if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 ∈ 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 ∈ 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 207   ∧ wa 396   ∨ wo 842   ∈ wcel 2083  ifcif 4387

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1781  ax-4 1795  ax-5 1892  ax-6 1951  ax-7 1996  ax-8 2085  ax-9 2093  ax-ext 2771

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-ex 1766  df-sb 2045  df-clab 2778  df-cleq 2790  df-clel 2865  df-if 4388

This theorem is referenced by:  clsk1indlem3  39899