Theorem ifel 4535

Description: Membership of a conditional operator. (Contributed by NM, 10-Sep-2005.)

Assertion

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

Proof of Theorem ifel

Step	Hyp	Ref	Expression
1		eleq1 2817	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐴 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐴 ∈ 𝐶))
2		eleq1 2817	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐵 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐵 ∈ 𝐶))
3	1, 2	elimif 4528	1 ⊢ (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ ((𝜑 ∧ 𝐴 ∈ 𝐶) ∨ (¬ 𝜑 ∧ 𝐵 ∈ 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 206   ∧ wa 395   ∨ wo 847   ∈ wcel 2109  ifcif 4490

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-if 4491

This theorem is referenced by:  clwlkclwwlklem2a  29933  smatrcl  33792  clsk1independent  44028