Theorem ifel 4500

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 2826	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐴 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐴 ∈ 𝐶))
2		eleq1 2826	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐵 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐵 ∈ 𝐶))
3	1, 2	elimif 4493	1 ⊢ (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ ((𝜑 ∧ 𝐴 ∈ 𝐶) ∨ (¬ 𝜑 ∧ 𝐵 ∈ 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 205   ∧ wa 395   ∨ wo 843   ∈ wcel 2108  ifcif 4456

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-if 4457

This theorem is referenced by:  clwlkclwwlklem2a  28263  smatrcl  31648  clsk1independent  41545