Theorem ifel 4526

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 2848	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐴 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐴 ∈ 𝐶))
2		eleq1 2848	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐵 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐵 ∈ 𝐶))
3	1, 2	elimif 4519	1 ⊢ (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ ((𝜑 ∧ 𝐴 ∈ 𝐶) ∨ (¬ 𝜑 ∧ 𝐵 ∈ 𝐶)))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 401   ∨ wo 861   ∈ wcel 2144  ifcif 4481

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-if 4482

This theorem is used by:  clwlkclwwlklem2a  30595  smatrcl  34475  clsk1independent  45140