Theorem ifel 4530

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 2850	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐴 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐴 ∈ 𝐶))
2		eleq1 2850	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐵 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐵 ∈ 𝐶))
3	1, 2	elimif 4523	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 2145  ifcif 4485

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-if 4486

This theorem is used by:  clwlkclwwlklem2a  30476  smatrcl  34314  clsk1independent  44894