Theorem ifel 4468

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 2877	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐴 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐴 ∈ 𝐶))
2		eleq1 2877	. 2 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐵 → (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ 𝐵 ∈ 𝐶))
3	1, 2	elimif 4461	1 ⊢ (if(𝜑, 𝐴, 𝐵) ∈ 𝐶 ↔ ((𝜑 ∧ 𝐴 ∈ 𝐶) ∨ (¬ 𝜑 ∧ 𝐵 ∈ 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 209   ∧ wa 399   ∨ wo 844   ∈ wcel 2111  ifcif 4425

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-if 4426

This theorem is referenced by:  clwlkclwwlklem2a  27783  smatrcl  31149  clsk1independent  40749