Theorem ifel 3698

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 2413	. 2 ⊢ ( if(φ, A, B) = A → ( if(φ, A, B) ∈ C ↔ A ∈ C))
2		eleq1 2413	. 2 ⊢ ( if(φ, A, B) = B → ( if(φ, A, B) ∈ C ↔ B ∈ C))
3	1, 2	elimif 3692	1 ⊢ ( if(φ, A, B) ∈ C ↔ ((φ ∧ A ∈ C) ∨ (¬ φ ∧ B ∈ C)))

Syntax hints:  ¬ wn 3   ↔ wb 176   ∨ wo 357   ∧ wa 358   ∈ wcel 1710   ifcif 3663

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-if 3664

This theorem is referenced by: (None)