Theorem eqif 4506

Description: Expansion of an equality with a conditional operator. (Contributed by NM, 14-Feb-2005.)

Assertion

Ref	Expression
eqif	⊢ (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 = 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 = 𝐶)))

Proof of Theorem eqif

Step	Hyp	Ref	Expression
1		eqeq2 2833	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2		eqeq2 2833	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
3	1, 2	elimif 4502	1 ⊢ (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 = 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 = 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 208   ∧ wa 398   ∨ wo 843   = wceq 1533  ifcif 4466

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1777  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-if 4467

This theorem is referenced by:  ifval  4507  xpima  6033  fin23lem19  9752  fin23lem28  9756  fin23lem29  9757  fin23lem30  9758  aalioulem3  24917  iocinif  30498  fsumcvg4  31188  ind1a  31273  esumsnf  31318  itg2addnclem2  34938  clsk1indlem4  40387  afvpcfv0  43339