Theorem eqif 4505

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 2751	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2		eqeq2 2751	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
3	1, 2	elimif 4501	1 ⊢ (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 = 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 = 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 205   ∧ wa 395   ∨ wo 843   = wceq 1541  ifcif 4464

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-ext 2710

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-ex 1786  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-if 4465

This theorem is referenced by:  ifval  4506  xpima  6082  fin23lem19  10076  fin23lem28  10080  fin23lem29  10081  fin23lem30  10082  aalioulem3  25475  iocinif  31081  fsumcvg4  31879  ind1a  31966  esumsnf  32011  itg2addnclem2  35808  clsk1indlem4  41607  afvpcfv0  44589