Theorem eqif 4570

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 2740	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2		eqeq2 2740	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
3	1, 2	elimif 4566	1 ⊢ (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 = 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 = 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 205   ∧ wa 395   ∨ wo 846   = wceq 1534  ifcif 4529

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-ex 1775  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-if 4530

This theorem is referenced by:  ifval  4571  xpima  6186  fin23lem19  10360  fin23lem28  10364  fin23lem29  10365  fin23lem30  10366  aalioulem3  26282  ifnebib  32353  iocinif  32562  fsumcvg4  33551  ind1a  33638  esumsnf  33683  itg2addnclem2  37145  clsk1indlem4  43474  afvpcfv0  46526