Theorem eqif 4530

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 2775	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2		eqeq2 2775	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
3	1, 2	elimif 4526	1 ⊢ (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 = 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 = 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 209   ∧ wa 400   ∨ wo 860   = wceq 1570  ifcif 4488

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-if 4489

This theorem is referenced by:  ifval  4531  xpima  6182  fin23lem19  10321  fin23lem28  10325  fin23lem29  10326  fin23lem30  10327  ind1a  12230  aalioulem3  26476  ifnebib  32873  iocinif  33104  fsumcvg4  34318  esumsnf  34432  itg2addnclem2  38301  clsk1indlem4  44750  afvpcfv0  47860