Theorem eqif 4509

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 2835	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2		eqeq2 2835	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
3	1, 2	elimif 4505	1 ⊢ (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 = 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 = 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 208   ∧ wa 398   ∨ wo 843   = wceq 1537  ifcif 4469

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1781  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-if 4470

This theorem is referenced by:  ifval  4510  xpima  6041  fin23lem19  9760  fin23lem28  9764  fin23lem29  9765  fin23lem30  9766  aalioulem3  24925  iocinif  30506  fsumcvg4  31195  ind1a  31280  esumsnf  31325  itg2addnclem2  34946  clsk1indlem4  40401  afvpcfv0  43352