Theorem eqif 4525

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 2777	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2		eqeq2 2777	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
3	1, 2	elimif 4521	1 ⊢ (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 = 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 = 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 209   ∧ wa 400   ∨ wo 860   = wceq 1563  ifcif 4483

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-if 4484

This theorem is referenced by:  ifval  4526  xpima  6172  fin23lem19  10308  fin23lem28  10312  fin23lem29  10313  fin23lem30  10314  ind1a  12220  aalioulem3  26456  ifnebib  32805  iocinif  33038  fsumcvg4  34257  esumsnf  34371  itg2addnclem2  38183  clsk1indlem4  44632  afvpcfv0  47738