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Theorem eqif 4568
Description: Expansion of an equality with a conditional operator. (Contributed by NM, 14-Feb-2005.)
Assertion
Ref Expression
eqif (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴 = 𝐵) ∨ (¬ 𝜑𝐴 = 𝐶)))

Proof of Theorem eqif
StepHypRef Expression
1 eqeq2 2744 . 2 (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2 eqeq2 2744 . 2 (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
31, 2elimif 4564 1 (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴 = 𝐵) ∨ (¬ 𝜑𝐴 = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 205  wa 396  wo 845   = wceq 1541  ifcif 4527
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2703
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-ex 1782  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-if 4528
This theorem is referenced by:  ifval  4569  xpima  6178  fin23lem19  10327  fin23lem28  10331  fin23lem29  10332  fin23lem30  10333  aalioulem3  25838  ifnebib  31768  iocinif  31979  fsumcvg4  32918  ind1a  33005  esumsnf  33050  itg2addnclem2  36528  clsk1indlem4  42780  afvpcfv0  45840
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