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Theorem eqif 4567
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 2749 . 2 (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2 eqeq2 2749 . 2 (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
31, 2elimif 4563 1 (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴 = 𝐵) ∨ (¬ 𝜑𝐴 = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 206  wa 395  wo 848   = wceq 1540  ifcif 4525
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-if 4526
This theorem is referenced by:  ifval  4568  xpima  6202  fin23lem19  10376  fin23lem28  10380  fin23lem29  10381  fin23lem30  10382  aalioulem3  26376  ifnebib  32562  iocinif  32783  ind1a  32844  fsumcvg4  33949  esumsnf  34065  itg2addnclem2  37679  clsk1indlem4  44057  afvpcfv0  47158
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