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Theorem eqif 4490
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 2836 . 2 (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2 eqeq2 2836 . 2 (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
31, 2elimif 4486 1 (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴 = 𝐵) ∨ (¬ 𝜑𝐴 = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 209  wa 399  wo 844   = wceq 1538  ifcif 4450
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 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-ext 2796
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-if 4451
This theorem is referenced by:  ifval  4491  xpima  6027  fin23lem19  9752  fin23lem28  9756  fin23lem29  9757  fin23lem30  9758  aalioulem3  24928  iocinif  30510  fsumcvg4  31220  ind1a  31305  esumsnf  31350  itg2addnclem2  35021  clsk1indlem4  40606  afvpcfv0  43568
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