Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  eqif Structured version   Visualization version   GIF version

Theorem eqif 4270
 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 2771 . 2 (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2 eqeq2 2771 . 2 (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
31, 2elimif 4266 1 (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑𝐴 = 𝐵) ∨ (¬ 𝜑𝐴 = 𝐶)))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ↔ wb 196   ∨ wo 382   ∧ wa 383   = wceq 1632  ifcif 4230 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740 This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-if 4231 This theorem is referenced by:  ifval  4271  xpima  5734  fin23lem19  9370  fin23lem28  9374  fin23lem29  9375  fin23lem30  9376  aalioulem3  24308  iocinif  29873  fsumcvg4  30326  ind1a  30411  esumsnf  30456  itg2addnclem2  33793  clsk1indlem4  38862  afvpcfv0  41750
 Copyright terms: Public domain W3C validator