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Theorem eqif 3695
 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
StepHypRef Expression
1 eqeq2 2362 . 2 ( if(φ, B, C) = B → (A = if(φ, B, C) ↔ A = B))
2 eqeq2 2362 . 2 ( if(φ, B, C) = C → (A = if(φ, B, C) ↔ A = C))
31, 2elimif 3691 1 (A = if(φ, B, C) ↔ ((φ A = B) φ A = C)))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ↔ wb 176   ∨ wo 357   ∧ wa 358   = wceq 1642   ifcif 3662 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-if 3663 This theorem is referenced by: (None)
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