Theorem eqif 4346

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

Step	Hyp	Ref	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))
3	1, 2	elimif 4342	1 ⊢ (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 = 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 = 𝐶)))

Syntax hints:  ¬ wn 3   ↔ wb 198   ∧ wa 386   ∨ wo 880   = wceq 1658  ifcif 4306

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1896  ax-4 1910  ax-5 2011  ax-6 2077  ax-7 2114  ax-9 2175  ax-10 2194  ax-11 2209  ax-12 2222  ax-ext 2803

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 881  df-tru 1662  df-ex 1881  df-nf 1885  df-sb 2070  df-clab 2812  df-cleq 2818  df-clel 2821  df-if 4307

This theorem is referenced by:  ifval  4347  xpima  5817  fin23lem19  9473  fin23lem28  9477  fin23lem29  9478  fin23lem30  9479  aalioulem3  24488  iocinif  30090  fsumcvg4  30541  ind1a  30626  esumsnf  30671  itg2addnclem2  34005  clsk1indlem4  39182  afvpcfv0  42048