Theorem eqif 4527

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 2774	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐵 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐵))
2		eqeq2 2774	. 2 ⊢ (if(𝜑, 𝐵, 𝐶) = 𝐶 → (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ 𝐴 = 𝐶))
3	1, 2	elimif 4523	1 ⊢ (𝐴 = if(𝜑, 𝐵, 𝐶) ↔ ((𝜑 ∧ 𝐴 = 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 = 𝐶)))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570  ifcif 4485

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-if 4486

This theorem is used by:  ifval  4528  xpima  6179  fin23lem19  10342  fin23lem28  10346  fin23lem29  10347  fin23lem30  10348  ind1a  12257  aalioulem3  26577  ifnebib  33032  iocinif  33260  fsumcvg4  34468  esumsnf  34582  itg2addnclem2  38429  clsk1indlem4  44892  afvpcfv0  48042