Nuprl Rule : divremEquality

This rule proved as lemma rule_arithop_equality_true3 in file rules/rules_arith.v
at https://github.com/vrahli/NuprlInCoq

H  ⊢ divrem(m1; n1) = divrem(m2; n2) ∈ (ℤ × ℤ)

  BY divremEquality ()

  H  ⊢ m1 = m2 ∈ ℤ
  H  ⊢ n1 = n2 ∈ ℤ
  H  ⊢ n1 ≠ 0

Definitions occuring in rule : product: x:A × B[x], divrem: divrem(n; m), equal: s = t ∈ T, nequal: a ≠ b ∈ T , int: ℤ, natural_number: $n, axiom: Ax

Latex:
H \mvdash{} divrem(m1; n1) = divrem(m2; n2) BY divremEquality () H \mvdash{} m1 = m2 H \mvdash{} n1 = n2 H \mvdash{} n1 \mneq{} 0 Date html generated: 2019_06_20-PM-04_12_24 Last ObjectModification: 2019_03_06-AM-10_42_08 Theory : rules Home Index