Nuprl Rule : divideEquality

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

H  ⊢ (m1 ÷ n1) = (m2 ÷ n2) ∈ ℤ

  BY divideEquality ()

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

Definitions occuring in rule : divide: n ÷ m, equal: s = t ∈ T, nequal: a ≠ b ∈ T , int: ℤ, natural_number: $n, axiom: Ax

Latex:
H \mvdash{} (m1 \mdiv{} n1) = (m2 \mdiv{} n2) BY divideEquality () H \mvdash{} m1 = m2 H \mvdash{} n1 = n2 H \mvdash{} n1 \mneq{} 0 Date html generated: 2019_06_20-PM-04_11_51 Last ObjectModification: 2016_07_08-PM-03_49_00 Theory : rules Home Index