Nuprl Rule : remainderEquality

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

H  ⊢ (m1 rem n1) (m2 rem n2) ∈ ℤ

  BY remainderEquality ()
  
  H  ⊢ m1 m2 ∈ ℤ
  H  ⊢ n1 n2 ∈ ℤ
  H  ⊢ n1 ≠ 0



Definitions occuring in rule :  remainder: rem m equal: t ∈ T nequal: a ≠ b ∈  int: natural_number: $n axiom: Ax

Latex:
H    \mvdash{}  (m1  rem  n1)  =  (m2  rem  n2)

    BY  remainderEquality  ()
   
    H    \mvdash{}  m1  =  m2
    H    \mvdash{}  n1  =  n2
    H    \mvdash{}  n1  \mneq{}  0



Date html generated: 2019_06_20-PM-04_12_02
Last ObjectModification: 2016_07_08-PM-03_48_49

Theory : rules


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