HepLean Documentation

Mathlib.Tactic.NormNum.DivMod

norm_num extension for integer div/mod and divides #

This file adds support for the %, /, and (divisibility) operators on to the norm_num tactic.

theorem Mathlib.Meta.NormNum.isInt_ediv {a b q m a' : } {b' r : } (ha : Mathlib.Meta.NormNum.IsInt a a') (hb : Mathlib.Meta.NormNum.IsNat b b') (hm : q * b' = m) (h : r + m = a') (h₂ : r.blt b' = true) :

The norm_num extension which identifies expressions of the form Int.ediv a b, such that norm_num successfully recognises both a and b.

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    def Mathlib.Meta.NormNum.evalIntDiv.core (a na : Q()) (za : ) (pa : Q(Mathlib.Meta.NormNum.IsInt «$a» «$na»)) (b : Q()) (nb : Q()) (pb : Q(Mathlib.Meta.NormNum.IsNat «$b» «$nb»)) :
    × (q : Q()) × Q(Mathlib.Meta.NormNum.IsInt («$a» / «$b») «$q»)

    Given a result for evaluating a b in where b > 0, evaluate a / b.

    Instances For
      theorem Mathlib.Meta.NormNum.isInt_emod {a b q m a' : } {b' r : } (ha : Mathlib.Meta.NormNum.IsInt a a') (hb : Mathlib.Meta.NormNum.IsNat b b') (hm : q * b' = m) (h : r + m = a') (h₂ : r.blt b' = true) :

      The norm_num extension which identifies expressions of the form Int.emod a b, such that norm_num successfully recognises both a and b.

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        def Mathlib.Meta.NormNum.evalIntMod.go (a na : Q()) (za : ) (pa : Q(Mathlib.Meta.NormNum.IsInt «$a» «$na»)) (b : Q()) :

        Given a result for evaluating a b in , evaluate a % b.

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          def Mathlib.Meta.NormNum.evalIntMod.core (a na : Q()) (za : ) (pa : Q(Mathlib.Meta.NormNum.IsInt «$a» «$na»)) (b : Q()) (nb : Q()) (pb : Q(Mathlib.Meta.NormNum.IsNat «$b» «$nb»)) :
          (r : Q()) × Q(Mathlib.Meta.NormNum.IsNat («$a» % «$b») «$r»)

          Given a result for evaluating a b in where b > 0, evaluate a % b.

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            theorem Mathlib.Meta.NormNum.isInt_dvd_true {a b a' b' c : } :
            Mathlib.Meta.NormNum.IsInt a a'Mathlib.Meta.NormNum.IsInt b b'a'.mul c = b'a b

            The norm_num extension which identifies expressions of the form (a : ℤ) ∣ b, such that norm_num successfully recognises both a and b.

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