(* (c) Copyright 2006-2016 Microsoft Corporation and Inria.                  *)
(* Distributed under the terms of CeCILL-B.                                  *)
From mathcomp
From mathcomp
From mathcomp
(******************************************************************************)
(* This file provides various results on divisibility of integers.            *)
(* It defines, for m, n, d : int,                                             *)
(*   (m %% d)%Z == the remainder of the Euclidean division of m by d; this is *)
(*                 the least non-negative element of the coset m + dZ when    *)
(*                 d != 0, and m if d = 0.                                    *)
(*   (m %/ d)%Z == the quotient of the Euclidean division of m by d, such     *)
(*                 that m = (m %/ d)%Z * d + (m %% d)%Z. Since for d != 0 the *)
(*                 remainder is non-negative, (m %/ d)%Z is non-zero for      *)
(*   (d %| m)%Z <=> m is divisible by d; dvdz d is the (collective) predicate *)
(*                 for integers divisible by d, and (d %| m)%Z is actually    *)
(*                 (transposing) notation for m \in dvdz d.                   *)
(* (m = n %[mod d])%Z, (m == n %[mod d])%Z, (m != n %[mod d])%Z               *)
(*                 m and n are (resp. compare, don't compare) equal mod d.    *)
(*     gcdz m n == the (non-negative) greatest common divisor of m and n,     *)
(*                 with gcdz 0 0 = 0.                                         *)
(* coprimez m n <=> m and n are coprime.                                      *)
(*    egcdz m n == the Bezout coefficients of the gcd of m and n: a pair      *)
(*                 (u, v) of coprime integers such that u*m + v*n = gcdz m n. *)
(*                 Alternatively, a Bezoutz lemma states such u and v exist.  *)
(* zchinese m1 m2 n1 n2 == for coprime m1 and m2, a solution to the Chinese   *)
(*                 remainder problem for n1 and n2, i.e., and integer n such  *)
(*                 that n = n1 %[mod m1] and n = n2 %[mod m2].                *)
(*  zcontents p == the contents of p : {poly int}, that is, the gcd of the    *)
(*                 coefficients of p, with the lead coefficient of p,         *)
(* zprimitive p == the primitive part of p : {poly int}, i.e., p divided by   *)
(*                 its contents.                                              *)
(* inIntSpan X v <-> v is an integral linear combination of elements of       *)
(*                 X : seq V, where V is a zmodType. We prove that this is a  *)
(*                 decidable property for Q-vector spaces.                    *)
(* int_Smith_normal_form :: a theorem asserting the existence of the Smith    *)
(*                 normal form for integer matrices.                          *)
(* Note that many of the concepts and results in this file could and perhaps  *)
(* sould be generalized to the more general setting of integral, unique       *)
(* factorization, principal ideal, or Euclidean domains.                      *)
(******************************************************************************)
Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.
Local Open Scope ring_scope.
Definition divz (m d : int) :=
let: (K, n) := match m with Posz n => (Posz, n) | Negz n => (Negz, n) end in
sgz d * K (n %/ `|d|)%N.
Definition modz (m d : int) : int := m - divz m d * d.
Definition dvdz d m := (`|d| %| `|m|)%N.
Definition gcdz m n := (gcdn `|m| `|n|)%:Z.
Definition egcdz m n : int * int :=
if m == 0 then (0, (-1) ^+ (n < 0)%R) else
let: (u, v) := egcdn `|m| `|n| in (sgz m * u, - (-1) ^+ (n < 0)%R * v%:Z).
Definition coprimez m n := (gcdz m n == 1).
Infix "%/" := divz : int_scope.
Infix "%%" := modz : int_scope.
Notation "d %| m" := (m \in dvdz d) : int_scope.
Notation "m = n %[mod d ]" := (modz m d = modz n d) : int_scope.
Notation "m == n %[mod d ]" := (modz m d == modz n d) : int_scope.
Notation "m <> n %[mod d ]" := (modz m d <> modz n d) : int_scope.
Notation "m != n %[mod d ]" := (modz m d != modz n d) : int_scope.

Lemma eqz_modDr p m n d : (m + p == n + p %[mod d])%Z = (m == n %[mod d])%Z.
Arguments dvdzP {d m}.