Question

我需要证明这个方程组没有解决方案（原因是它是过度确定的）。有没有一种简单的方法来实现 in Coq ？即策略或图书馆？

Require Import Reals.
Open Scope R.

Lemma no_solution:
  forall 
    b11 b12 b13 b14 b21 b22 b23 b24 b31 b32 b33 b34
    r r0 r1 r2 r3 r4 r5 r6 r7 r8 r9 r10 : R,
   1 = r * b11 + r0 * b21 + r1 * b31    ->
   0 = r * b12 + r0 * b22 + r1 * b32    ->
   0 = r * b13 + r0 * b23 + r1 * b33    ->
   0 = r * b14 + r0 * b24 + r1 * b34    ->
   0 = r2 * b11 + r3 * b21 + r4 * b31   ->
   1 = r2 * b12 + r3 * b22 + r4 * b32   ->
   0 = r2 * b13 + r3 * b23 + r4 * b33   ->
   0 = r2 * b14 + r3 * b24 + r4 * b34   ->
   0 = r5 * b11 + r6 * b21 + r7 * b31   ->
   0 = r5 * b12 + r6 * b22 + r7 * b32   ->
   1 = r5 * b13 + r6 * b23 + r7 * b33   ->
   0 = r5 * b14 + r6 * b24 + r7 * b34   ->
   0 = r8 * b11 + r9 * b21 + r10 * b31  ->
   0 = r8 * b12 + r9 * b22 + r10 * b32  ->
   0 = r8 * b13 + r9 * b23 + r10 * b33  ->
   1 = r8 * b14 + r9 * b24 + r10 * b34  ->
   False.

Answer 1

如果我理解得当，这组方程不能同时成立，因为它要求3x4矩阵的秩高于3。

结果的主要定理在mathematical components library中称为mulmx_max_rank。我有更多工作将您的非结构化问题的表示连接到使用矩阵构造一个而不是找到正确的定理。此实验在coq-8.7中进行，coq-mathcomp-ssreflect和coq-mathcomp-algebra通过opam（版本1.6.2加载）。

请注意，此结果适用于任何字段结构。

From mathcomp Require Import all_ssreflect all_algebra.

Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.

Import GRing.Theory Num.Theory.

Open Scope ring_scope.

Section Solving_linear_equation_systems_in_Coq.
Variable R : fieldType.

Definition seq2matrix m n (s : seq (seq R)) :=
  \matrix_(i < m, j < n)
     nth 0 (nth nil s i) j.

Lemma no_solution:
  forall 
    b11 b12 b13 b14 b21 b22 b23 b24 b31 b32 b33 b34
    r r0 r1 r2 r3 r4 r5 r6 r7 r8 r9 r10 : R,
   1 = r * b11 + r0 * b21 + r1 * b31    ->
   0 = r * b12 + r0 * b22 + r1 * b32    ->
   0 = r * b13 + r0 * b23 + r1 * b33    ->
   0 = r * b14 + r0 * b24 + r1 * b34    ->
   0 = r2 * b11 + r3 * b21 + r4 * b31   ->
   1 = r2 * b12 + r3 * b22 + r4 * b32   ->
   0 = r2 * b13 + r3 * b23 + r4 * b33   ->
   0 = r2 * b14 + r3 * b24 + r4 * b34   ->
   0 = r5 * b11 + r6 * b21 + r7 * b31   ->
   0 = r5 * b12 + r6 * b22 + r7 * b32   ->
   1 = r5 * b13 + r6 * b23 + r7 * b33   ->
   0 = r5 * b14 + r6 * b24 + r7 * b34   ->
   0 = r8 * b11 + r9 * b21 + r10 * b31  ->
   0 = r8 * b12 + r9 * b22 + r10 * b32  ->
   0 = r8 * b13 + r9 * b23 + r10 * b33  ->
   1 = r8 * b14 + r9 * b24 + r10 * b34  ->
   False.
Proof.
move => b11 b12 b13 b14 b21 b22 b23 b24 b31 b32 b33 b34
   r r0 r1 r2 r3 r4 r5 r6 r7 r8 r9 r10 eq1 eq2 eq3 eq4
   eq5 eq6 eq7 eq8 eq9 eq10 eq11 eq12 eq13 eq14 eq15 eq16.
set Inp := seq2matrix 4 3 
           [:: [:: r; r0; r1];
               [:: r2; r3; r4];
               [:: r5; r6; r7];
               [:: r8; r9; r10]].
set B := seq2matrix 3 4 [:: [:: b11; b12; b13; b14];
                            [:: b21; b22; b23; b24];
                            [:: b31; b32; b33; b34]].
suff abs: Inp *m B = 1%:M.
  have : (\rank (Inp *m B) <= 3)%N by apply: mulmx_max_rank.
  by rewrite abs mxrank1.
by apply/matrixP=> [[ [ | [ | [ | [ | ?]]]] pi]]
   [ [ | [ | [ | [ | ?]]]] pj] //;
 rewrite /Inp /seq2matrix /= !(mxE, big_ord_recr, big_ord0) //= add0r /=
  -?(eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9, eq10, eq11, eq12).
Qed.

End Solving_linear_equation_systems_in_Coq.

求解Coq中的线性方程组

1 个答案: