如何在Frama-C中使用真实公理证明代码 [英] How do I prove code with a real axiomatic in Frama-C

问题描述

我已从ACSL-by-Example的内部产品"代码中将int类型更改为float(具有int类型的代码对我有用)，现在我无法证明循环不变式inner.我为inf和NaN添加了一些检查，但没有成功.

I have changed the int type to float in the "Inner Product" code from the ACSL-by-Example book (the code with int type worked for me) and now I am not able to prove the loop invariant inner. I have added some checks for inf and NaN without any success.

#include "limits.h"

/*@
  predicate  Unchanged{K,L}(float* a, integer first, integer last) =
    \forall integer i; first <= i < last ==>
    \at(a[i],K) == \at(a[i],L);

  predicate  Unchanged{K,L}(float* a, integer n) =
    Unchanged{K,L}(a, 0, n);

  lemma UnchangedStep{K,L}:
    \forall float *a, integer n;
    0 <= n ==>
    Unchanged{K,L}(a, n) ==>
    \at(a[n],K) == \at(a[n],L) ==>
    Unchanged{K,L}(a, n+1);

  lemma UnchangedSection{K,L}:
    \forall float *a, integer m, n;
    0 <= m <= n ==>
    Unchanged{K,L}(a, n) ==>
    Unchanged{K,L}(a, m);
*/


/*@ axiomatic InnerProductAxiomatic
  {
  logic real InnerProduct{L}(float* a, float* b, integer n, float init)
  reads a[0..n-1], b[0..n-1];

  axiom InnerProductEmpty:
    \forall float *a, *b, init, integer n;
    n <= 0 ==> InnerProduct(a, b, n, init) == init;

  axiom InnerProductNext:
    \forall float *a, *b, init, integer n;
    n >= 0 ==>
    InnerProduct(a, b, n + 1, init) ==
    InnerProduct(a, b, n, init) + a[n] * b[n];

  axiom InnerProductRead{L1,L2}:
    \forall float *a, *b, init, integer n;
    Unchanged{L1,L2}(a, n) && Unchanged{L1,L2}(b, n) ==>
    InnerProduct{L1}(a, b, n, init) ==
    InnerProduct{L2}(a, b, n, init);
  }*/

/*@
  predicate ProductBounds(float* a, float* b, integer n) =
    \forall integer i; 0 <= i < n ==>
    (INT_MIN <= a[i] * b[i] <= INT_MAX) ;

  predicate InnerProductBounds(float* a, float* b, integer n, float init) =
    \forall integer i; 0 <= i <= n ==>
    INT_MIN <= InnerProduct(a, b, i, init) <= INT_MAX;
*/

/*@
  requires valid_a: \valid_read(a + (0..n-1));
  requires valid_b: \valid_read(b + (0..n-1));
  requires \is_finite(init);
  requires !\is_NaN(init);
  requires bounds: ProductBounds(a, b, n);
  requires bounds: InnerProductBounds(a, b, n, init);
  requires (n < 100) && (n>=0);
  requires \forall integer i; 0 <= i < n ==>  \is_finite(a[i]);
  requires \forall integer i; 0 <= i < n ==>  \is_finite(b[i]);
  requires \forall integer i; 0 <= i < n ==>  !\is_NaN(b[i]);
  requires \forall integer i; 0 <= i < n ==>  !\is_NaN(a[i]);

  assigns \nothing;
  ensures result: \result == InnerProduct(a, b, n, init);
  ensures unchanged: Unchanged{Here,Pre}(a, n);
  ensures unchanged: Unchanged{Here,Pre}(b, n);
*/

float inner_product(const float* a, const float* b, int n, float init)
{
  int i = 0;
  /*@
    loop invariant index: 0 <= i <= n;
    loop invariant inner: init == InnerProduct(a, b, i, \at(init,Pre));
    loop assigns i, init;
    loop variant n-i;
  */
  while (i < n) {
    init = init + a[i] * b[i];
    i++;
  }
  return init;
}

我如何通过?哪里有真实计算证明的好案例?

How can I pass? Where to get the good cases with proves of real computations?

坦白地说，我想证明Sine的循环不变.我为此创建了一个引理(有界Sine Taylor级数)并将其作为函数进行了测试.而且我不知道如何开始证明它.

And frankly speaking I would like then to prove my loop invariant for Sine. I created a lemma for it (bounded Sine Taylor series) and tested it as a function. And I don't know a way how to start proving it.

/*@
axiomatic SinNAxiomatic
{
logic real Sinnn {l} (real x, real sum, real current, integer i, integer i_max);
axiom SinnnEmpty: \forall real x, real sum, real current, integer i, integer i_max; (\abs(current) < 0.00001) || (i == i_max) ==> Sinnn(x, sum, current, i, i_max)
== sum + current;
axiom SinnnNext: \forall real x, real sum, real current, integer i, integer i_max; \abs(current) > 0.00001 ==> Sinnn(x, sum, current, i, i_max) ==
Sinnn(x, sum + current, current * (-1.0 * x * x / ((2 * i) * (2 * i + 1))), i + 1, i_max);

lemma  SinnnMemCmp{L1,L2}: \forall real x, real sum, real current, integer i, integer i_max;
\at(x, L1)==\at(x, L2) && \at(sum, L1)==\at(sum, L2) &&  \at(current, L1)==\at(current, L2) && \at(i, L1)==\at(i, L2) && \at(i_max, L1)==\at(i_max, L2)
  ==> Sinnn{L1}(x, sum, current, i, i_max) == Sinnn{L2}(x, sum, current, i, i_max);
}
*/
float SinTailor(float x) {
        float n = x;
        float sum = 0.0;
        int i = 1;
/*@
loop invariant over: \abs(sum - Sinnn(x, 0, x, 1, i - 1)) <= 0.001;
loop assigns sum, n, i;
*/
        do
        {
            sum += n;
            n *= -1.0 * x * x / ((2 * i) * (2 * i + 1));
            i++;
            //printf("sum my=%f recursion=%f\n", sum, TestSinnn(x, 0, x, 1, i - 1)); //prints the same values

        }
        while (fabs(n)> 0.00001);
return sum;
}

我注意到对于内部\sin，存在一些-1<=\sin(x)<=1, \cos^2(x)+\sin^2(x)==1等引理，但是对于sin(x)返回函数，我们无法证明\result==\sin(x).还是我在这里错了?

I noticed that for internal \sin there are a few lemmas like -1<=\sin(x)<=1, \cos^2(x)+\sin^2(x)==1, etc., but we cannot prove \result==\sin(x) for a sin(x) returning function. Or am I wrong here?

推荐答案

我将回答您问题的第一部分.问题出在公理InnerProductNext，更确切地说是InnerProduct(a, b, n + 1, init) == InnerProduct(a, b, n, init) + a[n] * b[n]. ACSL规范使用实数运算，而您的函数使用32位浮点计算.由于在C函数中发生舍入，因此无法获得证明.修复非常简单:在引理中适当舍入所有操作.

I'm going to answer the first part of your question. The problem lies in the axiom InnerProductNext, more precisely here InnerProduct(a, b, n + 1, init) == InnerProduct(a, b, n, init) + a[n] * b[n]. ACSL specification use real arithmetics, while your function use 32 bits floating point computation. Due to the rounding that occurs in the C function, the proof cannot be achieved. The fix is simple enough: round all operations appropriately in your lemma.

  axiom InnerProductNext:
    \forall float *a, *b, init, integer n;
    n >= 0 ==>
    InnerProduct(a, b, n + 1, init) ==
    (float)(InnerProduct(a, b, n, init) + (float)(a[n] * b[n]));

这足以证明成功.

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