3

我正在尝试对以下 LTL 属性的简单 Promela 模型进行模型检查：

ltl { M[0] U M[1] }

我得到一个错误，错误线索上的引导模拟产生以下输出：

ltl ltl_0: (M[0]) U (M[1])
spin: couldn't find claim 2 (ignored) 
0 :init ini M[0] = 1             
Process Statement                M[0]       M[1]       
0 :init ini M[1] = 0             1          0          
Starting net with pid 2
0 :init ini run net()            1          0          
spin: trail ends after 4 steps
#processes: 2
  4:    proc  1 (net) petri:11 (state 13)
  4:    proc  0 (:init:) petri:25 (state 5)
2 processes created
Exit-Status 0

现在我看不到“M [0]直到M [1]”在这里被违反。M[0] 在初始化过程中被设置为 1，并且一直如此，直到 M[1] 变为 1。并且跟踪结束得这么早，或者我可能完全误解了“stronguntil”的语义。我非常有信心是这种情况......但我做错了什么？在 Promela 文件中指定 LTL 可以吗？

有问题的模型如下（一个简单的 petri 网）：

#define nPlaces 2
#define nTransitions 2
#define inp1(x1) (x1>0) -> x1--
#define out1(x1) x1++

int M[nPlaces];
int T[nTransitions];

proctype net()
{
    do
    ::  d_step{inp1(M[0])->T[0]++;out1(M[1]);skip}
    ::  d_step{inp1(M[1])->T[1]++;out1(M[0]);skip}
    od
}
init
{
    atomic 
    {
        M[0] = 1;
        M[1] = 0;
    }
    run net();
}

ltl { M[0] U M[1] }
4

2 回答 2

2

Your claim is violated in the initial state (before the init use of atomic). Here is a SPIN verification run (pan -a) with output from the trail file:

ebg@ebg$ spin -a foo.pml
ltl ltl_0: (M[0]) U (M[1])
ebg@ebg$ gcc -o pan pan.c
ebg@ebg$ ./pan -a
pan:1: assertion violated  !(( !(M[0])&& !(M[1]))) (at depth 0)
pan: wrote foo.pml.trail

(Spin Version 6.2.4 -- 21 November 2012)
Warning: Search not completed
    + Partial Order Reduction

Full statespace search for:
    never claim             + (ltl_0)
    assertion violations    + (if within scope of claim)
    acceptance   cycles     + (fairness disabled)
    invalid end states  - (disabled by never claim)

State-vector 36 byte, depth reached 6, errors: 1
        4 states, stored (7 visited)
        1 states, matched
        8 transitions (= visited+matched)
        0 atomic steps
hash conflicts:         0 (resolved)

Stats on memory usage (in Megabytes):
    0.000   equivalent memory usage for states (stored*(State-vector + overhead))
    0.290   actual memory usage for states
  128.000   memory used for hash table (-w24)
    0.534   memory used for DFS stack (-m10000)
  128.730   total actual memory usage



pan: elapsed time 0 seconds
ebg@ebg$ spin -p -t foo.pml
ltl ltl_0: (M[0]) U (M[1])
starting claim 2
using statement merging
spin: _spin_nvr.tmp:5, Error: assertion violated
spin: text of failed assertion: assert(!((!(M[0])&&!(M[1]))))
Never claim moves to line 5 [D_STEP]
spin: trail ends after 1 steps
#processes: 1
        M[0] = 0
        M[1] = 0
        T[0] = 0
        T[1] = 0
  1:    proc  0 (:init:) foo.pml:18 (state 3)
  1:    proc  - (ltl_0) _spin_nvr.tmp:3 (state 6)
1 processes created

you can see that the ltl was translated to: assert(!(( !(M[0])&& !(M[1])))) which is:

  !(( !0 && !0))
  !((  1 &&  1))
  !((  1 ))
  0

and thus the assertion is violated.

The easiest way to avoid the problem is to change your arrays to just separate variables. Since your arrays are just size 2 it is easy to do:

int M0 = 1;
int M1 = 0;
int T0 = 0;
int T1 = 0;
/* then use as appropriate. */

With this, you can skip the init and just make the net proctype declared as active proctype net ()

于 2013-06-18T14:21:48.600 回答
1

您的 ltl 公式放置得很好。如果您使用 ispin 并验证（而不是模拟）您的程序，请确保选择“使用声明”选项。注意：默认值为“不使用从不声明或 ltl 属性”。

于 2013-06-17T13:33:24.903 回答