Problem AProVE 04 Liveness8

Tool Bounds

Execution Time	60.048985ms
Answer	TIMEOUT
Input	AProVE 04 Liveness8

stdout:

TIMEOUT

We consider the following Problem:

  Strict Trs:
    {  f(mark(x)) -> mark(f(x))
     , active(f(x)) -> f(active(x))
     , top(found(x)) -> top(active(x))
     , f(found(x)) -> found(f(x))
     , start(ok(x)) -> found(x)
     , f(ok(x)) -> ok(f(x))
     , proper(f(x)) -> f(proper(x))
     , proper(c()) -> ok(c())
     , match(X(), x) -> proper(x)
     , match(f(x), f(y)) -> f(match(x, y))
     , check(x) -> start(match(f(X()), x))
     , check(f(x)) -> f(check(x))
     , top(mark(x)) -> top(check(x))
     , top(active(c())) -> top(mark(c()))
     , active(f(x)) -> mark(x)}
  StartTerms: all
  Strategy: none

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	3.5478568ms
Answer	MAYBE
Input	AProVE 04 Liveness8

stdout:

MAYBE

Statistics:
Number of monomials: 948
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	60.035442ms
Answer	TIMEOUT
Input	AProVE 04 Liveness8

stdout:

Tool IDA

Execution Time	32.30698ms
Answer	YES(?,O(n^3))
Input	AProVE 04 Liveness8

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  f(mark(x)) -> mark(f(x))
     , active(f(x)) -> f(active(x))
     , top(found(x)) -> top(active(x))
     , f(found(x)) -> found(f(x))
     , start(ok(x)) -> found(x)
     , f(ok(x)) -> ok(f(x))
     , proper(f(x)) -> f(proper(x))
     , proper(c()) -> ok(c())
     , match(X(), x) -> proper(x)
     , match(f(x), f(y)) -> f(match(x, y))
     , check(x) -> start(match(f(X()), x))
     , check(f(x)) -> f(check(x))
     , top(mark(x)) -> top(check(x))
     , top(active(c())) -> top(mark(c()))
     , active(f(x)) -> mark(x)}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^3))

Proof:
  We have the following EDA-non-satisfying and IDA(3)-non-satisfying matrix interpretation:
  Interpretation Functions:
   f(x1) = [1 3 0] x1 + [0]
           [0 1 0]      [2]
           [0 0 0]      [0]
   active(x1) = [1 1 0] x1 + [0]
                [0 1 0]      [0]
                [0 0 1]      [0]
   mark(x1) = [1 2 0] x1 + [1]
              [0 1 0]      [2]
              [0 0 0]      [0]
   c() = [0]
         [2]
         [3]
   top(x1) = [1 2 3] x1 + [0]
             [0 0 0]      [0]
             [0 0 0]      [0]
   check(x1) = [1 2 0] x1 + [2]
               [0 1 0]      [1]
               [0 0 0]      [0]
   X() = [0]
         [0]
         [3]
   match(x1, x2) = [1 0 0] x1 + [1 2 0] x2 + [1]
                   [0 0 0]      [0 1 0]      [1]
                   [0 0 1]      [0 0 0]      [0]
   start(x1) = [1 0 0] x1 + [0]
               [0 1 0]      [0]
               [0 0 1]      [0]
   proper(x1) = [1 2 0] x1 + [0]
                [0 1 0]      [1]
                [0 0 0]      [3]
   ok(x1) = [1 1 0] x1 + [1]
            [0 1 0]      [1]
            [0 0 1]      [0]
   found(x1) = [1 1 0] x1 + [0]
               [0 1 0]      [1]
               [0 0 1]      [0]

Hurray, we answered YES(?,O(n^3))

Tool TRI

Execution Time	4.573349ms
Answer	YES(?,O(n^3))
Input	AProVE 04 Liveness8

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  f(mark(x)) -> mark(f(x))
     , active(f(x)) -> f(active(x))
     , top(found(x)) -> top(active(x))
     , f(found(x)) -> found(f(x))
     , start(ok(x)) -> found(x)
     , f(ok(x)) -> ok(f(x))
     , proper(f(x)) -> f(proper(x))
     , proper(c()) -> ok(c())
     , match(X(), x) -> proper(x)
     , match(f(x), f(y)) -> f(match(x, y))
     , check(x) -> start(match(f(X()), x))
     , check(f(x)) -> f(check(x))
     , top(mark(x)) -> top(check(x))
     , top(active(c())) -> top(mark(c()))
     , active(f(x)) -> mark(x)}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^3))

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   f(x1) = [1 3 0] x1 + [0]
           [0 1 0]      [2]
           [0 0 0]      [0]
   active(x1) = [1 1 0] x1 + [0]
                [0 1 0]      [0]
                [0 0 1]      [0]
   mark(x1) = [1 2 0] x1 + [1]
              [0 1 0]      [2]
              [0 0 0]      [0]
   c() = [0]
         [2]
         [3]
   top(x1) = [1 2 3] x1 + [0]
             [0 1 1]      [2]
             [0 0 0]      [0]
   check(x1) = [1 2 0] x1 + [2]
               [0 1 0]      [1]
               [0 0 0]      [0]
   X() = [0]
         [0]
         [3]
   match(x1, x2) = [1 0 0] x1 + [1 2 0] x2 + [1]
                   [0 0 0]      [0 1 0]      [1]
                   [0 0 1]      [0 0 0]      [0]
   start(x1) = [1 0 2] x1 + [0]
               [0 1 1]      [0]
               [0 0 0]      [0]
   proper(x1) = [1 2 0] x1 + [0]
                [0 1 0]      [1]
                [0 0 0]      [3]
   ok(x1) = [1 1 0] x1 + [1]
            [0 1 0]      [1]
            [0 0 1]      [0]
   found(x1) = [1 1 1] x1 + [0]
               [0 1 1]      [1]
               [0 0 0]      [0]

Hurray, we answered YES(?,O(n^3))

Tool TRI2

Execution Time	0.48886204ms
Answer	MAYBE
Input	AProVE 04 Liveness8

stdout:

MAYBE

We consider the following Problem:

  Strict Trs:
    {  f(mark(x)) -> mark(f(x))
     , active(f(x)) -> f(active(x))
     , top(found(x)) -> top(active(x))
     , f(found(x)) -> found(f(x))
     , start(ok(x)) -> found(x)
     , f(ok(x)) -> ok(f(x))
     , proper(f(x)) -> f(proper(x))
     , proper(c()) -> ok(c())
     , match(X(), x) -> proper(x)
     , match(f(x), f(y)) -> f(match(x, y))
     , check(x) -> start(match(f(X()), x))
     , check(f(x)) -> f(check(x))
     , top(mark(x)) -> top(check(x))
     , top(active(c())) -> top(mark(c()))
     , active(f(x)) -> mark(x)}
  StartTerms: all
  Strategy: none

Certificate: MAYBE

Proof:
  The input cannot be shown compatible

Arrrr..