Problem Transformed CSR 04 Ex9 Luc06 iGM

Tool Bounds

Execution Time	60.071934ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex9 Luc06 iGM

stdout:

TIMEOUT

We consider the following Problem:

  Strict Trs:
    {  f(X1, X2, active(X3)) -> f(X1, X2, X3)
     , f(X1, active(X2), X3) -> f(X1, X2, X3)
     , f(active(X1), X2, X3) -> f(X1, X2, X3)
     , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
     , f(X1, mark(X2), X3) -> f(X1, X2, X3)
     , f(mark(X1), X2, X3) -> f(X1, X2, X3)
     , mark(b()) -> active(b())
     , mark(a()) -> active(a())
     , mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
     , active(b()) -> mark(a())
     , active(f(a(), X, X)) -> mark(f(X, b(), b()))}
  StartTerms: all
  Strategy: none

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	37.860104ms
Answer	MAYBE
Input	Transformed CSR 04 Ex9 Luc06 iGM

stdout:

MAYBE

Statistics:
Number of monomials: 1800
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	12.691862ms
Answer	YES(?,O(n^4))
Input	Transformed CSR 04 Ex9 Luc06 iGM

stdout:

YES(?,O(n^4))

We consider the following Problem:

  Strict Trs:
    {  f(X1, X2, active(X3)) -> f(X1, X2, X3)
     , f(X1, active(X2), X3) -> f(X1, X2, X3)
     , f(active(X1), X2, X3) -> f(X1, X2, X3)
     , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
     , f(X1, mark(X2), X3) -> f(X1, X2, X3)
     , f(mark(X1), X2, X3) -> f(X1, X2, X3)
     , mark(b()) -> active(b())
     , mark(a()) -> active(a())
     , mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
     , active(b()) -> mark(a())
     , active(f(a(), X, X)) -> mark(f(X, b(), b()))}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^4))

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   a() = [0]
         [2]
         [0]
         [0]
   f(x1, x2, x3) = [1 3 0 0] x1 + [1 1 2 0] x2 + [1 2 0 0] x3 + [0]
                   [0 0 0 0]      [0 0 0 0]      [0 0 0 0]      [0]
                   [0 1 0 0]      [0 0 0 0]      [0 1 0 0]      [0]
                   [0 0 0 0]      [0 0 0 1]      [0 0 0 0]      [1]
   active(x1) = [1 0 1 0] x1 + [0]
                [0 1 1 0]      [1]
                [0 0 0 0]      [0]
                [0 0 0 1]      [0]
   b() = [0]
         [0]
         [2]
         [0]
   mark(x1) = [1 0 1 2] x1 + [1]
              [0 1 1 0]      [1]
              [0 0 0 0]      [0]
              [0 0 0 1]      [0]

Hurray, we answered YES(?,O(n^4))

Tool IDA

Execution Time	60.03579ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex9 Luc06 iGM

stdout:

Tool TRI

Execution Time	6.026172ms
Answer	YES(?,O(n^3))
Input	Transformed CSR 04 Ex9 Luc06 iGM

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  f(X1, X2, active(X3)) -> f(X1, X2, X3)
     , f(X1, active(X2), X3) -> f(X1, X2, X3)
     , f(active(X1), X2, X3) -> f(X1, X2, X3)
     , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
     , f(X1, mark(X2), X3) -> f(X1, X2, X3)
     , f(mark(X1), X2, X3) -> f(X1, X2, X3)
     , mark(b()) -> active(b())
     , mark(a()) -> active(a())
     , mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
     , active(b()) -> mark(a())
     , active(f(a(), X, X)) -> mark(f(X, b(), b()))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   a() = [0]
         [2]
         [2]
         [0]
         [0]
   f(x1, x2, x3) = [1 2 3 0 1] x1 + [1 0 0 2 1] x2 + [1 3 3 0 1] x3 + [0]
                   [0 1 0 0 0]      [0 1 0 1 0]      [0 0 0 0 0]      [3]
                   [0 0 0 0 0]      [0 0 0 0 0]      [0 0 0 0 0]      [0]
                   [0 0 0 0 0]      [0 0 0 0 0]      [0 0 0 0 0]      [0]
                   [0 0 0 0 0]      [0 0 0 0 0]      [0 0 0 0 0]      [1]
   active(x1) = [1 0 0 2 2] x1 + [0]
                [0 1 0 1 0]      [0]
                [0 0 1 1 0]      [0]
                [0 0 0 0 0]      [0]
                [0 0 0 0 0]      [1]
   b() = [0]
         [0]
         [0]
         [2]
         [0]
   mark(x1) = [1 1 0 3 1] x1 + [0]
              [0 1 0 1 0]      [0]
              [0 0 1 2 0]      [0]
              [0 0 0 0 0]      [0]
              [0 0 0 0 0]      [1]

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

Tool TRI2

Execution Time	0.2701521ms
Answer	MAYBE
Input	Transformed CSR 04 Ex9 Luc06 iGM

stdout:

MAYBE

We consider the following Problem:

  Strict Trs:
    {  f(X1, X2, active(X3)) -> f(X1, X2, X3)
     , f(X1, active(X2), X3) -> f(X1, X2, X3)
     , f(active(X1), X2, X3) -> f(X1, X2, X3)
     , f(X1, X2, mark(X3)) -> f(X1, X2, X3)
     , f(X1, mark(X2), X3) -> f(X1, X2, X3)
     , f(mark(X1), X2, X3) -> f(X1, X2, X3)
     , mark(b()) -> active(b())
     , mark(a()) -> active(a())
     , mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
     , active(b()) -> mark(a())
     , active(f(a(), X, X)) -> mark(f(X, b(), b()))}
  StartTerms: all
  Strategy: none

Certificate: MAYBE

Proof:
  The input cannot be shown compatible

Arrrr..