Problem Transformed CSR 04 Ex9 Luc06 GM

Tool Bounds

Execution Time	60.070522ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex9 Luc06 GM

stdout:

TIMEOUT

We consider the following Problem:

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

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	0.46067786ms
Answer	MAYBE
Input	Transformed CSR 04 Ex9 Luc06 GM

stdout:

MAYBE

Statistics:
Number of monomials: 306
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.329077ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex9 Luc06 GM

stdout:

YES(?,O(n^2))

We consider the following Problem:

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

Certificate: YES(?,O(n^2))

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

Hurray, we answered YES(?,O(n^2))

Tool IDA

Execution Time	0.6213808ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex9 Luc06 GM

stdout:

YES(?,O(n^2))

We consider the following Problem:

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

Certificate: YES(?,O(n^2))

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

Hurray, we answered YES(?,O(n^2))

Tool TRI

Execution Time	0.16180992ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex9 Luc06 GM

stdout:

YES(?,O(n^2))

We consider the following Problem:

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

Certificate: YES(?,O(n^2))

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

Hurray, we answered YES(?,O(n^2))

Tool TRI2

Execution Time	0.11725187ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex9 Luc06 GM

stdout:

YES(?,O(n^2))

We consider the following Problem:

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

Certificate: YES(?,O(n^2))

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

Hurray, we answered YES(?,O(n^2))