Problem Transformed CSR 04 Ex4 7 15 Bor03 GM

Tool Bounds

Execution Time	60.032814ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex4 7 15 Bor03 GM

stdout:

TIMEOUT

We consider the following Problem:

  Strict Trs:
    {  a__p(X) -> p(X)
     , a__f(X) -> f(X)
     , mark(s(X)) -> s(mark(X))
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(0()) -> 0()
     , mark(p(X)) -> a__p(mark(X))
     , mark(f(X)) -> a__f(mark(X))
     , a__p(s(0())) -> 0()
     , a__f(s(0())) -> a__f(a__p(s(0())))
     , a__f(0()) -> cons(0(), f(s(0())))}
  StartTerms: all
  Strategy: none

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	28.069267ms
Answer	MAYBE
Input	Transformed CSR 04 Ex4 7 15 Bor03 GM

stdout:

MAYBE

Statistics:
Number of monomials: 1464
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	3.7485461ms
Answer	YES(?,O(n^3))
Input	Transformed CSR 04 Ex4 7 15 Bor03 GM

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  a__p(X) -> p(X)
     , a__f(X) -> f(X)
     , mark(s(X)) -> s(mark(X))
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(0()) -> 0()
     , mark(p(X)) -> a__p(mark(X))
     , mark(f(X)) -> a__f(mark(X))
     , a__p(s(0())) -> 0()
     , a__f(s(0())) -> a__f(a__p(s(0())))
     , a__f(0()) -> cons(0(), f(s(0())))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   0() = [0]
         [0]
         [2]
   a__f(x1) = [1 0 2] x1 + [1]
              [0 1 2]      [1]
              [0 0 0]      [3]
   s(x1) = [1 0 0] x1 + [0]
           [0 1 0]      [0]
           [0 0 1]      [2]
   f(x1) = [1 0 1] x1 + [0]
           [0 1 2]      [1]
           [0 0 0]      [3]
   cons(x1, x2) = [1 0 0] x1 + [1 0 0] x2 + [0]
                  [0 1 2]      [0 0 0]      [0]
                  [0 0 0]      [0 0 0]      [1]
   a__p(x1) = [1 0 0] x1 + [1]
              [0 1 1]      [0]
              [0 0 0]      [2]
   mark(x1) = [1 1 1] x1 + [0]
              [0 1 0]      [0]
              [0 0 1]      [0]
   p(x1) = [1 0 0] x1 + [0]
           [0 1 1]      [0]
           [0 0 0]      [2]

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

Tool IDA

Execution Time	60.032608ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex4 7 15 Bor03 GM

stdout:

Tool TRI

Execution Time	1.137178ms
Answer	YES(?,O(n^3))
Input	Transformed CSR 04 Ex4 7 15 Bor03 GM

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  a__p(X) -> p(X)
     , a__f(X) -> f(X)
     , mark(s(X)) -> s(mark(X))
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(0()) -> 0()
     , mark(p(X)) -> a__p(mark(X))
     , mark(f(X)) -> a__f(mark(X))
     , a__p(s(0())) -> 0()
     , a__f(s(0())) -> a__f(a__p(s(0())))
     , a__f(0()) -> cons(0(), f(s(0())))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   0() = [0]
         [2]
         [2]
   a__f(x1) = [1 0 2] x1 + [1]
              [0 1 1]      [2]
              [0 0 0]      [0]
   s(x1) = [1 0 0] x1 + [0]
           [0 1 0]      [2]
           [0 0 1]      [2]
   f(x1) = [1 0 0] x1 + [0]
           [0 1 1]      [2]
           [0 0 0]      [0]
   cons(x1, x2) = [1 0 0] x1 + [1 0 0] x2 + [0]
                  [0 1 0]      [0 0 0]      [1]
                  [0 0 0]      [0 0 0]      [0]
   a__p(x1) = [1 0 0] x1 + [2]
              [0 1 0]      [1]
              [0 0 0]      [2]
   mark(x1) = [1 2 0] x1 + [0]
              [0 1 0]      [0]
              [0 0 1]      [0]
   p(x1) = [1 0 0] x1 + [1]
           [0 1 0]      [1]
           [0 0 0]      [2]

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

Tool TRI2

Execution Time	0.26111913ms
Answer	MAYBE
Input	Transformed CSR 04 Ex4 7 15 Bor03 GM

stdout:

MAYBE

We consider the following Problem:

  Strict Trs:
    {  a__p(X) -> p(X)
     , a__f(X) -> f(X)
     , mark(s(X)) -> s(mark(X))
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(0()) -> 0()
     , mark(p(X)) -> a__p(mark(X))
     , mark(f(X)) -> a__f(mark(X))
     , a__p(s(0())) -> 0()
     , a__f(s(0())) -> a__f(a__p(s(0())))
     , a__f(0()) -> cons(0(), f(s(0())))}
  StartTerms: all
  Strategy: none

Certificate: MAYBE

Proof:
  The input cannot be shown compatible

Arrrr..