Problem Transformed CSR 04 Ex4 7 77 Bor03 iGM

Tool Bounds

Execution Time	60.035553ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex4 7 77 Bor03 iGM

stdout:

TIMEOUT

We consider the following Problem:

  Strict Trs:
    {  tail(active(X)) -> tail(X)
     , tail(mark(X)) -> tail(X)
     , cons(X1, active(X2)) -> cons(X1, X2)
     , cons(active(X1), X2) -> cons(X1, X2)
     , cons(X1, mark(X2)) -> cons(X1, X2)
     , cons(mark(X1), X2) -> cons(X1, X2)
     , mark(tail(X)) -> active(tail(mark(X)))
     , mark(0()) -> active(0())
     , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
     , mark(zeros()) -> active(zeros())
     , active(tail(cons(X, XS))) -> mark(XS)
     , active(zeros()) -> mark(cons(0(), zeros()))}
  StartTerms: all
  Strategy: none

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	7.623403ms
Answer	MAYBE
Input	Transformed CSR 04 Ex4 7 77 Bor03 iGM

stdout:

MAYBE

Statistics:
Number of monomials: 941
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	59.014626ms
Answer	YES(?,O(n^4))
Input	Transformed CSR 04 Ex4 7 77 Bor03 iGM

stdout:

YES(?,O(n^4))

We consider the following Problem:

  Strict Trs:
    {  tail(active(X)) -> tail(X)
     , tail(mark(X)) -> tail(X)
     , cons(X1, active(X2)) -> cons(X1, X2)
     , cons(active(X1), X2) -> cons(X1, X2)
     , cons(X1, mark(X2)) -> cons(X1, X2)
     , cons(mark(X1), X2) -> cons(X1, X2)
     , mark(tail(X)) -> active(tail(mark(X)))
     , mark(0()) -> active(0())
     , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
     , mark(zeros()) -> active(zeros())
     , active(tail(cons(X, XS))) -> mark(XS)
     , active(zeros()) -> mark(cons(0(), zeros()))}
  StartTerms: all
  Strategy: none

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

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

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

Tool IDA

Execution Time	60.03522ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex4 7 77 Bor03 iGM

stdout:

Tool TRI

Execution Time	6.314977ms
Answer	YES(?,O(n^3))
Input	Transformed CSR 04 Ex4 7 77 Bor03 iGM

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  tail(active(X)) -> tail(X)
     , tail(mark(X)) -> tail(X)
     , cons(X1, active(X2)) -> cons(X1, X2)
     , cons(active(X1), X2) -> cons(X1, X2)
     , cons(X1, mark(X2)) -> cons(X1, X2)
     , cons(mark(X1), X2) -> cons(X1, X2)
     , mark(tail(X)) -> active(tail(mark(X)))
     , mark(0()) -> active(0())
     , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
     , mark(zeros()) -> active(zeros())
     , active(tail(cons(X, XS))) -> mark(XS)
     , active(zeros()) -> mark(cons(0(), zeros()))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   zeros() = [2]
             [2]
             [0]
             [3]
   active(x1) = [1 0 0 3] x1 + [1]
                [0 1 0 1]      [0]
                [0 0 1 2]      [0]
                [0 0 0 0]      [0]
   0() = [0]
         [1]
         [0]
         [0]
   cons(x1, x2) = [1 0 0 0] x1 + [1 0 0 0] x2 + [0]
                  [0 1 0 1]      [0 1 0 0]      [1]
                  [0 0 0 0]      [0 0 1 2]      [0]
                  [0 0 0 0]      [0 0 0 0]      [0]
   mark(x1) = [1 2 0 2] x1 + [1]
              [0 1 0 1]      [0]
              [0 0 1 2]      [0]
              [0 0 0 0]      [0]
   tail(x1) = [1 2 1 0] x1 + [0]
              [0 1 1 3]      [1]
              [0 0 1 2]      [0]
              [0 0 0 0]      [0]

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

Tool TRI2

Execution Time	0.261266ms
Answer	MAYBE
Input	Transformed CSR 04 Ex4 7 77 Bor03 iGM

stdout:

MAYBE

We consider the following Problem:

  Strict Trs:
    {  tail(active(X)) -> tail(X)
     , tail(mark(X)) -> tail(X)
     , cons(X1, active(X2)) -> cons(X1, X2)
     , cons(active(X1), X2) -> cons(X1, X2)
     , cons(X1, mark(X2)) -> cons(X1, X2)
     , cons(mark(X1), X2) -> cons(X1, X2)
     , mark(tail(X)) -> active(tail(mark(X)))
     , mark(0()) -> active(0())
     , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
     , mark(zeros()) -> active(zeros())
     , active(tail(cons(X, XS))) -> mark(XS)
     , active(zeros()) -> mark(cons(0(), zeros()))}
  StartTerms: all
  Strategy: none

Certificate: MAYBE

Proof:
  The input cannot be shown compatible

Arrrr..