Problem Transformed CSR 04 Ex18 Luc06 C

Tool Bounds

Execution Time	60.035675ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex18 Luc06 C

stdout:

TIMEOUT

We consider the following Problem:

  Strict Trs:
    {  top(ok(X)) -> top(active(X))
     , top(mark(X)) -> top(proper(X))
     , g(ok(X)) -> ok(g(X))
     , f(ok(X)) -> ok(f(X))
     , proper(g(X)) -> g(proper(X))
     , proper(a()) -> ok(a())
     , proper(f(X)) -> f(proper(X))
     , f(mark(X)) -> mark(f(X))
     , active(f(X)) -> f(active(X))
     , active(f(f(a()))) -> mark(f(g(f(a()))))}
  StartTerms: all
  Strategy: none

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	60.040962ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex18 Luc06 C

stdout:

TIMEOUT

Statistics:
Number of monomials: 0
Last formula building started for bound 0
Last SAT solving started for bound 0

Tool EDA

Execution Time	36.265884ms
Answer	YES(?,O(n^3))
Input	Transformed CSR 04 Ex18 Luc06 C

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  top(ok(X)) -> top(active(X))
     , top(mark(X)) -> top(proper(X))
     , g(ok(X)) -> ok(g(X))
     , f(ok(X)) -> ok(f(X))
     , proper(g(X)) -> g(proper(X))
     , proper(a()) -> ok(a())
     , proper(f(X)) -> f(proper(X))
     , f(mark(X)) -> mark(f(X))
     , active(f(X)) -> f(active(X))
     , active(f(f(a()))) -> mark(f(g(f(a()))))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   a() = [0]
         [0]
         [0]
   f(x1) = [1 3 0] x1 + [0]
           [0 1 0]      [2]
           [0 0 1]      [0]
   active(x1) = [1 1 0] x1 + [3]
                [0 1 0]      [0]
                [0 0 0]      [0]
   g(x1) = [1 0 3] x1 + [0]
           [0 0 1]      [0]
           [0 1 0]      [2]
   mark(x1) = [1 2 2] x1 + [0]
              [0 1 0]      [2]
              [0 0 0]      [0]
   proper(x1) = [1 2 2] x1 + [2]
                [0 1 0]      [1]
                [0 0 1]      [1]
   ok(x1) = [1 1 1] x1 + [1]
            [0 1 0]      [1]
            [0 0 1]      [1]
   top(x1) = [1 3 0] x1 + [0]
             [0 1 0]      [2]
             [0 0 0]      [0]

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

Tool IDA

Execution Time	60.04756ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex18 Luc06 C

stdout:

Tool TRI

Execution Time	24.635242ms
Answer	YES(?,O(n^4))
Input	Transformed CSR 04 Ex18 Luc06 C

stdout:

YES(?,O(n^4))

We consider the following Problem:

  Strict Trs:
    {  top(ok(X)) -> top(active(X))
     , top(mark(X)) -> top(proper(X))
     , g(ok(X)) -> ok(g(X))
     , f(ok(X)) -> ok(f(X))
     , proper(g(X)) -> g(proper(X))
     , proper(a()) -> ok(a())
     , proper(f(X)) -> f(proper(X))
     , f(mark(X)) -> mark(f(X))
     , active(f(X)) -> f(active(X))
     , active(f(f(a()))) -> mark(f(g(f(a()))))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   a() = [0]
         [0]
         [3]
         [0]
   f(x1) = [1 2 0 0] x1 + [0]
           [0 1 2 0]      [0]
           [0 0 0 0]      [0]
           [0 0 0 1]      [1]
   active(x1) = [1 0 0 1] x1 + [0]
                [0 1 0 0]      [0]
                [0 0 0 0]      [0]
                [0 0 0 1]      [0]
   g(x1) = [1 0 0 2] x1 + [0]
           [0 0 0 1]      [0]
           [0 0 0 0]      [0]
           [0 0 0 1]      [1]
   mark(x1) = [1 0 0 3] x1 + [0]
              [0 1 2 0]      [2]
              [0 0 0 0]      [0]
              [0 0 0 0]      [0]
   proper(x1) = [1 0 0 3] x1 + [1]
                [0 1 0 0]      [1]
                [0 0 1 0]      [0]
                [0 0 0 1]      [1]
   ok(x1) = [1 0 0 1] x1 + [0]
            [0 1 0 0]      [1]
            [0 0 1 0]      [0]
            [0 0 0 1]      [1]
   top(x1) = [1 3 0 0] x1 + [0]
             [0 1 0 0]      [2]
             [0 0 0 0]      [0]
             [0 0 0 0]      [0]

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

Tool TRI2

Execution Time	0.329926ms
Answer	MAYBE
Input	Transformed CSR 04 Ex18 Luc06 C

stdout:

MAYBE

We consider the following Problem:

  Strict Trs:
    {  top(ok(X)) -> top(active(X))
     , top(mark(X)) -> top(proper(X))
     , g(ok(X)) -> ok(g(X))
     , f(ok(X)) -> ok(f(X))
     , proper(g(X)) -> g(proper(X))
     , proper(a()) -> ok(a())
     , proper(f(X)) -> f(proper(X))
     , f(mark(X)) -> mark(f(X))
     , active(f(X)) -> f(active(X))
     , active(f(f(a()))) -> mark(f(g(f(a()))))}
  StartTerms: all
  Strategy: none

Certificate: MAYBE

Proof:
  The input cannot be shown compatible

Arrrr..