Problem Transformed CSR 04 Ex25 Luc06 C

Tool Bounds

Execution Time	60.04978ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex25 Luc06 C

stdout:

TIMEOUT

We consider the following Problem:

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

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	60.04112ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex25 Luc06 C

stdout:

TIMEOUT

Statistics:
Number of monomials: 5222
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	24.55575ms
Answer	YES(?,O(n^3))
Input	Transformed CSR 04 Ex25 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))
     , h(ok(X)) -> ok(h(X))
     , d(ok(X)) -> ok(d(X))
     , g(ok(X)) -> ok(g(X))
     , c(ok(X)) -> ok(c(X))
     , f(ok(X)) -> ok(f(X))
     , proper(h(X)) -> h(proper(X))
     , proper(d(X)) -> d(proper(X))
     , proper(g(X)) -> g(proper(X))
     , proper(c(X)) -> c(proper(X))
     , proper(f(X)) -> f(proper(X))
     , h(mark(X)) -> mark(h(X))
     , f(mark(X)) -> mark(f(X))
     , active(h(X)) -> h(active(X))
     , active(f(X)) -> f(active(X))
     , active(h(X)) -> mark(c(d(X)))
     , active(c(X)) -> mark(d(X))
     , active(f(f(X))) -> mark(c(f(g(f(X)))))}
  StartTerms: all
  Strategy: none

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

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

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

Tool IDA

Execution Time	60.063576ms
Answer	TIMEOUT
Input	Transformed CSR 04 Ex25 Luc06 C

stdout:

Tool TRI

Execution Time	3.820266ms
Answer	YES(?,O(n^3))
Input	Transformed CSR 04 Ex25 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))
     , h(ok(X)) -> ok(h(X))
     , d(ok(X)) -> ok(d(X))
     , g(ok(X)) -> ok(g(X))
     , c(ok(X)) -> ok(c(X))
     , f(ok(X)) -> ok(f(X))
     , proper(h(X)) -> h(proper(X))
     , proper(d(X)) -> d(proper(X))
     , proper(g(X)) -> g(proper(X))
     , proper(c(X)) -> c(proper(X))
     , proper(f(X)) -> f(proper(X))
     , h(mark(X)) -> mark(h(X))
     , f(mark(X)) -> mark(f(X))
     , active(h(X)) -> h(active(X))
     , active(f(X)) -> f(active(X))
     , active(h(X)) -> mark(c(d(X)))
     , active(c(X)) -> mark(d(X))
     , active(f(f(X))) -> mark(c(f(g(f(X)))))}
  StartTerms: all
  Strategy: none

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

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

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

Tool TRI2

Execution Time	0.591969ms
Answer	MAYBE
Input	Transformed CSR 04 Ex25 Luc06 C

stdout:

MAYBE

We consider the following Problem:

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

Certificate: MAYBE

Proof:
  The input cannot be shown compatible

Arrrr..