Problem SK90 4.51

Tool Bounds

Execution Time	5.6545973e-2ms
Answer	YES(?,O(n^1))
Input	SK90 4.51

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  k(f(x), y, x) -> f(x)
     , k(x, h(x), a()) -> h(x)
     , h(g(x)) -> g(h(f(x)))
     , f(a()) -> g(h(a()))}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^1))

Proof:
  The problem is match-bounded by 3.
  The enriched problem is compatible with the following automaton:
  {  a_0() -> 1
   , a_1() -> 1
   , a_2() -> 3
   , f_0(1) -> 1
   , f_1(1) -> 1
   , f_1(2) -> 1
   , f_2(1) -> 3
   , f_2(2) -> 3
   , f_3(2) -> 5
   , h_0(1) -> 1
   , h_1(1) -> 1
   , h_2(3) -> 2
   , h_3(5) -> 4
   , g_0(1) -> 1
   , g_1(1) -> 1
   , g_1(1) -> 3
   , g_2(2) -> 1
   , g_2(2) -> 2
   , g_2(2) -> 3
   , g_3(4) -> 2
   , k_0(1, 1, 1) -> 1}

Hurray, we answered YES(?,O(n^1))

Tool CDI

Execution Time	3.033406ms
Answer	MAYBE
Input	SK90 4.51

stdout:

MAYBE

Statistics:
Number of monomials: 364
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	1.575707ms
Answer	YES(?,O(n^3))
Input	SK90 4.51

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  k(f(x), y, x) -> f(x)
     , k(x, h(x), a()) -> h(x)
     , h(g(x)) -> g(h(f(x)))
     , f(a()) -> g(h(a()))}
  StartTerms: all
  Strategy: none

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

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

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

Tool IDA

Execution Time	4.34901ms
Answer	YES(?,O(n^3))
Input	SK90 4.51

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  k(f(x), y, x) -> f(x)
     , k(x, h(x), a()) -> h(x)
     , h(g(x)) -> g(h(f(x)))
     , f(a()) -> g(h(a()))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying and IDA(3)-non-satisfying matrix interpretation:
  Interpretation Functions:
   a() = [0]
         [0]
         [2]
   f(x1) = [1 0 2] x1 + [0]
           [0 0 1]      [0]
           [0 0 0]      [0]
   h(x1) = [1 2 0] x1 + [0]
           [0 0 0]      [1]
           [0 0 0]      [0]
   g(x1) = [1 0 0] x1 + [2]
           [0 0 2]      [1]
           [0 0 0]      [0]
   k(x1, x2, x3) = [1 0 1] x1 + [1 0 0] x2 + [1 0 1] x3 + [2]
                   [0 0 1]      [0 0 0]      [0 0 1]      [0]
                   [0 0 0]      [0 0 0]      [0 0 0]      [0]

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

Tool TRI

Execution Time	0.49087882ms
Answer	YES(?,O(n^3))
Input	SK90 4.51

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  k(f(x), y, x) -> f(x)
     , k(x, h(x), a()) -> h(x)
     , h(g(x)) -> g(h(f(x)))
     , f(a()) -> g(h(a()))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   a() = [0]
         [0]
         [2]
   f(x1) = [1 0 1] x1 + [0]
           [0 0 2]      [0]
           [0 0 0]      [0]
   h(x1) = [1 1 0] x1 + [0]
           [0 0 0]      [2]
           [0 0 0]      [0]
   g(x1) = [1 0 3] x1 + [0]
           [0 0 0]      [1]
           [0 0 0]      [0]
   k(x1, x2, x3) = [1 0 0] x1 + [1 0 3] x2 + [1 3 0] x3 + [2]
                   [0 1 0]      [0 0 3]      [0 1 2]      [0]
                   [0 0 0]      [0 0 1]      [0 0 0]      [0]

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

Tool TRI2

Execution Time	0.13393521ms
Answer	MAYBE
Input	SK90 4.51

stdout:

MAYBE

We consider the following Problem:

  Strict Trs:
    {  k(f(x), y, x) -> f(x)
     , k(x, h(x), a()) -> h(x)
     , h(g(x)) -> g(h(f(x)))
     , f(a()) -> g(h(a()))}
  StartTerms: all
  Strategy: none

Certificate: MAYBE

Proof:
  The input cannot be shown compatible

Arrrr..