Problem SK90 2.61

Tool Bounds

Execution Time	0.11207509ms
Answer	YES(?,O(n^1))
Input	SK90 2.61

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  k(h1(x, y)) -> h1(s(x), y)
     , k(h(x)) -> h1(0(), x)
     , i(h2(s(x), y, h1(x, z))) -> z
     , i(f(x, h(y))) -> y
     , h2(x, j(y, h1(z, u)), h1(z, u)) -> h2(s(x), y, h1(s(z), u))
     , g(h2(x, y, h1(z, u))) -> h2(s(x), y, h1(z, u))
     , f(x, h1(y, z)) -> h2(0(), x, h1(y, z))
     , f(j(x, y), y) -> g(f(x, k(y)))}
  StartTerms: all
  Strategy: none

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

Proof:
  The problem is match-bounded by 2.
  The enriched problem is compatible with the following automaton:
  {  j_0(1, 1) -> 1
   , f_0(1, 1) -> 1
   , f_1(1, 4) -> 3
   , k_0(1) -> 1
   , k_1(1) -> 4
   , g_0(1) -> 1
   , g_1(3) -> 1
   , g_1(3) -> 3
   , h1_0(1, 1) -> 1
   , h1_1(1, 1) -> 1
   , h1_1(2, 1) -> 1
   , h1_1(2, 1) -> 4
   , h1_1(5, 1) -> 1
   , h1_2(1, 1) -> 4
   , h1_2(2, 1) -> 7
   , h1_2(5, 1) -> 4
   , 0_0() -> 1
   , 0_1() -> 2
   , 0_2() -> 6
   , h2_0(1, 1, 1) -> 1
   , h2_1(1, 1, 1) -> 1
   , h2_1(1, 1, 1) -> 3
   , h2_1(1, 1, 4) -> 1
   , h2_1(1, 1, 4) -> 3
   , h2_1(2, 1, 1) -> 1
   , h2_1(2, 1, 1) -> 3
   , h2_1(2, 1, 4) -> 3
   , h2_1(8, 1, 1) -> 1
   , h2_1(8, 1, 1) -> 3
   , h2_1(8, 1, 4) -> 3
   , h2_2(5, 1, 4) -> 1
   , h2_2(5, 1, 4) -> 3
   , h2_2(5, 1, 7) -> 1
   , h2_2(5, 1, 7) -> 3
   , h2_2(6, 1, 4) -> 3
   , h2_2(6, 1, 7) -> 3
   , h2_2(9, 1, 4) -> 1
   , h2_2(9, 1, 4) -> 3
   , h2_2(9, 1, 7) -> 1
   , h2_2(9, 1, 7) -> 3
   , s_0(1) -> 1
   , s_1(1) -> 2
   , s_1(2) -> 2
   , s_1(5) -> 1
   , s_1(6) -> 8
   , s_1(8) -> 8
   , s_1(9) -> 2
   , s_2(1) -> 5
   , s_2(2) -> 5
   , s_2(5) -> 5
   , s_2(6) -> 9
   , s_2(8) -> 9
   , s_2(9) -> 5
   , h_0(1) -> 1
   , i_0(1) -> 1}

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

Tool CDI

Execution Time	27.767616ms
Answer	YES(?,O(n^2))
Input	SK90 2.61

stdout:

YES(?,O(n^2))
QUADRATIC upper bound on the derivational complexity

This TRS is terminating using the deltarestricted interpretation
h(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 0*delta
i(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 2*delta
s(delta, X0) =  + 0*X0 + 0 + 2*X0*delta + 0*delta
h1(delta, X1, X0) =  + 1*X0 + 0*X1 + 0 + 0*X0*delta + 2*X1*delta + 0*delta
0(delta) =  + 0 + 0*delta
h2(delta, X2, X1, X0) =  + 1*X0 + 1*X1 + 0*X2 + 0 + 0*X0*delta + 0*X1*delta + 2*X2*delta + 0*delta
j(delta, X1, X0) =  + 1*X0 + 1*X1 + 2 + 0*X0*delta + 0*X1*delta + 3*delta
k(delta, X0) =  + 1*X0 + 2 + 0*X0*delta + 1*delta
f(delta, X1, X0) =  + 1*X0 + 1*X1 + 0 + 0*X0*delta + 0*X1*delta + 2*delta
g(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 1*delta
h_tau_1(delta) = delta/(1 + 0 * delta)
i_tau_1(delta) = delta/(1 + 0 * delta)
s_tau_1(delta) = delta/(0 + 2 * delta)
h1_tau_1(delta) = delta/(0 + 2 * delta)
h1_tau_2(delta) = delta/(1 + 0 * delta)
h2_tau_1(delta) = delta/(0 + 2 * delta)
h2_tau_2(delta) = delta/(1 + 0 * delta)
h2_tau_3(delta) = delta/(1 + 0 * delta)
j_tau_1(delta) = delta/(1 + 0 * delta)
j_tau_2(delta) = delta/(1 + 0 * delta)
k_tau_1(delta) = delta/(1 + 0 * delta)
f_tau_1(delta) = delta/(1 + 0 * delta)
f_tau_2(delta) = delta/(1 + 0 * delta)
g_tau_1(delta) = delta/(1 + 0 * delta)

Time: 27.727868 seconds
Statistics:
Number of monomials: 2779
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.18326497ms
Answer	YES(?,O(n^1))
Input	SK90 2.61

stdout:

Tool IDA

Execution Time	0.534776ms
Answer	YES(?,O(n^1))
Input	SK90 2.61

stdout:

Tool TRI

Execution Time	0.146775ms
Answer	YES(?,O(n^1))
Input	SK90 2.61

stdout:

Tool TRI2

Execution Time	0.413409ms
Answer	YES(?,O(n^2))
Input	SK90 2.61

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  k(h1(x, y)) -> h1(s(x), y)
     , k(h(x)) -> h1(0(), x)
     , i(h2(s(x), y, h1(x, z))) -> z
     , i(f(x, h(y))) -> y
     , h2(x, j(y, h1(z, u)), h1(z, u)) -> h2(s(x), y, h1(s(z), u))
     , g(h2(x, y, h1(z, u))) -> h2(s(x), y, h1(z, u))
     , f(x, h1(y, z)) -> h2(0(), x, h1(y, z))
     , f(j(x, y), y) -> g(f(x, k(y)))}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^2))

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   j(x1, x2) = [1 1] x1 + [1 0] x2 + [3]
               [0 0]      [0 0]      [3]
   f(x1, x2) = [1 1] x1 + [1 0] x2 + [3]
               [0 0]      [0 1]      [0]
   k(x1) = [1 0] x1 + [3]
           [0 1]      [0]
   g(x1) = [1 0] x1 + [2]
           [0 1]      [0]
   h1(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                [0 0]      [0 1]      [0]
   0() = [0]
         [0]
   h2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 0] x3 + [0]
                    [0 0]      [0 0]      [0 1]      [0]
   s(x1) = [1 0] x1 + [0]
           [0 0]      [0]
   h(x1) = [1 3] x1 + [2]
           [0 1]      [0]
   i(x1) = [1 0] x1 + [0]
           [0 1]      [3]

Hurray, we answered YES(?,O(n^2))