Problem Transformed CSR 04 Ex4 7 15 Bor03 L

Tool Bounds

Execution Time	3.137088e-2ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 15 Bor03 L

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  p(s(0())) -> 0()
     , f(s(0())) -> f(p(s(0())))
     , f(0()) -> cons(0())}
  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:
  {  0_0() -> 1
   , 0_1() -> 5
   , 0_2() -> 6
   , f_0(1) -> 2
   , f_0(2) -> 2
   , f_0(3) -> 2
   , f_0(4) -> 2
   , f_0(5) -> 2
   , f_1(6) -> 2
   , cons_0(1) -> 3
   , cons_0(2) -> 3
   , cons_0(3) -> 3
   , cons_0(4) -> 3
   , cons_0(5) -> 3
   , cons_1(5) -> 2
   , cons_2(6) -> 2
   , s_0(1) -> 4
   , s_0(2) -> 4
   , s_0(3) -> 4
   , s_0(4) -> 4
   , s_0(5) -> 4
   , s_1(5) -> 7
   , p_0(1) -> 5
   , p_0(2) -> 5
   , p_0(3) -> 5
   , p_0(4) -> 5
   , p_0(5) -> 5
   , p_1(7) -> 6}

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

Tool CDI

Execution Time	3.869558ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex4 7 15 Bor03 L

stdout:

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

This TRS is terminating using the deltarestricted interpretation
s(delta, X0) =  + 0*X0 + 2 + 1*X0*delta + 1*delta
p(delta, X0) =  + 0*X0 + 0 + 1*X0*delta + 0*delta
f(delta, X0) =  + 1*X0 + 2 + 2*X0*delta + 2*delta
0(delta) =  + 0 + 0*delta
cons(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 0*delta
s_tau_1(delta) = delta/(0 + 1 * delta)
p_tau_1(delta) = delta/(0 + 1 * delta)
f_tau_1(delta) = delta/(1 + 2 * delta)
cons_tau_1(delta) = delta/(1 + 0 * delta)

Time: 3.830622 seconds
Statistics:
Number of monomials: 338
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.39064598ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex4 7 15 Bor03 L

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  p(s(0())) -> 0()
     , f(s(0())) -> f(p(s(0())))
     , f(0()) -> cons(0())}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   0() = [0]
         [2]
   f(x1) = [1 1] x1 + [0]
           [0 0]      [0]
   cons(x1) = [1 0] x1 + [1]
              [0 0]      [0]
   s(x1) = [1 0] x1 + [0]
           [0 1]      [2]
   p(x1) = [1 0] x1 + [1]
           [0 0]      [2]

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

Tool IDA

Execution Time	0.6203451ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex4 7 15 Bor03 L

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  p(s(0())) -> 0()
     , f(s(0())) -> f(p(s(0())))
     , f(0()) -> cons(0())}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying and IDA(2)-non-satisfying matrix interpretation:
  Interpretation Functions:
   0() = [0]
         [0]
   f(x1) = [1 2] x1 + [1]
           [0 0]      [0]
   cons(x1) = [1 0] x1 + [0]
              [0 0]      [0]
   s(x1) = [1 0] x1 + [1]
           [0 0]      [2]
   p(x1) = [1 0] x1 + [0]
           [0 0]      [1]

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

Tool TRI

Execution Time	0.18804598ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex4 7 15 Bor03 L

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  p(s(0())) -> 0()
     , f(s(0())) -> f(p(s(0())))
     , f(0()) -> cons(0())}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   0() = [0]
         [0]
   f(x1) = [1 1] x1 + [2]
           [0 0]      [1]
   cons(x1) = [1 3] x1 + [1]
              [0 1]      [1]
   s(x1) = [1 0] x1 + [2]
           [0 0]      [1]
   p(x1) = [1 0] x1 + [0]
           [0 0]      [0]

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

Tool TRI2

Execution Time	0.15193486ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex4 7 15 Bor03 L

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  p(s(0())) -> 0()
     , f(s(0())) -> f(p(s(0())))
     , f(0()) -> cons(0())}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   0() = [1]
         [0]
   f(x1) = [1 1] x1 + [1]
           [0 0]      [1]
   cons(x1) = [1 3] x1 + [0]
              [0 1]      [1]
   s(x1) = [1 0] x1 + [0]
           [0 0]      [3]
   p(x1) = [1 0] x1 + [1]
           [0 0]      [0]

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