Problem Transformed CSR 04 Ex4 7 77 Bor03 FR

Tool Bounds

Execution Time	7.858682e-2ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 77 Bor03 FR

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , activate(n__zeros()) -> zeros()
     , zeros() -> n__zeros()
     , tail(cons(X, XS)) -> activate(XS)
     , zeros() -> cons(0(), n__zeros())}
  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:
  {  zeros_0() -> 1
   , zeros_1() -> 1
   , zeros_2() -> 1
   , 0_0() -> 1
   , 0_1() -> 2
   , 0_2() -> 3
   , 0_3() -> 4
   , n__zeros_0() -> 1
   , n__zeros_1() -> 1
   , n__zeros_2() -> 1
   , n__zeros_3() -> 1
   , cons_0(1, 1) -> 1
   , cons_1(2, 1) -> 1
   , cons_2(3, 1) -> 1
   , cons_3(4, 1) -> 1
   , tail_0(1) -> 1
   , activate_0(1) -> 1
   , activate_1(1) -> 1}

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

Tool CDI

Execution Time	0.13175392ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex4 7 77 Bor03 FR

stdout:

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

This TRS is terminating using the deltarestricted interpretation
tail(delta, X0) =  + 1*X0 + 1 + 0*X0*delta + 3*delta
activate(delta, X0) =  + 1*X0 + 1 + 0*X0*delta + 2*delta
zeros(delta) =  + 1 + 1*delta
n__zeros(delta) =  + 0 + 0*delta
0(delta) =  + 0 + 0*delta
cons(delta, X1, X0) =  + 1*X0 + 1*X1 + 0 + 0*X0*delta + 0*X1*delta + 0*delta
tail_tau_1(delta) = delta/(1 + 0 * delta)
activate_tau_1(delta) = delta/(1 + 0 * delta)
cons_tau_1(delta) = delta/(1 + 0 * delta)
cons_tau_2(delta) = delta/(1 + 0 * delta)

Time: 0.093411 seconds
Statistics:
Number of monomials: 115
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.14826798ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 77 Bor03 FR

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , activate(n__zeros()) -> zeros()
     , zeros() -> n__zeros()
     , tail(cons(X, XS)) -> activate(XS)
     , zeros() -> cons(0(), n__zeros())}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   zeros() = [3]
   0() = [0]
   n__zeros() = [1]
   cons(x1, x2) = [1] x1 + [1] x2 + [1]
   tail(x1) = [1] x1 + [3]
   activate(x1) = [1] x1 + [3]

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

Tool IDA

Execution Time	0.14926815ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 77 Bor03 FR

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , activate(n__zeros()) -> zeros()
     , zeros() -> n__zeros()
     , tail(cons(X, XS)) -> activate(XS)
     , zeros() -> cons(0(), n__zeros())}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
  Interpretation Functions:
   zeros() = [3]
   0() = [0]
   n__zeros() = [1]
   cons(x1, x2) = [1] x1 + [1] x2 + [1]
   tail(x1) = [1] x1 + [3]
   activate(x1) = [1] x1 + [3]

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

Tool TRI

Execution Time	6.848788e-2ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 77 Bor03 FR

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , activate(n__zeros()) -> zeros()
     , zeros() -> n__zeros()
     , tail(cons(X, XS)) -> activate(XS)
     , zeros() -> cons(0(), n__zeros())}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   zeros() = [3]
   0() = [0]
   n__zeros() = [1]
   cons(x1, x2) = [1] x1 + [1] x2 + [1]
   tail(x1) = [1] x1 + [3]
   activate(x1) = [1] x1 + [3]

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

Tool TRI2

Execution Time	9.5052004e-2ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex4 7 77 Bor03 FR

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , activate(n__zeros()) -> zeros()
     , zeros() -> n__zeros()
     , tail(cons(X, XS)) -> activate(XS)
     , zeros() -> cons(0(), n__zeros())}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   zeros() = [1]
             [3]
   0() = [0]
         [1]
   n__zeros() = [0]
                [2]
   cons(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                  [0 1]      [0 1]      [0]
   tail(x1) = [1 1] x1 + [3]
              [0 1]      [3]
   activate(x1) = [1 1] x1 + [2]
                  [0 1]      [1]

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