Problem Transformed CSR 04 Ex6 Luc98 L

Tool Bounds

Execution Time	4.4703007e-2ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex6 Luc98 L

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  from(X) -> cons(X)
     , first(s(X), cons(Y)) -> cons(Y)
     , first(0(), X) -> nil()}
  StartTerms: all
  Strategy: none

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

Proof:
  The problem is match-bounded by 1.
  The enriched problem is compatible with the following automaton:
  {  0_0() -> 1
   , first_0(1, 1) -> 1
   , nil_0() -> 1
   , nil_1() -> 1
   , s_0(1) -> 1
   , cons_0(1) -> 1
   , cons_1(1) -> 1
   , from_0(1) -> 1}

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

Tool CDI

Execution Time	0.10657811ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex6 Luc98 L

stdout:

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

This TRS is terminating using the deltarestricted interpretation
from(delta, X0) =  + 0*X0 + 0 + 2*X0*delta + 2*delta
s(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 2*delta
cons(delta, X0) =  + 0*X0 + 0 + 2*X0*delta + 0*delta
0(delta) =  + 2 + 0*delta
first(delta, X1, X0) =  + 0*X0 + 1*X1 + 0 + 1*X0*delta + 2*X1*delta + 0*delta
nil(delta) =  + 0 + 0*delta
from_tau_1(delta) = delta/(0 + 2 * delta)
s_tau_1(delta) = delta/(1 + 0 * delta)
cons_tau_1(delta) = delta/(0 + 2 * delta)
first_tau_1(delta) = delta/(1 + 2 * delta)
first_tau_2(delta) = delta/(0 + 1 * delta)

Time: 0.067345 seconds
Statistics:
Number of monomials: 102
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.10888696ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex6 Luc98 L

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  from(X) -> cons(X)
     , first(s(X), cons(Y)) -> cons(Y)
     , first(0(), X) -> nil()}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   0() = [3]
   first(x1, x2) = [1] x1 + [1] x2 + [3]
   nil() = [1]
   s(x1) = [1] x1 + [1]
   cons(x1) = [1] x1 + [2]
   from(x1) = [1] x1 + [3]

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

Tool IDA

Execution Time	0.19846702ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex6 Luc98 L

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  from(X) -> cons(X)
     , first(s(X), cons(Y)) -> cons(Y)
     , first(0(), X) -> nil()}
  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:
   0() = [3]
   first(x1, x2) = [1] x1 + [1] x2 + [3]
   nil() = [1]
   s(x1) = [1] x1 + [1]
   cons(x1) = [1] x1 + [2]
   from(x1) = [1] x1 + [3]

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

Tool TRI

Execution Time	7.737994e-2ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex6 Luc98 L

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  from(X) -> cons(X)
     , first(s(X), cons(Y)) -> cons(Y)
     , first(0(), X) -> nil()}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   0() = [3]
   first(x1, x2) = [1] x1 + [1] x2 + [3]
   nil() = [1]
   s(x1) = [1] x1 + [1]
   cons(x1) = [1] x1 + [2]
   from(x1) = [1] x1 + [3]

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

Tool TRI2

Execution Time	7.80251e-2ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex6 Luc98 L

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  from(X) -> cons(X)
     , first(s(X), cons(Y)) -> cons(Y)
     , first(0(), X) -> nil()}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   0() = [3]
         [3]
   first(x1, x2) = [1 3] x1 + [1 3] x2 + [3]
                   [0 1]      [0 1]      [3]
   nil() = [2]
           [1]
   s(x1) = [1 0] x1 + [3]
           [0 0]      [3]
   cons(x1) = [1 3] x1 + [0]
              [0 1]      [1]
   from(x1) = [1 3] x1 + [3]
              [0 1]      [3]

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