Problem CSR 04 Ex25 Luc06

Tool Bounds

Execution Time	6.4175844e-2ms
Answer	YES(?,O(n^1))
Input	CSR 04 Ex25 Luc06

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  h(X) -> c(d(X))
     , c(X) -> d(X)
     , f(f(X)) -> c(f(g(f(X))))}
  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:
  {  f_0(1) -> 1
   , f_1(1) -> 4
   , f_1(3) -> 2
   , c_0(1) -> 1
   , c_1(2) -> 1
   , c_1(2) -> 4
   , g_0(1) -> 1
   , g_1(4) -> 3
   , d_0(1) -> 1
   , d_1(1) -> 1
   , d_1(1) -> 2
   , d_2(2) -> 1
   , d_2(2) -> 4
   , h_0(1) -> 1}

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

Tool CDI

Execution Time	5.802847ms
Answer	YES(?,O(n^2))
Input	CSR 04 Ex25 Luc06

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 + 2 + 0*X0*delta + 3*delta
d(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 0*delta
g(delta, X0) =  + 0*X0 + 0 + 1*X0*delta + 0*delta
f(delta, X0) =  + 0*X0 + 1 + 3*X0*delta + 0*delta
c(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 1*delta
h_tau_1(delta) = delta/(1 + 0 * delta)
d_tau_1(delta) = delta/(1 + 0 * delta)
g_tau_1(delta) = delta/(0 + 1 * delta)
f_tau_1(delta) = delta/(0 + 3 * delta)
c_tau_1(delta) = delta/(1 + 0 * delta)

Time: 5.764680 seconds
Statistics:
Number of monomials: 470
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.5185571ms
Answer	YES(?,O(n^2))
Input	CSR 04 Ex25 Luc06

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  h(X) -> c(d(X))
     , c(X) -> d(X)
     , f(f(X)) -> c(f(g(f(X))))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   f(x1) = [1 1] x1 + [1]
           [0 0]      [3]
   c(x1) = [1 0] x1 + [1]
           [0 0]      [3]
   g(x1) = [1 0] x1 + [0]
           [0 0]      [1]
   d(x1) = [1 0] x1 + [0]
           [0 0]      [3]
   h(x1) = [1 0] x1 + [3]
           [0 0]      [3]

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

Tool IDA

Execution Time	0.668972ms
Answer	YES(?,O(n^2))
Input	CSR 04 Ex25 Luc06

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  h(X) -> c(d(X))
     , c(X) -> d(X)
     , f(f(X)) -> c(f(g(f(X))))}
  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:
   f(x1) = [1 1] x1 + [1]
           [0 0]      [3]
   c(x1) = [1 0] x1 + [2]
           [0 0]      [3]
   g(x1) = [1 0] x1 + [0]
           [0 0]      [0]
   d(x1) = [1 0] x1 + [0]
           [0 0]      [3]
   h(x1) = [1 0] x1 + [3]
           [0 0]      [3]

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

Tool TRI

Execution Time	0.14567304ms
Answer	YES(?,O(n^2))
Input	CSR 04 Ex25 Luc06

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  h(X) -> c(d(X))
     , c(X) -> d(X)
     , f(f(X)) -> c(f(g(f(X))))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   f(x1) = [1 1] x1 + [0]
           [0 0]      [2]
   c(x1) = [1 0] x1 + [1]
           [0 0]      [2]
   g(x1) = [1 0] x1 + [0]
           [0 0]      [0]
   d(x1) = [1 0] x1 + [0]
           [0 0]      [1]
   h(x1) = [1 3] x1 + [3]
           [0 1]      [3]

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

Tool TRI2

Execution Time	0.12440801ms
Answer	YES(?,O(n^2))
Input	CSR 04 Ex25 Luc06

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  h(X) -> c(d(X))
     , c(X) -> d(X)
     , f(f(X)) -> c(f(g(f(X))))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   f(x1) = [1 1] x1 + [0]
           [0 0]      [2]
   c(x1) = [1 0] x1 + [1]
           [0 0]      [2]
   g(x1) = [1 0] x1 + [0]
           [0 0]      [0]
   d(x1) = [1 0] x1 + [0]
           [0 0]      [1]
   h(x1) = [1 3] x1 + [3]
           [0 1]      [3]

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