Problem Transformed CSR 04 Ex5 Zan97 Z

Tool Bounds

Execution Time	3.8136005e-2ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 Z

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , activate(n__f(X)) -> f(X)
     , f(X) -> n__f(X)
     , if(false(), X, Y) -> activate(Y)
     , if(true(), X, Y) -> X
     , f(X) -> if(X, c(), n__f(true()))}
  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:
  {  f_0(1) -> 1
   , f_1(1) -> 1
   , f_1(4) -> 1
   , f_1(7) -> 1
   , f_1(10) -> 1
   , f_2(1) -> 1
   , f_2(4) -> 1
   , f_2(7) -> 1
   , f_2(10) -> 1
   , c_0() -> 1
   , c_1() -> 1
   , c_1() -> 2
   , c_2() -> 1
   , c_2() -> 5
   , c_3() -> 1
   , c_3() -> 8
   , true_0() -> 1
   , true_1() -> 4
   , true_2() -> 7
   , true_3() -> 10
   , n__f_0(1) -> 1
   , n__f_1(1) -> 1
   , n__f_1(4) -> 1
   , n__f_1(4) -> 3
   , n__f_2(1) -> 1
   , n__f_2(4) -> 1
   , n__f_2(7) -> 1
   , n__f_2(7) -> 6
   , n__f_2(10) -> 1
   , n__f_3(1) -> 1
   , n__f_3(4) -> 1
   , n__f_3(7) -> 1
   , n__f_3(10) -> 1
   , n__f_3(10) -> 9
   , if_0(1, 1, 1) -> 1
   , if_1(1, 2, 3) -> 1
   , if_2(1, 5, 6) -> 1
   , if_2(4, 5, 6) -> 1
   , if_2(7, 5, 6) -> 1
   , if_2(10, 5, 6) -> 1
   , if_3(1, 8, 9) -> 1
   , if_3(4, 8, 9) -> 1
   , if_3(7, 8, 9) -> 1
   , if_3(10, 8, 9) -> 1
   , false_0() -> 1
   , activate_0(1) -> 1
   , activate_1(1) -> 1
   , activate_1(3) -> 1
   , activate_1(6) -> 1
   , activate_1(9) -> 1}

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

Tool CDI

Execution Time	0.637264ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex5 Zan97 Z

stdout:

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

This TRS is terminating using the deltarestricted interpretation
false(delta) =  + 2 + 0*delta
activate(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 3*delta
f(delta, X0) =  + 0*X0 + 0 + 2*X0*delta + 2*delta
true(delta) =  + 0 + 0*delta
n__f(delta, X0) =  + 0*X0 + 0 + 2*X0*delta + 0*delta
c(delta) =  + 0 + 0*delta
if(delta, X2, X1, X0) =  + 1*X0 + 1*X1 + 0*X2 + 0 + 0*X0*delta + 0*X1*delta + 2*X2*delta + 1*delta
activate_tau_1(delta) = delta/(1 + 0 * delta)
f_tau_1(delta) = delta/(0 + 2 * delta)
n__f_tau_1(delta) = delta/(0 + 2 * delta)
if_tau_1(delta) = delta/(0 + 2 * delta)
if_tau_2(delta) = delta/(1 + 0 * delta)
if_tau_3(delta) = delta/(1 + 0 * delta)

Time: 0.599110 seconds
Statistics:
Number of monomials: 295
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.13787794ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 Z

stdout:

Tool IDA

Execution Time	0.21422696ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 Z

stdout:

Tool TRI

Execution Time	8.5483074e-2ms
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 Z

stdout:

Tool TRI2

Execution Time	0.14687896ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex5 Zan97 Z

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , activate(n__f(X)) -> f(X)
     , f(X) -> n__f(X)
     , if(false(), X, Y) -> activate(Y)
     , if(true(), X, Y) -> X
     , f(X) -> if(X, c(), n__f(true()))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   f(x1) = [1 1] x1 + [3]
           [0 1]      [0]
   c() = [0]
         [0]
   true() = [1]
            [0]
   n__f(x1) = [1 0] x1 + [1]
              [0 1]      [0]
   if(x1, x2, x3) = [1 1] x1 + [1 3] x2 + [1 2] x3 + [0]
                    [0 0]      [0 1]      [0 1]      [0]
   false() = [2]
             [2]
   activate(x1) = [1 1] x1 + [3]
                  [0 1]      [0]

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