MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { primes() -> sieve(from(s(s(0()))))
  , sieve(X) -> n__sieve(X)
  , sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y))))
  , from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , s(X) -> n__s(X)
  , cons(X1, X2) -> n__cons(X1, X2)
  , head(cons(X, Y)) -> X
  , tail(cons(X, Y)) -> activate(Y)
  , activate(X) -> X
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X))
  , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
  , activate(n__sieve(X)) -> sieve(activate(X))
  , if(true(), X, Y) -> activate(X)
  , if(false(), X, Y) -> activate(Y)
  , filter(X1, X2) -> n__filter(X1, X2)
  , filter(s(s(X)), cons(Y, Z)) ->
    if(divides(s(s(X)), Y),
       n__filter(n__s(n__s(X)), activate(Z)),
       n__cons(Y, n__filter(X, n__sieve(Y)))) }
Obligation:
  runtime complexity
Answer:
  MAYBE

None of the processors succeeded.

Details of failed attempt(s):
-----------------------------
1) 'WithProblem (timeout of 60 seconds)' failed due to the
   following reason:
   
   Computation stopped due to timeout after 60.0 seconds.

2) 'Best' failed due to the following reason:
   
   None of the processors succeeded.
   
   Details of failed attempt(s):
   -----------------------------
   1) 'WithProblem (timeout of 30 seconds) (timeout of 60 seconds)'
      failed due to the following reason:
      
      Computation stopped due to timeout after 30.0 seconds.
   
   2) 'Best' failed due to the following reason:
      
      None of the processors succeeded.
      
      Details of failed attempt(s):
      -----------------------------
      1) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due
         to the following reason:
         
         The processor is inapplicable, reason:
           Processor only applicable for innermost runtime complexity analysis
      
      2) 'bsearch-popstar (timeout of 60 seconds)' failed due to the
         following reason:
         
         The processor is inapplicable, reason:
           Processor only applicable for innermost runtime complexity analysis
      
   
   3) 'Fastest (timeout of 5 seconds) (timeout of 60 seconds)' failed
      due to the following reason:
      
      None of the processors succeeded.
      
      Details of failed attempt(s):
      -----------------------------
      1) 'Bounds with perSymbol-enrichment and initial automaton 'match''
         failed due to the following reason:
         
         match-boundness of the problem could not be verified.
      
      2) 'Bounds with minimal-enrichment and initial automaton 'match''
         failed due to the following reason:
         
         match-boundness of the problem could not be verified.
      
   

3) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed
   due to the following reason:
   
   We add the following weak dependency pairs:
   
   Strict DPs:
     { primes^#() -> c_1(sieve^#(from(s(s(0())))))
     , sieve^#(X) -> c_2(X)
     , sieve^#(cons(X, Y)) ->
       c_3(cons^#(X, n__filter(X, n__sieve(activate(Y)))))
     , cons^#(X1, X2) -> c_7(X1, X2)
     , from^#(X) -> c_4(cons^#(X, n__from(n__s(X))))
     , from^#(X) -> c_5(X)
     , s^#(X) -> c_6(X)
     , head^#(cons(X, Y)) -> c_8(X)
     , tail^#(cons(X, Y)) -> c_9(activate^#(Y))
     , activate^#(X) -> c_10(X)
     , activate^#(n__from(X)) -> c_11(from^#(activate(X)))
     , activate^#(n__s(X)) -> c_12(s^#(activate(X)))
     , activate^#(n__filter(X1, X2)) ->
       c_13(filter^#(activate(X1), activate(X2)))
     , activate^#(n__cons(X1, X2)) -> c_14(cons^#(activate(X1), X2))
     , activate^#(n__sieve(X)) -> c_15(sieve^#(activate(X)))
     , filter^#(X1, X2) -> c_18(X1, X2)
     , filter^#(s(s(X)), cons(Y, Z)) ->
       c_19(if^#(divides(s(s(X)), Y),
                 n__filter(n__s(n__s(X)), activate(Z)),
                 n__cons(Y, n__filter(X, n__sieve(Y)))))
     , if^#(true(), X, Y) -> c_16(activate^#(X))
     , if^#(false(), X, Y) -> c_17(activate^#(Y)) }
   
   and mark the set of starting terms.
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { primes^#() -> c_1(sieve^#(from(s(s(0())))))
     , sieve^#(X) -> c_2(X)
     , sieve^#(cons(X, Y)) ->
       c_3(cons^#(X, n__filter(X, n__sieve(activate(Y)))))
     , cons^#(X1, X2) -> c_7(X1, X2)
     , from^#(X) -> c_4(cons^#(X, n__from(n__s(X))))
     , from^#(X) -> c_5(X)
     , s^#(X) -> c_6(X)
     , head^#(cons(X, Y)) -> c_8(X)
     , tail^#(cons(X, Y)) -> c_9(activate^#(Y))
     , activate^#(X) -> c_10(X)
     , activate^#(n__from(X)) -> c_11(from^#(activate(X)))
     , activate^#(n__s(X)) -> c_12(s^#(activate(X)))
     , activate^#(n__filter(X1, X2)) ->
       c_13(filter^#(activate(X1), activate(X2)))
     , activate^#(n__cons(X1, X2)) -> c_14(cons^#(activate(X1), X2))
     , activate^#(n__sieve(X)) -> c_15(sieve^#(activate(X)))
     , filter^#(X1, X2) -> c_18(X1, X2)
     , filter^#(s(s(X)), cons(Y, Z)) ->
       c_19(if^#(divides(s(s(X)), Y),
                 n__filter(n__s(n__s(X)), activate(Z)),
                 n__cons(Y, n__filter(X, n__sieve(Y)))))
     , if^#(true(), X, Y) -> c_16(activate^#(X))
     , if^#(false(), X, Y) -> c_17(activate^#(Y)) }
   Strict Trs:
     { primes() -> sieve(from(s(s(0()))))
     , sieve(X) -> n__sieve(X)
     , sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y))))
     , from(X) -> cons(X, n__from(n__s(X)))
     , from(X) -> n__from(X)
     , s(X) -> n__s(X)
     , cons(X1, X2) -> n__cons(X1, X2)
     , head(cons(X, Y)) -> X
     , tail(cons(X, Y)) -> activate(Y)
     , activate(X) -> X
     , activate(n__from(X)) -> from(activate(X))
     , activate(n__s(X)) -> s(activate(X))
     , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
     , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
     , activate(n__sieve(X)) -> sieve(activate(X))
     , if(true(), X, Y) -> activate(X)
     , if(false(), X, Y) -> activate(Y)
     , filter(X1, X2) -> n__filter(X1, X2)
     , filter(s(s(X)), cons(Y, Z)) ->
       if(divides(s(s(X)), Y),
          n__filter(n__s(n__s(X)), activate(Z)),
          n__cons(Y, n__filter(X, n__sieve(Y)))) }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   We estimate the number of application of {17} by applications of
   Pre({17}) = {2,4,6,7,8,10,13,16}. Here rules are labeled as
   follows:
   
     DPs:
       { 1: primes^#() -> c_1(sieve^#(from(s(s(0())))))
       , 2: sieve^#(X) -> c_2(X)
       , 3: sieve^#(cons(X, Y)) ->
            c_3(cons^#(X, n__filter(X, n__sieve(activate(Y)))))
       , 4: cons^#(X1, X2) -> c_7(X1, X2)
       , 5: from^#(X) -> c_4(cons^#(X, n__from(n__s(X))))
       , 6: from^#(X) -> c_5(X)
       , 7: s^#(X) -> c_6(X)
       , 8: head^#(cons(X, Y)) -> c_8(X)
       , 9: tail^#(cons(X, Y)) -> c_9(activate^#(Y))
       , 10: activate^#(X) -> c_10(X)
       , 11: activate^#(n__from(X)) -> c_11(from^#(activate(X)))
       , 12: activate^#(n__s(X)) -> c_12(s^#(activate(X)))
       , 13: activate^#(n__filter(X1, X2)) ->
             c_13(filter^#(activate(X1), activate(X2)))
       , 14: activate^#(n__cons(X1, X2)) -> c_14(cons^#(activate(X1), X2))
       , 15: activate^#(n__sieve(X)) -> c_15(sieve^#(activate(X)))
       , 16: filter^#(X1, X2) -> c_18(X1, X2)
       , 17: filter^#(s(s(X)), cons(Y, Z)) ->
             c_19(if^#(divides(s(s(X)), Y),
                       n__filter(n__s(n__s(X)), activate(Z)),
                       n__cons(Y, n__filter(X, n__sieve(Y)))))
       , 18: if^#(true(), X, Y) -> c_16(activate^#(X))
       , 19: if^#(false(), X, Y) -> c_17(activate^#(Y)) }
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { primes^#() -> c_1(sieve^#(from(s(s(0())))))
     , sieve^#(X) -> c_2(X)
     , sieve^#(cons(X, Y)) ->
       c_3(cons^#(X, n__filter(X, n__sieve(activate(Y)))))
     , cons^#(X1, X2) -> c_7(X1, X2)
     , from^#(X) -> c_4(cons^#(X, n__from(n__s(X))))
     , from^#(X) -> c_5(X)
     , s^#(X) -> c_6(X)
     , head^#(cons(X, Y)) -> c_8(X)
     , tail^#(cons(X, Y)) -> c_9(activate^#(Y))
     , activate^#(X) -> c_10(X)
     , activate^#(n__from(X)) -> c_11(from^#(activate(X)))
     , activate^#(n__s(X)) -> c_12(s^#(activate(X)))
     , activate^#(n__filter(X1, X2)) ->
       c_13(filter^#(activate(X1), activate(X2)))
     , activate^#(n__cons(X1, X2)) -> c_14(cons^#(activate(X1), X2))
     , activate^#(n__sieve(X)) -> c_15(sieve^#(activate(X)))
     , filter^#(X1, X2) -> c_18(X1, X2)
     , if^#(true(), X, Y) -> c_16(activate^#(X))
     , if^#(false(), X, Y) -> c_17(activate^#(Y)) }
   Strict Trs:
     { primes() -> sieve(from(s(s(0()))))
     , sieve(X) -> n__sieve(X)
     , sieve(cons(X, Y)) -> cons(X, n__filter(X, n__sieve(activate(Y))))
     , from(X) -> cons(X, n__from(n__s(X)))
     , from(X) -> n__from(X)
     , s(X) -> n__s(X)
     , cons(X1, X2) -> n__cons(X1, X2)
     , head(cons(X, Y)) -> X
     , tail(cons(X, Y)) -> activate(Y)
     , activate(X) -> X
     , activate(n__from(X)) -> from(activate(X))
     , activate(n__s(X)) -> s(activate(X))
     , activate(n__filter(X1, X2)) -> filter(activate(X1), activate(X2))
     , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
     , activate(n__sieve(X)) -> sieve(activate(X))
     , if(true(), X, Y) -> activate(X)
     , if(false(), X, Y) -> activate(Y)
     , filter(X1, X2) -> n__filter(X1, X2)
     , filter(s(s(X)), cons(Y, Z)) ->
       if(divides(s(s(X)), Y),
          n__filter(n__s(n__s(X)), activate(Z)),
          n__cons(Y, n__filter(X, n__sieve(Y)))) }
   Weak DPs:
     { filter^#(s(s(X)), cons(Y, Z)) ->
       c_19(if^#(divides(s(s(X)), Y),
                 n__filter(n__s(n__s(X)), activate(Z)),
                 n__cons(Y, n__filter(X, n__sieve(Y))))) }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   Empty strict component of the problem is NOT empty.


Arrrr..