MAYBE

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

Strict Trs:
  { pairNs() -> cons(0(), n__incr(n__oddNs()))
  , cons(X1, X2) -> n__cons(X1, X2)
  , oddNs() -> n__oddNs()
  , oddNs() -> incr(pairNs())
  , incr(X) -> n__incr(X)
  , incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
  , activate(X) -> X
  , activate(n__incr(X)) -> incr(activate(X))
  , activate(n__oddNs()) -> oddNs()
  , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
  , activate(n__zip(X1, X2)) -> zip(activate(X1), activate(X2))
  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
  , activate(n__repItems(X)) -> repItems(activate(X))
  , take(X1, X2) -> n__take(X1, X2)
  , take(0(), XS) -> nil()
  , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
  , zip(X1, X2) -> n__zip(X1, X2)
  , zip(X, nil()) -> nil()
  , zip(cons(X, XS), cons(Y, YS)) ->
    cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
  , zip(nil(), XS) -> nil()
  , tail(cons(X, XS)) -> activate(XS)
  , repItems(X) -> n__repItems(X)
  , repItems(cons(X, XS)) ->
    cons(X, n__cons(X, n__repItems(activate(XS))))
  , repItems(nil()) -> nil() }
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) '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
      
      2) '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
      
   
   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:
     { pairNs^#() -> c_1(cons^#(0(), n__incr(n__oddNs())))
     , cons^#(X1, X2) -> c_2(X1, X2)
     , oddNs^#() -> c_3()
     , oddNs^#() -> c_4(incr^#(pairNs()))
     , incr^#(X) -> c_5(X)
     , incr^#(cons(X, XS)) -> c_6(cons^#(s(X), n__incr(activate(XS))))
     , activate^#(X) -> c_7(X)
     , activate^#(n__incr(X)) -> c_8(incr^#(activate(X)))
     , activate^#(n__oddNs()) -> c_9(oddNs^#())
     , activate^#(n__take(X1, X2)) ->
       c_10(take^#(activate(X1), activate(X2)))
     , activate^#(n__zip(X1, X2)) ->
       c_11(zip^#(activate(X1), activate(X2)))
     , activate^#(n__cons(X1, X2)) -> c_12(cons^#(activate(X1), X2))
     , activate^#(n__repItems(X)) -> c_13(repItems^#(activate(X)))
     , take^#(X1, X2) -> c_14(X1, X2)
     , take^#(0(), XS) -> c_15()
     , take^#(s(N), cons(X, XS)) ->
       c_16(cons^#(X, n__take(N, activate(XS))))
     , zip^#(X1, X2) -> c_17(X1, X2)
     , zip^#(X, nil()) -> c_18()
     , zip^#(cons(X, XS), cons(Y, YS)) ->
       c_19(cons^#(pair(X, Y), n__zip(activate(XS), activate(YS))))
     , zip^#(nil(), XS) -> c_20()
     , repItems^#(X) -> c_22(X)
     , repItems^#(cons(X, XS)) ->
       c_23(cons^#(X, n__cons(X, n__repItems(activate(XS)))))
     , repItems^#(nil()) -> c_24()
     , tail^#(cons(X, XS)) -> c_21(activate^#(XS)) }
   
   and mark the set of starting terms.
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { pairNs^#() -> c_1(cons^#(0(), n__incr(n__oddNs())))
     , cons^#(X1, X2) -> c_2(X1, X2)
     , oddNs^#() -> c_3()
     , oddNs^#() -> c_4(incr^#(pairNs()))
     , incr^#(X) -> c_5(X)
     , incr^#(cons(X, XS)) -> c_6(cons^#(s(X), n__incr(activate(XS))))
     , activate^#(X) -> c_7(X)
     , activate^#(n__incr(X)) -> c_8(incr^#(activate(X)))
     , activate^#(n__oddNs()) -> c_9(oddNs^#())
     , activate^#(n__take(X1, X2)) ->
       c_10(take^#(activate(X1), activate(X2)))
     , activate^#(n__zip(X1, X2)) ->
       c_11(zip^#(activate(X1), activate(X2)))
     , activate^#(n__cons(X1, X2)) -> c_12(cons^#(activate(X1), X2))
     , activate^#(n__repItems(X)) -> c_13(repItems^#(activate(X)))
     , take^#(X1, X2) -> c_14(X1, X2)
     , take^#(0(), XS) -> c_15()
     , take^#(s(N), cons(X, XS)) ->
       c_16(cons^#(X, n__take(N, activate(XS))))
     , zip^#(X1, X2) -> c_17(X1, X2)
     , zip^#(X, nil()) -> c_18()
     , zip^#(cons(X, XS), cons(Y, YS)) ->
       c_19(cons^#(pair(X, Y), n__zip(activate(XS), activate(YS))))
     , zip^#(nil(), XS) -> c_20()
     , repItems^#(X) -> c_22(X)
     , repItems^#(cons(X, XS)) ->
       c_23(cons^#(X, n__cons(X, n__repItems(activate(XS)))))
     , repItems^#(nil()) -> c_24()
     , tail^#(cons(X, XS)) -> c_21(activate^#(XS)) }
   Strict Trs:
     { pairNs() -> cons(0(), n__incr(n__oddNs()))
     , cons(X1, X2) -> n__cons(X1, X2)
     , oddNs() -> n__oddNs()
     , oddNs() -> incr(pairNs())
     , incr(X) -> n__incr(X)
     , incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
     , activate(X) -> X
     , activate(n__incr(X)) -> incr(activate(X))
     , activate(n__oddNs()) -> oddNs()
     , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
     , activate(n__zip(X1, X2)) -> zip(activate(X1), activate(X2))
     , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
     , activate(n__repItems(X)) -> repItems(activate(X))
     , take(X1, X2) -> n__take(X1, X2)
     , take(0(), XS) -> nil()
     , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
     , zip(X1, X2) -> n__zip(X1, X2)
     , zip(X, nil()) -> nil()
     , zip(cons(X, XS), cons(Y, YS)) ->
       cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
     , zip(nil(), XS) -> nil()
     , tail(cons(X, XS)) -> activate(XS)
     , repItems(X) -> n__repItems(X)
     , repItems(cons(X, XS)) ->
       cons(X, n__cons(X, n__repItems(activate(XS))))
     , repItems(nil()) -> nil() }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   We estimate the number of application of {3,15,18,20,23} by
   applications of Pre({3,15,18,20,23}) = {2,5,7,9,10,11,13,14,17,21}.
   Here rules are labeled as follows:
   
     DPs:
       { 1: pairNs^#() -> c_1(cons^#(0(), n__incr(n__oddNs())))
       , 2: cons^#(X1, X2) -> c_2(X1, X2)
       , 3: oddNs^#() -> c_3()
       , 4: oddNs^#() -> c_4(incr^#(pairNs()))
       , 5: incr^#(X) -> c_5(X)
       , 6: incr^#(cons(X, XS)) ->
            c_6(cons^#(s(X), n__incr(activate(XS))))
       , 7: activate^#(X) -> c_7(X)
       , 8: activate^#(n__incr(X)) -> c_8(incr^#(activate(X)))
       , 9: activate^#(n__oddNs()) -> c_9(oddNs^#())
       , 10: activate^#(n__take(X1, X2)) ->
             c_10(take^#(activate(X1), activate(X2)))
       , 11: activate^#(n__zip(X1, X2)) ->
             c_11(zip^#(activate(X1), activate(X2)))
       , 12: activate^#(n__cons(X1, X2)) -> c_12(cons^#(activate(X1), X2))
       , 13: activate^#(n__repItems(X)) -> c_13(repItems^#(activate(X)))
       , 14: take^#(X1, X2) -> c_14(X1, X2)
       , 15: take^#(0(), XS) -> c_15()
       , 16: take^#(s(N), cons(X, XS)) ->
             c_16(cons^#(X, n__take(N, activate(XS))))
       , 17: zip^#(X1, X2) -> c_17(X1, X2)
       , 18: zip^#(X, nil()) -> c_18()
       , 19: zip^#(cons(X, XS), cons(Y, YS)) ->
             c_19(cons^#(pair(X, Y), n__zip(activate(XS), activate(YS))))
       , 20: zip^#(nil(), XS) -> c_20()
       , 21: repItems^#(X) -> c_22(X)
       , 22: repItems^#(cons(X, XS)) ->
             c_23(cons^#(X, n__cons(X, n__repItems(activate(XS)))))
       , 23: repItems^#(nil()) -> c_24()
       , 24: tail^#(cons(X, XS)) -> c_21(activate^#(XS)) }
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { pairNs^#() -> c_1(cons^#(0(), n__incr(n__oddNs())))
     , cons^#(X1, X2) -> c_2(X1, X2)
     , oddNs^#() -> c_4(incr^#(pairNs()))
     , incr^#(X) -> c_5(X)
     , incr^#(cons(X, XS)) -> c_6(cons^#(s(X), n__incr(activate(XS))))
     , activate^#(X) -> c_7(X)
     , activate^#(n__incr(X)) -> c_8(incr^#(activate(X)))
     , activate^#(n__oddNs()) -> c_9(oddNs^#())
     , activate^#(n__take(X1, X2)) ->
       c_10(take^#(activate(X1), activate(X2)))
     , activate^#(n__zip(X1, X2)) ->
       c_11(zip^#(activate(X1), activate(X2)))
     , activate^#(n__cons(X1, X2)) -> c_12(cons^#(activate(X1), X2))
     , activate^#(n__repItems(X)) -> c_13(repItems^#(activate(X)))
     , take^#(X1, X2) -> c_14(X1, X2)
     , take^#(s(N), cons(X, XS)) ->
       c_16(cons^#(X, n__take(N, activate(XS))))
     , zip^#(X1, X2) -> c_17(X1, X2)
     , zip^#(cons(X, XS), cons(Y, YS)) ->
       c_19(cons^#(pair(X, Y), n__zip(activate(XS), activate(YS))))
     , repItems^#(X) -> c_22(X)
     , repItems^#(cons(X, XS)) ->
       c_23(cons^#(X, n__cons(X, n__repItems(activate(XS)))))
     , tail^#(cons(X, XS)) -> c_21(activate^#(XS)) }
   Strict Trs:
     { pairNs() -> cons(0(), n__incr(n__oddNs()))
     , cons(X1, X2) -> n__cons(X1, X2)
     , oddNs() -> n__oddNs()
     , oddNs() -> incr(pairNs())
     , incr(X) -> n__incr(X)
     , incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
     , activate(X) -> X
     , activate(n__incr(X)) -> incr(activate(X))
     , activate(n__oddNs()) -> oddNs()
     , activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
     , activate(n__zip(X1, X2)) -> zip(activate(X1), activate(X2))
     , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
     , activate(n__repItems(X)) -> repItems(activate(X))
     , take(X1, X2) -> n__take(X1, X2)
     , take(0(), XS) -> nil()
     , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
     , zip(X1, X2) -> n__zip(X1, X2)
     , zip(X, nil()) -> nil()
     , zip(cons(X, XS), cons(Y, YS)) ->
       cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
     , zip(nil(), XS) -> nil()
     , tail(cons(X, XS)) -> activate(XS)
     , repItems(X) -> n__repItems(X)
     , repItems(cons(X, XS)) ->
       cons(X, n__cons(X, n__repItems(activate(XS))))
     , repItems(nil()) -> nil() }
   Weak DPs:
     { oddNs^#() -> c_3()
     , take^#(0(), XS) -> c_15()
     , zip^#(X, nil()) -> c_18()
     , zip^#(nil(), XS) -> c_20()
     , repItems^#(nil()) -> c_24() }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   Empty strict component of the problem is NOT empty.


Arrrr..