MAYBE

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

Strict Trs:
  { a__pairNs() -> cons(0(), incr(oddNs()))
  , a__pairNs() -> pairNs()
  , a__oddNs() -> oddNs()
  , a__oddNs() -> a__incr(a__pairNs())
  , a__incr(X) -> incr(X)
  , a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(0()) -> 0()
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(oddNs()) -> a__oddNs()
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
  , mark(pair(X1, X2)) -> pair(mark(X1), mark(X2))
  , mark(zip(X1, X2)) -> a__zip(mark(X1), mark(X2))
  , mark(repItems(X)) -> a__repItems(mark(X))
  , mark(pairNs()) -> a__pairNs()
  , mark(tail(X)) -> a__tail(mark(X))
  , a__take(X1, X2) -> take(X1, X2)
  , a__take(0(), XS) -> nil()
  , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS))
  , a__zip(X1, X2) -> zip(X1, X2)
  , a__zip(X, nil()) -> nil()
  , a__zip(cons(X, XS), cons(Y, YS)) ->
    cons(pair(mark(X), mark(Y)), zip(XS, YS))
  , a__zip(nil(), XS) -> nil()
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, XS)) -> mark(XS)
  , a__repItems(X) -> repItems(X)
  , a__repItems(cons(X, XS)) -> cons(mark(X), cons(X, repItems(XS)))
  , a__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) '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) '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) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed
   due to the following reason:
   
   We add the following weak dependency pairs:
   
   Strict DPs:
     { a__pairNs^#() -> c_1()
     , a__pairNs^#() -> c_2()
     , a__oddNs^#() -> c_3()
     , a__oddNs^#() -> c_4(a__incr^#(a__pairNs()))
     , a__incr^#(X) -> c_5(X)
     , a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS)
     , mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2)
     , mark^#(0()) -> c_8()
     , mark^#(incr(X)) -> c_9(a__incr^#(mark(X)))
     , mark^#(oddNs()) -> c_10(a__oddNs^#())
     , mark^#(s(X)) -> c_11(mark^#(X))
     , mark^#(nil()) -> c_12()
     , mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2)))
     , mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2))
     , mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2)))
     , mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X)))
     , mark^#(pairNs()) -> c_17(a__pairNs^#())
     , mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
     , a__take^#(X1, X2) -> c_19(X1, X2)
     , a__take^#(0(), XS) -> c_20()
     , a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS)
     , a__zip^#(X1, X2) -> c_22(X1, X2)
     , a__zip^#(X, nil()) -> c_23()
     , a__zip^#(cons(X, XS), cons(Y, YS)) ->
       c_24(mark^#(X), mark^#(Y), XS, YS)
     , a__zip^#(nil(), XS) -> c_25()
     , a__repItems^#(X) -> c_28(X)
     , a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS)
     , a__repItems^#(nil()) -> c_30()
     , a__tail^#(X) -> c_26(X)
     , a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) }
   
   and mark the set of starting terms.
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { a__pairNs^#() -> c_1()
     , a__pairNs^#() -> c_2()
     , a__oddNs^#() -> c_3()
     , a__oddNs^#() -> c_4(a__incr^#(a__pairNs()))
     , a__incr^#(X) -> c_5(X)
     , a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS)
     , mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2)
     , mark^#(0()) -> c_8()
     , mark^#(incr(X)) -> c_9(a__incr^#(mark(X)))
     , mark^#(oddNs()) -> c_10(a__oddNs^#())
     , mark^#(s(X)) -> c_11(mark^#(X))
     , mark^#(nil()) -> c_12()
     , mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2)))
     , mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2))
     , mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2)))
     , mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X)))
     , mark^#(pairNs()) -> c_17(a__pairNs^#())
     , mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
     , a__take^#(X1, X2) -> c_19(X1, X2)
     , a__take^#(0(), XS) -> c_20()
     , a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS)
     , a__zip^#(X1, X2) -> c_22(X1, X2)
     , a__zip^#(X, nil()) -> c_23()
     , a__zip^#(cons(X, XS), cons(Y, YS)) ->
       c_24(mark^#(X), mark^#(Y), XS, YS)
     , a__zip^#(nil(), XS) -> c_25()
     , a__repItems^#(X) -> c_28(X)
     , a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS)
     , a__repItems^#(nil()) -> c_30()
     , a__tail^#(X) -> c_26(X)
     , a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) }
   Strict Trs:
     { a__pairNs() -> cons(0(), incr(oddNs()))
     , a__pairNs() -> pairNs()
     , a__oddNs() -> oddNs()
     , a__oddNs() -> a__incr(a__pairNs())
     , a__incr(X) -> incr(X)
     , a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(0()) -> 0()
     , mark(incr(X)) -> a__incr(mark(X))
     , mark(oddNs()) -> a__oddNs()
     , mark(s(X)) -> s(mark(X))
     , mark(nil()) -> nil()
     , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
     , mark(pair(X1, X2)) -> pair(mark(X1), mark(X2))
     , mark(zip(X1, X2)) -> a__zip(mark(X1), mark(X2))
     , mark(repItems(X)) -> a__repItems(mark(X))
     , mark(pairNs()) -> a__pairNs()
     , mark(tail(X)) -> a__tail(mark(X))
     , a__take(X1, X2) -> take(X1, X2)
     , a__take(0(), XS) -> nil()
     , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS))
     , a__zip(X1, X2) -> zip(X1, X2)
     , a__zip(X, nil()) -> nil()
     , a__zip(cons(X, XS), cons(Y, YS)) ->
       cons(pair(mark(X), mark(Y)), zip(XS, YS))
     , a__zip(nil(), XS) -> nil()
     , a__tail(X) -> tail(X)
     , a__tail(cons(X, XS)) -> mark(XS)
     , a__repItems(X) -> repItems(X)
     , a__repItems(cons(X, XS)) -> cons(mark(X), cons(X, repItems(XS)))
     , a__repItems(nil()) -> nil() }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   We estimate the number of application of {1,2,3,8,12,20,23,25,28}
   by applications of Pre({1,2,3,8,12,20,23,25,28}) =
   {5,6,7,10,11,13,14,15,16,17,19,21,22,24,26,27,29,30}. Here rules
   are labeled as follows:
   
     DPs:
       { 1: a__pairNs^#() -> c_1()
       , 2: a__pairNs^#() -> c_2()
       , 3: a__oddNs^#() -> c_3()
       , 4: a__oddNs^#() -> c_4(a__incr^#(a__pairNs()))
       , 5: a__incr^#(X) -> c_5(X)
       , 6: a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS)
       , 7: mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2)
       , 8: mark^#(0()) -> c_8()
       , 9: mark^#(incr(X)) -> c_9(a__incr^#(mark(X)))
       , 10: mark^#(oddNs()) -> c_10(a__oddNs^#())
       , 11: mark^#(s(X)) -> c_11(mark^#(X))
       , 12: mark^#(nil()) -> c_12()
       , 13: mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2)))
       , 14: mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2))
       , 15: mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2)))
       , 16: mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X)))
       , 17: mark^#(pairNs()) -> c_17(a__pairNs^#())
       , 18: mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
       , 19: a__take^#(X1, X2) -> c_19(X1, X2)
       , 20: a__take^#(0(), XS) -> c_20()
       , 21: a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS)
       , 22: a__zip^#(X1, X2) -> c_22(X1, X2)
       , 23: a__zip^#(X, nil()) -> c_23()
       , 24: a__zip^#(cons(X, XS), cons(Y, YS)) ->
             c_24(mark^#(X), mark^#(Y), XS, YS)
       , 25: a__zip^#(nil(), XS) -> c_25()
       , 26: a__repItems^#(X) -> c_28(X)
       , 27: a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS)
       , 28: a__repItems^#(nil()) -> c_30()
       , 29: a__tail^#(X) -> c_26(X)
       , 30: a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) }
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { a__oddNs^#() -> c_4(a__incr^#(a__pairNs()))
     , a__incr^#(X) -> c_5(X)
     , a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS)
     , mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2)
     , mark^#(incr(X)) -> c_9(a__incr^#(mark(X)))
     , mark^#(oddNs()) -> c_10(a__oddNs^#())
     , mark^#(s(X)) -> c_11(mark^#(X))
     , mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2)))
     , mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2))
     , mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2)))
     , mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X)))
     , mark^#(pairNs()) -> c_17(a__pairNs^#())
     , mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
     , a__take^#(X1, X2) -> c_19(X1, X2)
     , a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS)
     , a__zip^#(X1, X2) -> c_22(X1, X2)
     , a__zip^#(cons(X, XS), cons(Y, YS)) ->
       c_24(mark^#(X), mark^#(Y), XS, YS)
     , a__repItems^#(X) -> c_28(X)
     , a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS)
     , a__tail^#(X) -> c_26(X)
     , a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) }
   Strict Trs:
     { a__pairNs() -> cons(0(), incr(oddNs()))
     , a__pairNs() -> pairNs()
     , a__oddNs() -> oddNs()
     , a__oddNs() -> a__incr(a__pairNs())
     , a__incr(X) -> incr(X)
     , a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(0()) -> 0()
     , mark(incr(X)) -> a__incr(mark(X))
     , mark(oddNs()) -> a__oddNs()
     , mark(s(X)) -> s(mark(X))
     , mark(nil()) -> nil()
     , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
     , mark(pair(X1, X2)) -> pair(mark(X1), mark(X2))
     , mark(zip(X1, X2)) -> a__zip(mark(X1), mark(X2))
     , mark(repItems(X)) -> a__repItems(mark(X))
     , mark(pairNs()) -> a__pairNs()
     , mark(tail(X)) -> a__tail(mark(X))
     , a__take(X1, X2) -> take(X1, X2)
     , a__take(0(), XS) -> nil()
     , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS))
     , a__zip(X1, X2) -> zip(X1, X2)
     , a__zip(X, nil()) -> nil()
     , a__zip(cons(X, XS), cons(Y, YS)) ->
       cons(pair(mark(X), mark(Y)), zip(XS, YS))
     , a__zip(nil(), XS) -> nil()
     , a__tail(X) -> tail(X)
     , a__tail(cons(X, XS)) -> mark(XS)
     , a__repItems(X) -> repItems(X)
     , a__repItems(cons(X, XS)) -> cons(mark(X), cons(X, repItems(XS)))
     , a__repItems(nil()) -> nil() }
   Weak DPs:
     { a__pairNs^#() -> c_1()
     , a__pairNs^#() -> c_2()
     , a__oddNs^#() -> c_3()
     , mark^#(0()) -> c_8()
     , mark^#(nil()) -> c_12()
     , a__take^#(0(), XS) -> c_20()
     , a__zip^#(X, nil()) -> c_23()
     , a__zip^#(nil(), XS) -> c_25()
     , a__repItems^#(nil()) -> c_30() }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   We estimate the number of application of {12} by applications of
   Pre({12}) = {2,3,4,7,9,14,15,16,17,18,19,20,21}. Here rules are
   labeled as follows:
   
     DPs:
       { 1: a__oddNs^#() -> c_4(a__incr^#(a__pairNs()))
       , 2: a__incr^#(X) -> c_5(X)
       , 3: a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS)
       , 4: mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2)
       , 5: mark^#(incr(X)) -> c_9(a__incr^#(mark(X)))
       , 6: mark^#(oddNs()) -> c_10(a__oddNs^#())
       , 7: mark^#(s(X)) -> c_11(mark^#(X))
       , 8: mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2)))
       , 9: mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2))
       , 10: mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2)))
       , 11: mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X)))
       , 12: mark^#(pairNs()) -> c_17(a__pairNs^#())
       , 13: mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
       , 14: a__take^#(X1, X2) -> c_19(X1, X2)
       , 15: a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS)
       , 16: a__zip^#(X1, X2) -> c_22(X1, X2)
       , 17: a__zip^#(cons(X, XS), cons(Y, YS)) ->
             c_24(mark^#(X), mark^#(Y), XS, YS)
       , 18: a__repItems^#(X) -> c_28(X)
       , 19: a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS)
       , 20: a__tail^#(X) -> c_26(X)
       , 21: a__tail^#(cons(X, XS)) -> c_27(mark^#(XS))
       , 22: a__pairNs^#() -> c_1()
       , 23: a__pairNs^#() -> c_2()
       , 24: a__oddNs^#() -> c_3()
       , 25: mark^#(0()) -> c_8()
       , 26: mark^#(nil()) -> c_12()
       , 27: a__take^#(0(), XS) -> c_20()
       , 28: a__zip^#(X, nil()) -> c_23()
       , 29: a__zip^#(nil(), XS) -> c_25()
       , 30: a__repItems^#(nil()) -> c_30() }
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { a__oddNs^#() -> c_4(a__incr^#(a__pairNs()))
     , a__incr^#(X) -> c_5(X)
     , a__incr^#(cons(X, XS)) -> c_6(mark^#(X), XS)
     , mark^#(cons(X1, X2)) -> c_7(mark^#(X1), X2)
     , mark^#(incr(X)) -> c_9(a__incr^#(mark(X)))
     , mark^#(oddNs()) -> c_10(a__oddNs^#())
     , mark^#(s(X)) -> c_11(mark^#(X))
     , mark^#(take(X1, X2)) -> c_13(a__take^#(mark(X1), mark(X2)))
     , mark^#(pair(X1, X2)) -> c_14(mark^#(X1), mark^#(X2))
     , mark^#(zip(X1, X2)) -> c_15(a__zip^#(mark(X1), mark(X2)))
     , mark^#(repItems(X)) -> c_16(a__repItems^#(mark(X)))
     , mark^#(tail(X)) -> c_18(a__tail^#(mark(X)))
     , a__take^#(X1, X2) -> c_19(X1, X2)
     , a__take^#(s(N), cons(X, XS)) -> c_21(mark^#(X), N, XS)
     , a__zip^#(X1, X2) -> c_22(X1, X2)
     , a__zip^#(cons(X, XS), cons(Y, YS)) ->
       c_24(mark^#(X), mark^#(Y), XS, YS)
     , a__repItems^#(X) -> c_28(X)
     , a__repItems^#(cons(X, XS)) -> c_29(mark^#(X), X, XS)
     , a__tail^#(X) -> c_26(X)
     , a__tail^#(cons(X, XS)) -> c_27(mark^#(XS)) }
   Strict Trs:
     { a__pairNs() -> cons(0(), incr(oddNs()))
     , a__pairNs() -> pairNs()
     , a__oddNs() -> oddNs()
     , a__oddNs() -> a__incr(a__pairNs())
     , a__incr(X) -> incr(X)
     , a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(0()) -> 0()
     , mark(incr(X)) -> a__incr(mark(X))
     , mark(oddNs()) -> a__oddNs()
     , mark(s(X)) -> s(mark(X))
     , mark(nil()) -> nil()
     , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2))
     , mark(pair(X1, X2)) -> pair(mark(X1), mark(X2))
     , mark(zip(X1, X2)) -> a__zip(mark(X1), mark(X2))
     , mark(repItems(X)) -> a__repItems(mark(X))
     , mark(pairNs()) -> a__pairNs()
     , mark(tail(X)) -> a__tail(mark(X))
     , a__take(X1, X2) -> take(X1, X2)
     , a__take(0(), XS) -> nil()
     , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS))
     , a__zip(X1, X2) -> zip(X1, X2)
     , a__zip(X, nil()) -> nil()
     , a__zip(cons(X, XS), cons(Y, YS)) ->
       cons(pair(mark(X), mark(Y)), zip(XS, YS))
     , a__zip(nil(), XS) -> nil()
     , a__tail(X) -> tail(X)
     , a__tail(cons(X, XS)) -> mark(XS)
     , a__repItems(X) -> repItems(X)
     , a__repItems(cons(X, XS)) -> cons(mark(X), cons(X, repItems(XS)))
     , a__repItems(nil()) -> nil() }
   Weak DPs:
     { a__pairNs^#() -> c_1()
     , a__pairNs^#() -> c_2()
     , a__oddNs^#() -> c_3()
     , mark^#(0()) -> c_8()
     , mark^#(nil()) -> c_12()
     , mark^#(pairNs()) -> c_17(a__pairNs^#())
     , a__take^#(0(), XS) -> c_20()
     , a__zip^#(X, nil()) -> c_23()
     , a__zip^#(nil(), XS) -> c_25()
     , a__repItems^#(nil()) -> c_30() }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   Empty strict component of the problem is NOT empty.


Arrrr..