MAYBE

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

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


Arrrr..