MAYBE

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

Strict Trs:
  { a__fib(N) -> a__sel(mark(N), a__fib1(s(0()), s(0())))
  , a__fib(X) -> fib(X)
  , 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)
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(fib1(X1, X2)) -> a__fib1(mark(X1), mark(X2))
  , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
  , mark(fib(X)) -> a__fib(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__fib1(X, Y) -> cons(mark(X), fib1(Y, add(X, Y)))
  , a__fib1(X1, X2) -> fib1(X1, X2)
  , a__add(X1, X2) -> add(X1, X2)
  , a__add(s(X), Y) -> s(a__add(mark(X), mark(Y)))
  , a__add(0(), X) -> mark(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__fib^#(N) ->
    c_1(a__sel^#(mark(N), a__fib1(s(0()), s(0()))),
        mark^#(N),
        a__fib1^#(s(0()), s(0())))
  , a__fib^#(X) -> c_2()
  , a__sel^#(X1, X2) -> c_3()
  , a__sel^#(s(N), cons(X, XS)) ->
    c_4(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__sel^#(0(), cons(X, XS)) -> c_5(mark^#(X))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(0()) -> c_7()
  , mark^#(cons(X1, X2)) -> c_8(mark^#(X1))
  , mark^#(fib1(X1, X2)) ->
    c_9(a__fib1^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(add(X1, X2)) ->
    c_10(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(fib(X)) -> c_11(a__fib^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_12(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__fib1^#(X, Y) -> c_13(mark^#(X))
  , a__fib1^#(X1, X2) -> c_14()
  , a__add^#(X1, X2) -> c_15()
  , a__add^#(s(X), Y) ->
    c_16(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y))
  , a__add^#(0(), X) -> c_17(mark^#(X)) }

and mark the set of starting terms.

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

Strict DPs:
  { a__fib^#(N) ->
    c_1(a__sel^#(mark(N), a__fib1(s(0()), s(0()))),
        mark^#(N),
        a__fib1^#(s(0()), s(0())))
  , a__fib^#(X) -> c_2()
  , a__sel^#(X1, X2) -> c_3()
  , a__sel^#(s(N), cons(X, XS)) ->
    c_4(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__sel^#(0(), cons(X, XS)) -> c_5(mark^#(X))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(0()) -> c_7()
  , mark^#(cons(X1, X2)) -> c_8(mark^#(X1))
  , mark^#(fib1(X1, X2)) ->
    c_9(a__fib1^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(add(X1, X2)) ->
    c_10(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(fib(X)) -> c_11(a__fib^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_12(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__fib1^#(X, Y) -> c_13(mark^#(X))
  , a__fib1^#(X1, X2) -> c_14()
  , a__add^#(X1, X2) -> c_15()
  , a__add^#(s(X), Y) ->
    c_16(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y))
  , a__add^#(0(), X) -> c_17(mark^#(X)) }
Weak Trs:
  { a__fib(N) -> a__sel(mark(N), a__fib1(s(0()), s(0())))
  , a__fib(X) -> fib(X)
  , 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)
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(fib1(X1, X2)) -> a__fib1(mark(X1), mark(X2))
  , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
  , mark(fib(X)) -> a__fib(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__fib1(X, Y) -> cons(mark(X), fib1(Y, add(X, Y)))
  , a__fib1(X1, X2) -> fib1(X1, X2)
  , a__add(X1, X2) -> add(X1, X2)
  , a__add(s(X), Y) -> s(a__add(mark(X), mark(Y)))
  , a__add(0(), X) -> mark(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {2,3,7,14,15} by
applications of Pre({2,3,7,14,15}) =
{1,4,5,6,8,9,10,11,12,13,16,17}. Here rules are labeled as follows:

  DPs:
    { 1: a__fib^#(N) ->
         c_1(a__sel^#(mark(N), a__fib1(s(0()), s(0()))),
             mark^#(N),
             a__fib1^#(s(0()), s(0())))
    , 2: a__fib^#(X) -> c_2()
    , 3: a__sel^#(X1, X2) -> c_3()
    , 4: a__sel^#(s(N), cons(X, XS)) ->
         c_4(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
    , 5: a__sel^#(0(), cons(X, XS)) -> c_5(mark^#(X))
    , 6: mark^#(s(X)) -> c_6(mark^#(X))
    , 7: mark^#(0()) -> c_7()
    , 8: mark^#(cons(X1, X2)) -> c_8(mark^#(X1))
    , 9: mark^#(fib1(X1, X2)) ->
         c_9(a__fib1^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 10: mark^#(add(X1, X2)) ->
          c_10(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 11: mark^#(fib(X)) -> c_11(a__fib^#(mark(X)), mark^#(X))
    , 12: mark^#(sel(X1, X2)) ->
          c_12(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 13: a__fib1^#(X, Y) -> c_13(mark^#(X))
    , 14: a__fib1^#(X1, X2) -> c_14()
    , 15: a__add^#(X1, X2) -> c_15()
    , 16: a__add^#(s(X), Y) ->
          c_16(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y))
    , 17: a__add^#(0(), X) -> c_17(mark^#(X)) }

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

Strict DPs:
  { a__fib^#(N) ->
    c_1(a__sel^#(mark(N), a__fib1(s(0()), s(0()))),
        mark^#(N),
        a__fib1^#(s(0()), s(0())))
  , a__sel^#(s(N), cons(X, XS)) ->
    c_4(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__sel^#(0(), cons(X, XS)) -> c_5(mark^#(X))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_8(mark^#(X1))
  , mark^#(fib1(X1, X2)) ->
    c_9(a__fib1^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(add(X1, X2)) ->
    c_10(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(fib(X)) -> c_11(a__fib^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_12(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__fib1^#(X, Y) -> c_13(mark^#(X))
  , a__add^#(s(X), Y) ->
    c_16(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y))
  , a__add^#(0(), X) -> c_17(mark^#(X)) }
Weak DPs:
  { a__fib^#(X) -> c_2()
  , a__sel^#(X1, X2) -> c_3()
  , mark^#(0()) -> c_7()
  , a__fib1^#(X1, X2) -> c_14()
  , a__add^#(X1, X2) -> c_15() }
Weak Trs:
  { a__fib(N) -> a__sel(mark(N), a__fib1(s(0()), s(0())))
  , a__fib(X) -> fib(X)
  , 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)
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(fib1(X1, X2)) -> a__fib1(mark(X1), mark(X2))
  , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
  , mark(fib(X)) -> a__fib(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__fib1(X, Y) -> cons(mark(X), fib1(Y, add(X, Y)))
  , a__fib1(X1, X2) -> fib1(X1, X2)
  , a__add(X1, X2) -> add(X1, X2)
  , a__add(s(X), Y) -> s(a__add(mark(X), mark(Y)))
  , a__add(0(), X) -> mark(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ a__fib^#(X) -> c_2()
, a__sel^#(X1, X2) -> c_3()
, mark^#(0()) -> c_7()
, a__fib1^#(X1, X2) -> c_14()
, a__add^#(X1, X2) -> c_15() }

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

Strict DPs:
  { a__fib^#(N) ->
    c_1(a__sel^#(mark(N), a__fib1(s(0()), s(0()))),
        mark^#(N),
        a__fib1^#(s(0()), s(0())))
  , a__sel^#(s(N), cons(X, XS)) ->
    c_4(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__sel^#(0(), cons(X, XS)) -> c_5(mark^#(X))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_8(mark^#(X1))
  , mark^#(fib1(X1, X2)) ->
    c_9(a__fib1^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(add(X1, X2)) ->
    c_10(a__add^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(fib(X)) -> c_11(a__fib^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_12(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__fib1^#(X, Y) -> c_13(mark^#(X))
  , a__add^#(s(X), Y) ->
    c_16(a__add^#(mark(X), mark(Y)), mark^#(X), mark^#(Y))
  , a__add^#(0(), X) -> c_17(mark^#(X)) }
Weak Trs:
  { a__fib(N) -> a__sel(mark(N), a__fib1(s(0()), s(0())))
  , a__fib(X) -> fib(X)
  , 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)
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(fib1(X1, X2)) -> a__fib1(mark(X1), mark(X2))
  , mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
  , mark(fib(X)) -> a__fib(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__fib1(X, Y) -> cons(mark(X), fib1(Y, add(X, Y)))
  , a__fib1(X1, X2) -> fib1(X1, X2)
  , a__add(X1, X2) -> add(X1, X2)
  , a__add(s(X), Y) -> s(a__add(mark(X), mark(Y)))
  , a__add(0(), X) -> mark(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

None of the processors succeeded.

Details of failed attempt(s):
-----------------------------
1) 'matrices' failed due to the following reason:
   
   None of the processors succeeded.
   
   Details of failed attempt(s):
   -----------------------------
   1) 'matrix interpretation of dimension 4' failed due to the
      following reason:
      
      Following exception was raised:
        stack overflow
   
   2) 'matrix interpretation of dimension 3' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   3) 'matrix interpretation of dimension 3' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   4) 'matrix interpretation of dimension 2' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   5) 'matrix interpretation of dimension 2' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   6) 'matrix interpretation of dimension 1' failed due to the
      following reason:
      
      The input cannot be shown compatible
   

2) 'empty' failed due to the following reason:
   
   Empty strict component of the problem is NOT empty.


Arrrr..