MAYBE

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

Strict Trs:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(X)) -> mark(X)
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(0()) -> c_2()
  , a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0())))
  , a__p^#(X) -> c_4()
  , a__p^#(s(X)) -> c_5(mark^#(X))
  , mark^#(0()) -> c_6()
  , mark^#(cons(X1, X2)) -> c_7(mark^#(X1))
  , mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_9(mark^#(X))
  , mark^#(p(X)) -> c_10(a__p^#(mark(X)), 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__f^#(X) -> c_1()
  , a__f^#(0()) -> c_2()
  , a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0())))
  , a__p^#(X) -> c_4()
  , a__p^#(s(X)) -> c_5(mark^#(X))
  , mark^#(0()) -> c_6()
  , mark^#(cons(X1, X2)) -> c_7(mark^#(X1))
  , mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_9(mark^#(X))
  , mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(X)) -> mark(X)
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__f^#(X) -> c_1()
    , 2: a__f^#(0()) -> c_2()
    , 3: a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0())))
    , 4: a__p^#(X) -> c_4()
    , 5: a__p^#(s(X)) -> c_5(mark^#(X))
    , 6: mark^#(0()) -> c_6()
    , 7: mark^#(cons(X1, X2)) -> c_7(mark^#(X1))
    , 8: mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X))
    , 9: mark^#(s(X)) -> c_9(mark^#(X))
    , 10: mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) }

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

Strict DPs:
  { a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0())))
  , a__p^#(s(X)) -> c_5(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_7(mark^#(X1))
  , mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_9(mark^#(X))
  , mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) }
Weak DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(0()) -> c_2()
  , a__p^#(X) -> c_4()
  , mark^#(0()) -> c_6() }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(X)) -> mark(X)
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(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__f^#(X) -> c_1()
, a__f^#(0()) -> c_2()
, a__p^#(X) -> c_4()
, mark^#(0()) -> c_6() }

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

Strict DPs:
  { a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0())))
  , a__p^#(s(X)) -> c_5(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_7(mark^#(X1))
  , mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_9(mark^#(X))
  , mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(X)) -> mark(X)
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..