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(after(X1, X2)) -> a__after(mark(X1), mark(X2))
  , a__after(X1, X2) -> after(X1, X2)
  , a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS))
  , a__after(0(), XS) -> mark(XS) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__from^#(X) -> c_1(mark^#(X))
  , a__from^#(X) -> c_2()
  , mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
  , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_5(mark^#(X))
  , mark^#(0()) -> c_6()
  , mark^#(after(X1, X2)) ->
    c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__after^#(X1, X2) -> c_8()
  , a__after^#(s(N), cons(X, XS)) ->
    c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__after^#(0(), XS) -> c_10(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__from^#(X) -> c_1(mark^#(X))
  , a__from^#(X) -> c_2()
  , mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
  , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_5(mark^#(X))
  , mark^#(0()) -> c_6()
  , mark^#(after(X1, X2)) ->
    c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__after^#(X1, X2) -> c_8()
  , a__after^#(s(N), cons(X, XS)) ->
    c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__after^#(0(), XS) -> c_10(mark^#(XS)) }
Weak 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(after(X1, X2)) -> a__after(mark(X1), mark(X2))
  , a__after(X1, X2) -> after(X1, X2)
  , a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS))
  , a__after(0(), XS) -> mark(XS) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__from^#(X) -> c_1(mark^#(X))
    , 2: a__from^#(X) -> c_2()
    , 3: mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
    , 4: mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
    , 5: mark^#(s(X)) -> c_5(mark^#(X))
    , 6: mark^#(0()) -> c_6()
    , 7: mark^#(after(X1, X2)) ->
         c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 8: a__after^#(X1, X2) -> c_8()
    , 9: a__after^#(s(N), cons(X, XS)) ->
         c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
    , 10: a__after^#(0(), XS) -> c_10(mark^#(XS)) }

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

Strict DPs:
  { a__from^#(X) -> c_1(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
  , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_5(mark^#(X))
  , mark^#(after(X1, X2)) ->
    c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__after^#(s(N), cons(X, XS)) ->
    c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__after^#(0(), XS) -> c_10(mark^#(XS)) }
Weak DPs:
  { a__from^#(X) -> c_2()
  , mark^#(0()) -> c_6()
  , a__after^#(X1, X2) -> c_8() }
Weak 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(after(X1, X2)) -> a__after(mark(X1), mark(X2))
  , a__after(X1, X2) -> after(X1, X2)
  , a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS))
  , a__after(0(), XS) -> mark(XS) }
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__from^#(X) -> c_2()
, mark^#(0()) -> c_6()
, a__after^#(X1, X2) -> c_8() }

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

Strict DPs:
  { a__from^#(X) -> c_1(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
  , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_5(mark^#(X))
  , mark^#(after(X1, X2)) ->
    c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__after^#(s(N), cons(X, XS)) ->
    c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__after^#(0(), XS) -> c_10(mark^#(XS)) }
Weak 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(after(X1, X2)) -> a__after(mark(X1), mark(X2))
  , a__after(X1, X2) -> after(X1, X2)
  , a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS))
  , a__after(0(), XS) -> mark(XS) }
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:
      
      The input cannot be shown compatible
   
   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..