MAYBE

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

Strict Trs:
  { a__2nd(X) -> 2nd(X)
  , a__2nd(cons1(X, cons(Y, Z))) -> mark(Y)
  , a__2nd(cons(X, X1)) -> a__2nd(cons1(mark(X), mark(X1)))
  , mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(2nd(X)) -> a__2nd(mark(X))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__2nd^#(X) -> c_1()
  , a__2nd^#(cons1(X, cons(Y, Z))) -> c_2(mark^#(Y))
  , a__2nd^#(cons(X, X1)) ->
    c_3(a__2nd^#(cons1(mark(X), mark(X1))), mark^#(X), mark^#(X1))
  , mark^#(cons1(X1, X2)) -> c_4(mark^#(X1), mark^#(X2))
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(from(X)) -> c_6(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_7(mark^#(X))
  , mark^#(2nd(X)) -> c_8(a__2nd^#(mark(X)), mark^#(X))
  , a__from^#(X) -> c_9(mark^#(X))
  , a__from^#(X) -> c_10() }

and mark the set of starting terms.

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

Strict DPs:
  { a__2nd^#(X) -> c_1()
  , a__2nd^#(cons1(X, cons(Y, Z))) -> c_2(mark^#(Y))
  , a__2nd^#(cons(X, X1)) ->
    c_3(a__2nd^#(cons1(mark(X), mark(X1))), mark^#(X), mark^#(X1))
  , mark^#(cons1(X1, X2)) -> c_4(mark^#(X1), mark^#(X2))
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(from(X)) -> c_6(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_7(mark^#(X))
  , mark^#(2nd(X)) -> c_8(a__2nd^#(mark(X)), mark^#(X))
  , a__from^#(X) -> c_9(mark^#(X))
  , a__from^#(X) -> c_10() }
Weak Trs:
  { a__2nd(X) -> 2nd(X)
  , a__2nd(cons1(X, cons(Y, Z))) -> mark(Y)
  , a__2nd(cons(X, X1)) -> a__2nd(cons1(mark(X), mark(X1)))
  , mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(2nd(X)) -> a__2nd(mark(X))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__2nd^#(X) -> c_1()
    , 2: a__2nd^#(cons1(X, cons(Y, Z))) -> c_2(mark^#(Y))
    , 3: a__2nd^#(cons(X, X1)) ->
         c_3(a__2nd^#(cons1(mark(X), mark(X1))), mark^#(X), mark^#(X1))
    , 4: mark^#(cons1(X1, X2)) -> c_4(mark^#(X1), mark^#(X2))
    , 5: mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
    , 6: mark^#(from(X)) -> c_6(a__from^#(mark(X)), mark^#(X))
    , 7: mark^#(s(X)) -> c_7(mark^#(X))
    , 8: mark^#(2nd(X)) -> c_8(a__2nd^#(mark(X)), mark^#(X))
    , 9: a__from^#(X) -> c_9(mark^#(X))
    , 10: a__from^#(X) -> c_10() }

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

Strict DPs:
  { a__2nd^#(cons1(X, cons(Y, Z))) -> c_2(mark^#(Y))
  , a__2nd^#(cons(X, X1)) ->
    c_3(a__2nd^#(cons1(mark(X), mark(X1))), mark^#(X), mark^#(X1))
  , mark^#(cons1(X1, X2)) -> c_4(mark^#(X1), mark^#(X2))
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(from(X)) -> c_6(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_7(mark^#(X))
  , mark^#(2nd(X)) -> c_8(a__2nd^#(mark(X)), mark^#(X))
  , a__from^#(X) -> c_9(mark^#(X)) }
Weak DPs:
  { a__2nd^#(X) -> c_1()
  , a__from^#(X) -> c_10() }
Weak Trs:
  { a__2nd(X) -> 2nd(X)
  , a__2nd(cons1(X, cons(Y, Z))) -> mark(Y)
  , a__2nd(cons(X, X1)) -> a__2nd(cons1(mark(X), mark(X1)))
  , mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(2nd(X)) -> a__2nd(mark(X))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(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__2nd^#(X) -> c_1()
, a__from^#(X) -> c_10() }

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

Strict DPs:
  { a__2nd^#(cons1(X, cons(Y, Z))) -> c_2(mark^#(Y))
  , a__2nd^#(cons(X, X1)) ->
    c_3(a__2nd^#(cons1(mark(X), mark(X1))), mark^#(X), mark^#(X1))
  , mark^#(cons1(X1, X2)) -> c_4(mark^#(X1), mark^#(X2))
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(from(X)) -> c_6(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_7(mark^#(X))
  , mark^#(2nd(X)) -> c_8(a__2nd^#(mark(X)), mark^#(X))
  , a__from^#(X) -> c_9(mark^#(X)) }
Weak Trs:
  { a__2nd(X) -> 2nd(X)
  , a__2nd(cons1(X, cons(Y, Z))) -> mark(Y)
  , a__2nd(cons(X, X1)) -> a__2nd(cons1(mark(X), mark(X1)))
  , mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2))
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(2nd(X)) -> a__2nd(mark(X))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(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..