MAYBE

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

Strict Trs:
  { double(0()) -> 0()
  , double(s(x)) -> s(s(double(x)))
  , del(x, nil()) -> nil()
  , del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
  , if(true(), x, y, xs) -> xs
  , if(false(), x, y, xs) -> cons(y, del(x, xs))
  , eq(0(), 0()) -> true()
  , eq(0(), s(y)) -> false()
  , eq(s(x), 0()) -> false()
  , eq(s(x), s(y)) -> eq(x, y)
  , first(nil()) -> 0()
  , first(cons(x, xs)) -> x
  , doublelist(nil()) -> nil()
  , doublelist(cons(x, xs)) ->
    cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { double^#(0()) -> c_1()
  , double^#(s(x)) -> c_2(double^#(x))
  , del^#(x, nil()) -> c_3()
  , del^#(x, cons(y, xs)) ->
    c_4(if^#(eq(x, y), x, y, xs), eq^#(x, y))
  , if^#(true(), x, y, xs) -> c_5()
  , if^#(false(), x, y, xs) -> c_6(del^#(x, xs))
  , eq^#(0(), 0()) -> c_7()
  , eq^#(0(), s(y)) -> c_8()
  , eq^#(s(x), 0()) -> c_9()
  , eq^#(s(x), s(y)) -> c_10(eq^#(x, y))
  , first^#(nil()) -> c_11()
  , first^#(cons(x, xs)) -> c_12()
  , doublelist^#(nil()) -> c_13()
  , doublelist^#(cons(x, xs)) ->
    c_14(double^#(x),
         doublelist^#(del(first(cons(x, xs)), cons(x, xs))),
         del^#(first(cons(x, xs)), cons(x, xs)),
         first^#(cons(x, xs))) }

and mark the set of starting terms.

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

Strict DPs:
  { double^#(0()) -> c_1()
  , double^#(s(x)) -> c_2(double^#(x))
  , del^#(x, nil()) -> c_3()
  , del^#(x, cons(y, xs)) ->
    c_4(if^#(eq(x, y), x, y, xs), eq^#(x, y))
  , if^#(true(), x, y, xs) -> c_5()
  , if^#(false(), x, y, xs) -> c_6(del^#(x, xs))
  , eq^#(0(), 0()) -> c_7()
  , eq^#(0(), s(y)) -> c_8()
  , eq^#(s(x), 0()) -> c_9()
  , eq^#(s(x), s(y)) -> c_10(eq^#(x, y))
  , first^#(nil()) -> c_11()
  , first^#(cons(x, xs)) -> c_12()
  , doublelist^#(nil()) -> c_13()
  , doublelist^#(cons(x, xs)) ->
    c_14(double^#(x),
         doublelist^#(del(first(cons(x, xs)), cons(x, xs))),
         del^#(first(cons(x, xs)), cons(x, xs)),
         first^#(cons(x, xs))) }
Weak Trs:
  { double(0()) -> 0()
  , double(s(x)) -> s(s(double(x)))
  , del(x, nil()) -> nil()
  , del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
  , if(true(), x, y, xs) -> xs
  , if(false(), x, y, xs) -> cons(y, del(x, xs))
  , eq(0(), 0()) -> true()
  , eq(0(), s(y)) -> false()
  , eq(s(x), 0()) -> false()
  , eq(s(x), s(y)) -> eq(x, y)
  , first(nil()) -> 0()
  , first(cons(x, xs)) -> x
  , doublelist(nil()) -> nil()
  , doublelist(cons(x, xs)) ->
    cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: double^#(0()) -> c_1()
    , 2: double^#(s(x)) -> c_2(double^#(x))
    , 3: del^#(x, nil()) -> c_3()
    , 4: del^#(x, cons(y, xs)) ->
         c_4(if^#(eq(x, y), x, y, xs), eq^#(x, y))
    , 5: if^#(true(), x, y, xs) -> c_5()
    , 6: if^#(false(), x, y, xs) -> c_6(del^#(x, xs))
    , 7: eq^#(0(), 0()) -> c_7()
    , 8: eq^#(0(), s(y)) -> c_8()
    , 9: eq^#(s(x), 0()) -> c_9()
    , 10: eq^#(s(x), s(y)) -> c_10(eq^#(x, y))
    , 11: first^#(nil()) -> c_11()
    , 12: first^#(cons(x, xs)) -> c_12()
    , 13: doublelist^#(nil()) -> c_13()
    , 14: doublelist^#(cons(x, xs)) ->
          c_14(double^#(x),
               doublelist^#(del(first(cons(x, xs)), cons(x, xs))),
               del^#(first(cons(x, xs)), cons(x, xs)),
               first^#(cons(x, xs))) }

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

Strict DPs:
  { double^#(s(x)) -> c_2(double^#(x))
  , del^#(x, cons(y, xs)) ->
    c_4(if^#(eq(x, y), x, y, xs), eq^#(x, y))
  , if^#(false(), x, y, xs) -> c_6(del^#(x, xs))
  , eq^#(s(x), s(y)) -> c_10(eq^#(x, y))
  , doublelist^#(cons(x, xs)) ->
    c_14(double^#(x),
         doublelist^#(del(first(cons(x, xs)), cons(x, xs))),
         del^#(first(cons(x, xs)), cons(x, xs)),
         first^#(cons(x, xs))) }
Weak DPs:
  { double^#(0()) -> c_1()
  , del^#(x, nil()) -> c_3()
  , if^#(true(), x, y, xs) -> c_5()
  , eq^#(0(), 0()) -> c_7()
  , eq^#(0(), s(y)) -> c_8()
  , eq^#(s(x), 0()) -> c_9()
  , first^#(nil()) -> c_11()
  , first^#(cons(x, xs)) -> c_12()
  , doublelist^#(nil()) -> c_13() }
Weak Trs:
  { double(0()) -> 0()
  , double(s(x)) -> s(s(double(x)))
  , del(x, nil()) -> nil()
  , del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
  , if(true(), x, y, xs) -> xs
  , if(false(), x, y, xs) -> cons(y, del(x, xs))
  , eq(0(), 0()) -> true()
  , eq(0(), s(y)) -> false()
  , eq(s(x), 0()) -> false()
  , eq(s(x), s(y)) -> eq(x, y)
  , first(nil()) -> 0()
  , first(cons(x, xs)) -> x
  , doublelist(nil()) -> nil()
  , doublelist(cons(x, xs)) ->
    cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, 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.

{ double^#(0()) -> c_1()
, del^#(x, nil()) -> c_3()
, if^#(true(), x, y, xs) -> c_5()
, eq^#(0(), 0()) -> c_7()
, eq^#(0(), s(y)) -> c_8()
, eq^#(s(x), 0()) -> c_9()
, first^#(nil()) -> c_11()
, first^#(cons(x, xs)) -> c_12()
, doublelist^#(nil()) -> c_13() }

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

Strict DPs:
  { double^#(s(x)) -> c_2(double^#(x))
  , del^#(x, cons(y, xs)) ->
    c_4(if^#(eq(x, y), x, y, xs), eq^#(x, y))
  , if^#(false(), x, y, xs) -> c_6(del^#(x, xs))
  , eq^#(s(x), s(y)) -> c_10(eq^#(x, y))
  , doublelist^#(cons(x, xs)) ->
    c_14(double^#(x),
         doublelist^#(del(first(cons(x, xs)), cons(x, xs))),
         del^#(first(cons(x, xs)), cons(x, xs)),
         first^#(cons(x, xs))) }
Weak Trs:
  { double(0()) -> 0()
  , double(s(x)) -> s(s(double(x)))
  , del(x, nil()) -> nil()
  , del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
  , if(true(), x, y, xs) -> xs
  , if(false(), x, y, xs) -> cons(y, del(x, xs))
  , eq(0(), 0()) -> true()
  , eq(0(), s(y)) -> false()
  , eq(s(x), 0()) -> false()
  , eq(s(x), s(y)) -> eq(x, y)
  , first(nil()) -> 0()
  , first(cons(x, xs)) -> x
  , doublelist(nil()) -> nil()
  , doublelist(cons(x, xs)) ->
    cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { doublelist^#(cons(x, xs)) ->
    c_14(double^#(x),
         doublelist^#(del(first(cons(x, xs)), cons(x, xs))),
         del^#(first(cons(x, xs)), cons(x, xs)),
         first^#(cons(x, xs))) }

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

Strict DPs:
  { double^#(s(x)) -> c_1(double^#(x))
  , del^#(x, cons(y, xs)) ->
    c_2(if^#(eq(x, y), x, y, xs), eq^#(x, y))
  , if^#(false(), x, y, xs) -> c_3(del^#(x, xs))
  , eq^#(s(x), s(y)) -> c_4(eq^#(x, y))
  , doublelist^#(cons(x, xs)) ->
    c_5(double^#(x),
        doublelist^#(del(first(cons(x, xs)), cons(x, xs))),
        del^#(first(cons(x, xs)), cons(x, xs))) }
Weak Trs:
  { double(0()) -> 0()
  , double(s(x)) -> s(s(double(x)))
  , del(x, nil()) -> nil()
  , del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
  , if(true(), x, y, xs) -> xs
  , if(false(), x, y, xs) -> cons(y, del(x, xs))
  , eq(0(), 0()) -> true()
  , eq(0(), s(y)) -> false()
  , eq(s(x), 0()) -> false()
  , eq(s(x), s(y)) -> eq(x, y)
  , first(nil()) -> 0()
  , first(cons(x, xs)) -> x
  , doublelist(nil()) -> nil()
  , doublelist(cons(x, xs)) ->
    cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { del(x, nil()) -> nil()
    , del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
    , if(true(), x, y, xs) -> xs
    , if(false(), x, y, xs) -> cons(y, del(x, xs))
    , eq(0(), 0()) -> true()
    , eq(0(), s(y)) -> false()
    , eq(s(x), 0()) -> false()
    , eq(s(x), s(y)) -> eq(x, y)
    , first(nil()) -> 0()
    , first(cons(x, xs)) -> x }

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

Strict DPs:
  { double^#(s(x)) -> c_1(double^#(x))
  , del^#(x, cons(y, xs)) ->
    c_2(if^#(eq(x, y), x, y, xs), eq^#(x, y))
  , if^#(false(), x, y, xs) -> c_3(del^#(x, xs))
  , eq^#(s(x), s(y)) -> c_4(eq^#(x, y))
  , doublelist^#(cons(x, xs)) ->
    c_5(double^#(x),
        doublelist^#(del(first(cons(x, xs)), cons(x, xs))),
        del^#(first(cons(x, xs)), cons(x, xs))) }
Weak Trs:
  { del(x, nil()) -> nil()
  , del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
  , if(true(), x, y, xs) -> xs
  , if(false(), x, y, xs) -> cons(y, del(x, xs))
  , eq(0(), 0()) -> true()
  , eq(0(), s(y)) -> false()
  , eq(s(x), 0()) -> false()
  , eq(s(x), s(y)) -> eq(x, y)
  , first(nil()) -> 0()
  , first(cons(x, xs)) -> 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..