MAYBE

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

Strict Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , lastbit(0()) -> 0()
  , lastbit(s(0())) -> s(0())
  , lastbit(s(s(x))) -> lastbit(x)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , conv(x) -> conviter(x, cons(0(), nil()))
  , conviter(x, l) -> if(zero(x), x, l)
  , if(true(), x, l) -> l
  , if(false(), x, l) -> conviter(half(x), cons(lastbit(x), l)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , half^#(s(s(x))) -> c_3(half^#(x))
  , lastbit^#(0()) -> c_4()
  , lastbit^#(s(0())) -> c_5()
  , lastbit^#(s(s(x))) -> c_6(lastbit^#(x))
  , zero^#(0()) -> c_7()
  , zero^#(s(x)) -> c_8()
  , conv^#(x) -> c_9(conviter^#(x, cons(0(), nil())))
  , conviter^#(x, l) -> c_10(if^#(zero(x), x, l), zero^#(x))
  , if^#(true(), x, l) -> c_11()
  , if^#(false(), x, l) ->
    c_12(conviter^#(half(x), cons(lastbit(x), l)),
         half^#(x),
         lastbit^#(x)) }

and mark the set of starting terms.

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

Strict DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , half^#(s(s(x))) -> c_3(half^#(x))
  , lastbit^#(0()) -> c_4()
  , lastbit^#(s(0())) -> c_5()
  , lastbit^#(s(s(x))) -> c_6(lastbit^#(x))
  , zero^#(0()) -> c_7()
  , zero^#(s(x)) -> c_8()
  , conv^#(x) -> c_9(conviter^#(x, cons(0(), nil())))
  , conviter^#(x, l) -> c_10(if^#(zero(x), x, l), zero^#(x))
  , if^#(true(), x, l) -> c_11()
  , if^#(false(), x, l) ->
    c_12(conviter^#(half(x), cons(lastbit(x), l)),
         half^#(x),
         lastbit^#(x)) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , lastbit(0()) -> 0()
  , lastbit(s(0())) -> s(0())
  , lastbit(s(s(x))) -> lastbit(x)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , conv(x) -> conviter(x, cons(0(), nil()))
  , conviter(x, l) -> if(zero(x), x, l)
  , if(true(), x, l) -> l
  , if(false(), x, l) -> conviter(half(x), cons(lastbit(x), l)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: half^#(0()) -> c_1()
    , 2: half^#(s(0())) -> c_2()
    , 3: half^#(s(s(x))) -> c_3(half^#(x))
    , 4: lastbit^#(0()) -> c_4()
    , 5: lastbit^#(s(0())) -> c_5()
    , 6: lastbit^#(s(s(x))) -> c_6(lastbit^#(x))
    , 7: zero^#(0()) -> c_7()
    , 8: zero^#(s(x)) -> c_8()
    , 9: conv^#(x) -> c_9(conviter^#(x, cons(0(), nil())))
    , 10: conviter^#(x, l) -> c_10(if^#(zero(x), x, l), zero^#(x))
    , 11: if^#(true(), x, l) -> c_11()
    , 12: if^#(false(), x, l) ->
          c_12(conviter^#(half(x), cons(lastbit(x), l)),
               half^#(x),
               lastbit^#(x)) }

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

Strict DPs:
  { half^#(s(s(x))) -> c_3(half^#(x))
  , lastbit^#(s(s(x))) -> c_6(lastbit^#(x))
  , conv^#(x) -> c_9(conviter^#(x, cons(0(), nil())))
  , conviter^#(x, l) -> c_10(if^#(zero(x), x, l), zero^#(x))
  , if^#(false(), x, l) ->
    c_12(conviter^#(half(x), cons(lastbit(x), l)),
         half^#(x),
         lastbit^#(x)) }
Weak DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , lastbit^#(0()) -> c_4()
  , lastbit^#(s(0())) -> c_5()
  , zero^#(0()) -> c_7()
  , zero^#(s(x)) -> c_8()
  , if^#(true(), x, l) -> c_11() }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , lastbit(0()) -> 0()
  , lastbit(s(0())) -> s(0())
  , lastbit(s(s(x))) -> lastbit(x)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , conv(x) -> conviter(x, cons(0(), nil()))
  , conviter(x, l) -> if(zero(x), x, l)
  , if(true(), x, l) -> l
  , if(false(), x, l) -> conviter(half(x), cons(lastbit(x), l)) }
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.

{ half^#(0()) -> c_1()
, half^#(s(0())) -> c_2()
, lastbit^#(0()) -> c_4()
, lastbit^#(s(0())) -> c_5()
, zero^#(0()) -> c_7()
, zero^#(s(x)) -> c_8()
, if^#(true(), x, l) -> c_11() }

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

Strict DPs:
  { half^#(s(s(x))) -> c_3(half^#(x))
  , lastbit^#(s(s(x))) -> c_6(lastbit^#(x))
  , conv^#(x) -> c_9(conviter^#(x, cons(0(), nil())))
  , conviter^#(x, l) -> c_10(if^#(zero(x), x, l), zero^#(x))
  , if^#(false(), x, l) ->
    c_12(conviter^#(half(x), cons(lastbit(x), l)),
         half^#(x),
         lastbit^#(x)) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , lastbit(0()) -> 0()
  , lastbit(s(0())) -> s(0())
  , lastbit(s(s(x))) -> lastbit(x)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , conv(x) -> conviter(x, cons(0(), nil()))
  , conviter(x, l) -> if(zero(x), x, l)
  , if(true(), x, l) -> l
  , if(false(), x, l) -> conviter(half(x), cons(lastbit(x), l)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { conviter^#(x, l) -> c_10(if^#(zero(x), x, l), zero^#(x)) }

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

Strict DPs:
  { half^#(s(s(x))) -> c_1(half^#(x))
  , lastbit^#(s(s(x))) -> c_2(lastbit^#(x))
  , conv^#(x) -> c_3(conviter^#(x, cons(0(), nil())))
  , conviter^#(x, l) -> c_4(if^#(zero(x), x, l))
  , if^#(false(), x, l) ->
    c_5(conviter^#(half(x), cons(lastbit(x), l)),
        half^#(x),
        lastbit^#(x)) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , lastbit(0()) -> 0()
  , lastbit(s(0())) -> s(0())
  , lastbit(s(s(x))) -> lastbit(x)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , conv(x) -> conviter(x, cons(0(), nil()))
  , conviter(x, l) -> if(zero(x), x, l)
  , if(true(), x, l) -> l
  , if(false(), x, l) -> conviter(half(x), cons(lastbit(x), l)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { half(0()) -> 0()
    , half(s(0())) -> 0()
    , half(s(s(x))) -> s(half(x))
    , lastbit(0()) -> 0()
    , lastbit(s(0())) -> s(0())
    , lastbit(s(s(x))) -> lastbit(x)
    , zero(0()) -> true()
    , zero(s(x)) -> false() }

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

Strict DPs:
  { half^#(s(s(x))) -> c_1(half^#(x))
  , lastbit^#(s(s(x))) -> c_2(lastbit^#(x))
  , conv^#(x) -> c_3(conviter^#(x, cons(0(), nil())))
  , conviter^#(x, l) -> c_4(if^#(zero(x), x, l))
  , if^#(false(), x, l) ->
    c_5(conviter^#(half(x), cons(lastbit(x), l)),
        half^#(x),
        lastbit^#(x)) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , lastbit(0()) -> 0()
  , lastbit(s(0())) -> s(0())
  , lastbit(s(s(x))) -> lastbit(x)
  , zero(0()) -> true()
  , zero(s(x)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: half^#(s(s(x))) -> c_1(half^#(x))
     -->_1 half^#(s(s(x))) -> c_1(half^#(x)) :1
  
  2: lastbit^#(s(s(x))) -> c_2(lastbit^#(x))
     -->_1 lastbit^#(s(s(x))) -> c_2(lastbit^#(x)) :2
  
  3: conv^#(x) -> c_3(conviter^#(x, cons(0(), nil())))
     -->_1 conviter^#(x, l) -> c_4(if^#(zero(x), x, l)) :4
  
  4: conviter^#(x, l) -> c_4(if^#(zero(x), x, l))
     -->_1 if^#(false(), x, l) ->
           c_5(conviter^#(half(x), cons(lastbit(x), l)),
               half^#(x),
               lastbit^#(x)) :5
  
  5: if^#(false(), x, l) ->
     c_5(conviter^#(half(x), cons(lastbit(x), l)),
         half^#(x),
         lastbit^#(x))
     -->_1 conviter^#(x, l) -> c_4(if^#(zero(x), x, l)) :4
     -->_3 lastbit^#(s(s(x))) -> c_2(lastbit^#(x)) :2
     -->_2 half^#(s(s(x))) -> c_1(half^#(x)) :1
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { conv^#(x) -> c_3(conviter^#(x, cons(0(), nil()))) }


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

Strict DPs:
  { half^#(s(s(x))) -> c_1(half^#(x))
  , lastbit^#(s(s(x))) -> c_2(lastbit^#(x))
  , conviter^#(x, l) -> c_4(if^#(zero(x), x, l))
  , if^#(false(), x, l) ->
    c_5(conviter^#(half(x), cons(lastbit(x), l)),
        half^#(x),
        lastbit^#(x)) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , lastbit(0()) -> 0()
  , lastbit(s(0())) -> s(0())
  , lastbit(s(s(x))) -> lastbit(x)
  , zero(0()) -> true()
  , zero(s(x)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..