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))
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , logarithm(x) -> logIter(x, 0())
  , logIter(x, y) ->
    if(le(s(0()), x), le(s(s(0())), x), half(x), inc(y))
  , if(true(), true(), x, y) -> logIter(x, y)
  , if(true(), false(), x, s(y)) -> y
  , if(false(), b, x, y) -> logZeroError()
  , f() -> g()
  , f() -> h() }
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))
  , le^#(0(), y) -> c_4()
  , le^#(s(x), 0()) -> c_5()
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , inc^#(0()) -> c_7()
  , inc^#(s(x)) -> c_8(inc^#(x))
  , logarithm^#(x) -> c_9(logIter^#(x, 0()))
  , logIter^#(x, y) ->
    c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
         le^#(s(0()), x),
         le^#(s(s(0())), x),
         half^#(x),
         inc^#(y))
  , if^#(true(), true(), x, y) -> c_11(logIter^#(x, y))
  , if^#(true(), false(), x, s(y)) -> c_12()
  , if^#(false(), b, x, y) -> c_13()
  , f^#() -> c_14()
  , f^#() -> c_15() }

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))
  , le^#(0(), y) -> c_4()
  , le^#(s(x), 0()) -> c_5()
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , inc^#(0()) -> c_7()
  , inc^#(s(x)) -> c_8(inc^#(x))
  , logarithm^#(x) -> c_9(logIter^#(x, 0()))
  , logIter^#(x, y) ->
    c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
         le^#(s(0()), x),
         le^#(s(s(0())), x),
         half^#(x),
         inc^#(y))
  , if^#(true(), true(), x, y) -> c_11(logIter^#(x, y))
  , if^#(true(), false(), x, s(y)) -> c_12()
  , if^#(false(), b, x, y) -> c_13()
  , f^#() -> c_14()
  , f^#() -> c_15() }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , logarithm(x) -> logIter(x, 0())
  , logIter(x, y) ->
    if(le(s(0()), x), le(s(s(0())), x), half(x), inc(y))
  , if(true(), true(), x, y) -> logIter(x, y)
  , if(true(), false(), x, s(y)) -> y
  , if(false(), b, x, y) -> logZeroError()
  , f() -> g()
  , f() -> h() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,4,5,7,12,13,14,15} by
applications of Pre({1,2,4,5,7,12,13,14,15}) = {3,6,8,10}. 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: le^#(0(), y) -> c_4()
    , 5: le^#(s(x), 0()) -> c_5()
    , 6: le^#(s(x), s(y)) -> c_6(le^#(x, y))
    , 7: inc^#(0()) -> c_7()
    , 8: inc^#(s(x)) -> c_8(inc^#(x))
    , 9: logarithm^#(x) -> c_9(logIter^#(x, 0()))
    , 10: logIter^#(x, y) ->
          c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
               le^#(s(0()), x),
               le^#(s(s(0())), x),
               half^#(x),
               inc^#(y))
    , 11: if^#(true(), true(), x, y) -> c_11(logIter^#(x, y))
    , 12: if^#(true(), false(), x, s(y)) -> c_12()
    , 13: if^#(false(), b, x, y) -> c_13()
    , 14: f^#() -> c_14()
    , 15: f^#() -> c_15() }

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

Strict DPs:
  { half^#(s(s(x))) -> c_3(half^#(x))
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , inc^#(s(x)) -> c_8(inc^#(x))
  , logarithm^#(x) -> c_9(logIter^#(x, 0()))
  , logIter^#(x, y) ->
    c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
         le^#(s(0()), x),
         le^#(s(s(0())), x),
         half^#(x),
         inc^#(y))
  , if^#(true(), true(), x, y) -> c_11(logIter^#(x, y)) }
Weak DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , le^#(0(), y) -> c_4()
  , le^#(s(x), 0()) -> c_5()
  , inc^#(0()) -> c_7()
  , if^#(true(), false(), x, s(y)) -> c_12()
  , if^#(false(), b, x, y) -> c_13()
  , f^#() -> c_14()
  , f^#() -> c_15() }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , logarithm(x) -> logIter(x, 0())
  , logIter(x, y) ->
    if(le(s(0()), x), le(s(s(0())), x), half(x), inc(y))
  , if(true(), true(), x, y) -> logIter(x, y)
  , if(true(), false(), x, s(y)) -> y
  , if(false(), b, x, y) -> logZeroError()
  , f() -> g()
  , f() -> h() }
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()
, le^#(0(), y) -> c_4()
, le^#(s(x), 0()) -> c_5()
, inc^#(0()) -> c_7()
, if^#(true(), false(), x, s(y)) -> c_12()
, if^#(false(), b, x, y) -> c_13()
, f^#() -> c_14()
, f^#() -> c_15() }

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

Strict DPs:
  { half^#(s(s(x))) -> c_3(half^#(x))
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , inc^#(s(x)) -> c_8(inc^#(x))
  , logarithm^#(x) -> c_9(logIter^#(x, 0()))
  , logIter^#(x, y) ->
    c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
         le^#(s(0()), x),
         le^#(s(s(0())), x),
         half^#(x),
         inc^#(y))
  , if^#(true(), true(), x, y) -> c_11(logIter^#(x, y)) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , logarithm(x) -> logIter(x, 0())
  , logIter(x, y) ->
    if(le(s(0()), x), le(s(s(0())), x), half(x), inc(y))
  , if(true(), true(), x, y) -> logIter(x, y)
  , if(true(), false(), x, s(y)) -> y
  , if(false(), b, x, y) -> logZeroError()
  , f() -> g()
  , f() -> h() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: half^#(s(s(x))) -> c_3(half^#(x))
     -->_1 half^#(s(s(x))) -> c_3(half^#(x)) :1
  
  2: le^#(s(x), s(y)) -> c_6(le^#(x, y))
     -->_1 le^#(s(x), s(y)) -> c_6(le^#(x, y)) :2
  
  3: inc^#(s(x)) -> c_8(inc^#(x))
     -->_1 inc^#(s(x)) -> c_8(inc^#(x)) :3
  
  4: logarithm^#(x) -> c_9(logIter^#(x, 0()))
     -->_1 logIter^#(x, y) ->
           c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
                le^#(s(0()), x),
                le^#(s(s(0())), x),
                half^#(x),
                inc^#(y)) :5
  
  5: logIter^#(x, y) ->
     c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
          le^#(s(0()), x),
          le^#(s(s(0())), x),
          half^#(x),
          inc^#(y))
     -->_1 if^#(true(), true(), x, y) -> c_11(logIter^#(x, y)) :6
     -->_5 inc^#(s(x)) -> c_8(inc^#(x)) :3
     -->_3 le^#(s(x), s(y)) -> c_6(le^#(x, y)) :2
     -->_2 le^#(s(x), s(y)) -> c_6(le^#(x, y)) :2
     -->_4 half^#(s(s(x))) -> c_3(half^#(x)) :1
  
  6: if^#(true(), true(), x, y) -> c_11(logIter^#(x, y))
     -->_1 logIter^#(x, y) ->
           c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
                le^#(s(0()), x),
                le^#(s(s(0())), x),
                half^#(x),
                inc^#(y)) :5
  

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).

  { logarithm^#(x) -> c_9(logIter^#(x, 0())) }


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

Strict DPs:
  { half^#(s(s(x))) -> c_3(half^#(x))
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , inc^#(s(x)) -> c_8(inc^#(x))
  , logIter^#(x, y) ->
    c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
         le^#(s(0()), x),
         le^#(s(s(0())), x),
         half^#(x),
         inc^#(y))
  , if^#(true(), true(), x, y) -> c_11(logIter^#(x, y)) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , logarithm(x) -> logIter(x, 0())
  , logIter(x, y) ->
    if(le(s(0()), x), le(s(s(0())), x), half(x), inc(y))
  , if(true(), true(), x, y) -> logIter(x, y)
  , if(true(), false(), x, s(y)) -> y
  , if(false(), b, x, y) -> logZeroError()
  , f() -> g()
  , f() -> h() }
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))
    , le(0(), y) -> true()
    , le(s(x), 0()) -> false()
    , le(s(x), s(y)) -> le(x, y)
    , inc(0()) -> s(0())
    , inc(s(x)) -> s(inc(x)) }

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

Strict DPs:
  { half^#(s(s(x))) -> c_3(half^#(x))
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , inc^#(s(x)) -> c_8(inc^#(x))
  , logIter^#(x, y) ->
    c_10(if^#(le(s(0()), x), le(s(s(0())), x), half(x), inc(y)),
         le^#(s(0()), x),
         le^#(s(s(0())), x),
         half^#(x),
         inc^#(y))
  , if^#(true(), true(), x, y) -> c_11(logIter^#(x, y)) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..