MAYBE

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

Strict Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , double(s(x)) -> s(s(double(x)))
  , double(0()) -> 0()
  , log(s(x)) -> loop(s(x), s(0()), 0())
  , log(0()) -> logError()
  , loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
  , if(false(), x, y, z) -> loop(x, double(y), s(z))
  , if(true(), x, y, z) -> z }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { le^#(s(x), s(y)) -> c_1(le^#(x, y))
  , le^#(s(x), 0()) -> c_2()
  , le^#(0(), y) -> c_3()
  , double^#(s(x)) -> c_4(double^#(x))
  , double^#(0()) -> c_5()
  , log^#(s(x)) -> c_6(loop^#(s(x), s(0()), 0()))
  , log^#(0()) -> c_7()
  , loop^#(x, s(y), z) ->
    c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y)))
  , if^#(false(), x, y, z) ->
    c_9(loop^#(x, double(y), s(z)), double^#(y))
  , if^#(true(), x, y, z) -> c_10() }

and mark the set of starting terms.

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_1(le^#(x, y))
  , le^#(s(x), 0()) -> c_2()
  , le^#(0(), y) -> c_3()
  , double^#(s(x)) -> c_4(double^#(x))
  , double^#(0()) -> c_5()
  , log^#(s(x)) -> c_6(loop^#(s(x), s(0()), 0()))
  , log^#(0()) -> c_7()
  , loop^#(x, s(y), z) ->
    c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y)))
  , if^#(false(), x, y, z) ->
    c_9(loop^#(x, double(y), s(z)), double^#(y))
  , if^#(true(), x, y, z) -> c_10() }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , double(s(x)) -> s(s(double(x)))
  , double(0()) -> 0()
  , log(s(x)) -> loop(s(x), s(0()), 0())
  , log(0()) -> logError()
  , loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
  , if(false(), x, y, z) -> loop(x, double(y), s(z))
  , if(true(), x, y, z) -> z }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: le^#(s(x), s(y)) -> c_1(le^#(x, y))
    , 2: le^#(s(x), 0()) -> c_2()
    , 3: le^#(0(), y) -> c_3()
    , 4: double^#(s(x)) -> c_4(double^#(x))
    , 5: double^#(0()) -> c_5()
    , 6: log^#(s(x)) -> c_6(loop^#(s(x), s(0()), 0()))
    , 7: log^#(0()) -> c_7()
    , 8: loop^#(x, s(y), z) ->
         c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y)))
    , 9: if^#(false(), x, y, z) ->
         c_9(loop^#(x, double(y), s(z)), double^#(y))
    , 10: if^#(true(), x, y, z) -> c_10() }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_1(le^#(x, y))
  , double^#(s(x)) -> c_4(double^#(x))
  , log^#(s(x)) -> c_6(loop^#(s(x), s(0()), 0()))
  , loop^#(x, s(y), z) ->
    c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y)))
  , if^#(false(), x, y, z) ->
    c_9(loop^#(x, double(y), s(z)), double^#(y)) }
Weak DPs:
  { le^#(s(x), 0()) -> c_2()
  , le^#(0(), y) -> c_3()
  , double^#(0()) -> c_5()
  , log^#(0()) -> c_7()
  , if^#(true(), x, y, z) -> c_10() }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , double(s(x)) -> s(s(double(x)))
  , double(0()) -> 0()
  , log(s(x)) -> loop(s(x), s(0()), 0())
  , log(0()) -> logError()
  , loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
  , if(false(), x, y, z) -> loop(x, double(y), s(z))
  , if(true(), x, y, z) -> z }
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.

{ le^#(s(x), 0()) -> c_2()
, le^#(0(), y) -> c_3()
, double^#(0()) -> c_5()
, log^#(0()) -> c_7()
, if^#(true(), x, y, z) -> c_10() }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_1(le^#(x, y))
  , double^#(s(x)) -> c_4(double^#(x))
  , log^#(s(x)) -> c_6(loop^#(s(x), s(0()), 0()))
  , loop^#(x, s(y), z) ->
    c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y)))
  , if^#(false(), x, y, z) ->
    c_9(loop^#(x, double(y), s(z)), double^#(y)) }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , double(s(x)) -> s(s(double(x)))
  , double(0()) -> 0()
  , log(s(x)) -> loop(s(x), s(0()), 0())
  , log(0()) -> logError()
  , loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
  , if(false(), x, y, z) -> loop(x, double(y), s(z))
  , if(true(), x, y, z) -> z }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: le^#(s(x), s(y)) -> c_1(le^#(x, y))
     -->_1 le^#(s(x), s(y)) -> c_1(le^#(x, y)) :1
  
  2: double^#(s(x)) -> c_4(double^#(x))
     -->_1 double^#(s(x)) -> c_4(double^#(x)) :2
  
  3: log^#(s(x)) -> c_6(loop^#(s(x), s(0()), 0()))
     -->_1 loop^#(x, s(y), z) ->
           c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y))) :4
  
  4: loop^#(x, s(y), z) ->
     c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y)))
     -->_1 if^#(false(), x, y, z) ->
           c_9(loop^#(x, double(y), s(z)), double^#(y)) :5
     -->_2 le^#(s(x), s(y)) -> c_1(le^#(x, y)) :1
  
  5: if^#(false(), x, y, z) ->
     c_9(loop^#(x, double(y), s(z)), double^#(y))
     -->_1 loop^#(x, s(y), z) ->
           c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y))) :4
     -->_2 double^#(s(x)) -> c_4(double^#(x)) :2
  

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).

  { log^#(s(x)) -> c_6(loop^#(s(x), s(0()), 0())) }


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

Strict DPs:
  { le^#(s(x), s(y)) -> c_1(le^#(x, y))
  , double^#(s(x)) -> c_4(double^#(x))
  , loop^#(x, s(y), z) ->
    c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y)))
  , if^#(false(), x, y, z) ->
    c_9(loop^#(x, double(y), s(z)), double^#(y)) }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , double(s(x)) -> s(s(double(x)))
  , double(0()) -> 0()
  , log(s(x)) -> loop(s(x), s(0()), 0())
  , log(0()) -> logError()
  , loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
  , if(false(), x, y, z) -> loop(x, double(y), s(z))
  , if(true(), x, y, z) -> z }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { le(s(x), s(y)) -> le(x, y)
    , le(s(x), 0()) -> false()
    , le(0(), y) -> true()
    , double(s(x)) -> s(s(double(x)))
    , double(0()) -> 0() }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_1(le^#(x, y))
  , double^#(s(x)) -> c_4(double^#(x))
  , loop^#(x, s(y), z) ->
    c_8(if^#(le(x, s(y)), x, s(y), z), le^#(x, s(y)))
  , if^#(false(), x, y, z) ->
    c_9(loop^#(x, double(y), s(z)), double^#(y)) }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , double(s(x)) -> s(s(double(x)))
  , double(0()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..