MAYBE

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

Strict Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(0(), y) -> 0()
  , minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
  , if_minus(true(), s(x), y) -> 0()
  , if_minus(false(), s(x), y) -> s(minus(x, y))
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(quot(x, s(s(0())))))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , minus^#(0(), y) -> c_4()
  , minus^#(s(x), y) ->
    c_5(if_minus^#(le(s(x), y), s(x), y), le^#(s(x), y))
  , if_minus^#(true(), s(x), y) -> c_6()
  , if_minus^#(false(), s(x), y) -> c_7(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_8()
  , quot^#(s(x), s(y)) ->
    c_9(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(s(0())) -> c_10()
  , log^#(s(s(x))) ->
    c_11(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) }

and mark the set of starting terms.

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

Strict DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , minus^#(0(), y) -> c_4()
  , minus^#(s(x), y) ->
    c_5(if_minus^#(le(s(x), y), s(x), y), le^#(s(x), y))
  , if_minus^#(true(), s(x), y) -> c_6()
  , if_minus^#(false(), s(x), y) -> c_7(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_8()
  , quot^#(s(x), s(y)) ->
    c_9(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(s(0())) -> c_10()
  , log^#(s(s(x))) ->
    c_11(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(0(), y) -> 0()
  , minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
  , if_minus(true(), s(x), y) -> 0()
  , if_minus(false(), s(x), y) -> s(minus(x, y))
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(quot(x, s(s(0())))))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: le^#(0(), y) -> c_1()
    , 2: le^#(s(x), 0()) -> c_2()
    , 3: le^#(s(x), s(y)) -> c_3(le^#(x, y))
    , 4: minus^#(0(), y) -> c_4()
    , 5: minus^#(s(x), y) ->
         c_5(if_minus^#(le(s(x), y), s(x), y), le^#(s(x), y))
    , 6: if_minus^#(true(), s(x), y) -> c_6()
    , 7: if_minus^#(false(), s(x), y) -> c_7(minus^#(x, y))
    , 8: quot^#(0(), s(y)) -> c_8()
    , 9: quot^#(s(x), s(y)) ->
         c_9(quot^#(minus(x, y), s(y)), minus^#(x, y))
    , 10: log^#(s(0())) -> c_10()
    , 11: log^#(s(s(x))) ->
          c_11(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , minus^#(s(x), y) ->
    c_5(if_minus^#(le(s(x), y), s(x), y), le^#(s(x), y))
  , if_minus^#(false(), s(x), y) -> c_7(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_9(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(s(s(x))) ->
    c_11(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) }
Weak DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , minus^#(0(), y) -> c_4()
  , if_minus^#(true(), s(x), y) -> c_6()
  , quot^#(0(), s(y)) -> c_8()
  , log^#(s(0())) -> c_10() }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(0(), y) -> 0()
  , minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
  , if_minus(true(), s(x), y) -> 0()
  , if_minus(false(), s(x), y) -> s(minus(x, y))
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(quot(x, s(s(0())))))) }
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^#(0(), y) -> c_1()
, le^#(s(x), 0()) -> c_2()
, minus^#(0(), y) -> c_4()
, if_minus^#(true(), s(x), y) -> c_6()
, quot^#(0(), s(y)) -> c_8()
, log^#(s(0())) -> c_10() }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , minus^#(s(x), y) ->
    c_5(if_minus^#(le(s(x), y), s(x), y), le^#(s(x), y))
  , if_minus^#(false(), s(x), y) -> c_7(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_9(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(s(s(x))) ->
    c_11(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(0(), y) -> 0()
  , minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
  , if_minus(true(), s(x), y) -> 0()
  , if_minus(false(), s(x), y) -> s(minus(x, y))
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(quot(x, s(s(0())))))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

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

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , minus^#(s(x), y) ->
    c_5(if_minus^#(le(s(x), y), s(x), y), le^#(s(x), y))
  , if_minus^#(false(), s(x), y) -> c_7(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_9(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(s(s(x))) ->
    c_11(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(0(), y) -> 0()
  , minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
  , if_minus(true(), s(x), y) -> 0()
  , if_minus(false(), s(x), y) -> s(minus(x, y))
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..