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)
  , inc(0()) -> 0()
  , inc(s(x)) -> s(inc(x))
  , minus(x, 0()) -> x
  , minus(0(), y) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , log(x) -> log2(x, 0())
  , log2(x, y) -> if(le(x, 0()), le(x, s(0())), x, inc(y))
  , if(true(), b, x, y) -> log_undefined()
  , if(false(), b, x, y) -> if2(b, x, y)
  , if2(true(), x, s(y)) -> y
  , if2(false(), x, y) -> log2(quot(x, s(s(0()))), y) }
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))
  , inc^#(0()) -> c_4()
  , inc^#(s(x)) -> c_5(inc^#(x))
  , minus^#(x, 0()) -> c_6()
  , minus^#(0(), y) -> c_7()
  , minus^#(s(x), s(y)) -> c_8(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_9()
  , quot^#(s(x), s(y)) ->
    c_10(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(x) -> c_11(log2^#(x, 0()))
  , log2^#(x, y) ->
    c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
         le^#(x, 0()),
         le^#(x, s(0())),
         inc^#(y))
  , if^#(true(), b, x, y) -> c_13()
  , if^#(false(), b, x, y) -> c_14(if2^#(b, x, y))
  , if2^#(true(), x, s(y)) -> c_15()
  , if2^#(false(), x, y) ->
    c_16(log2^#(quot(x, s(s(0()))), y), 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))
  , inc^#(0()) -> c_4()
  , inc^#(s(x)) -> c_5(inc^#(x))
  , minus^#(x, 0()) -> c_6()
  , minus^#(0(), y) -> c_7()
  , minus^#(s(x), s(y)) -> c_8(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_9()
  , quot^#(s(x), s(y)) ->
    c_10(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(x) -> c_11(log2^#(x, 0()))
  , log2^#(x, y) ->
    c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
         le^#(x, 0()),
         le^#(x, s(0())),
         inc^#(y))
  , if^#(true(), b, x, y) -> c_13()
  , if^#(false(), b, x, y) -> c_14(if2^#(b, x, y))
  , if2^#(true(), x, s(y)) -> c_15()
  , if2^#(false(), x, y) ->
    c_16(log2^#(quot(x, s(s(0()))), y), quot^#(x, s(s(0())))) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> 0()
  , inc(s(x)) -> s(inc(x))
  , minus(x, 0()) -> x
  , minus(0(), y) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , log(x) -> log2(x, 0())
  , log2(x, y) -> if(le(x, 0()), le(x, s(0())), x, inc(y))
  , if(true(), b, x, y) -> log_undefined()
  , if(false(), b, x, y) -> if2(b, x, y)
  , if2(true(), x, s(y)) -> y
  , if2(false(), x, y) -> log2(quot(x, s(s(0()))), y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,4,6,7,9,13,15} by
applications of Pre({1,2,4,6,7,9,13,15}) = {3,5,8,10,12,14,16}.
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: inc^#(0()) -> c_4()
    , 5: inc^#(s(x)) -> c_5(inc^#(x))
    , 6: minus^#(x, 0()) -> c_6()
    , 7: minus^#(0(), y) -> c_7()
    , 8: minus^#(s(x), s(y)) -> c_8(minus^#(x, y))
    , 9: quot^#(0(), s(y)) -> c_9()
    , 10: quot^#(s(x), s(y)) ->
          c_10(quot^#(minus(x, y), s(y)), minus^#(x, y))
    , 11: log^#(x) -> c_11(log2^#(x, 0()))
    , 12: log2^#(x, y) ->
          c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
               le^#(x, 0()),
               le^#(x, s(0())),
               inc^#(y))
    , 13: if^#(true(), b, x, y) -> c_13()
    , 14: if^#(false(), b, x, y) -> c_14(if2^#(b, x, y))
    , 15: if2^#(true(), x, s(y)) -> c_15()
    , 16: if2^#(false(), x, y) ->
          c_16(log2^#(quot(x, s(s(0()))), y), 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))
  , inc^#(s(x)) -> c_5(inc^#(x))
  , minus^#(s(x), s(y)) -> c_8(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_10(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(x) -> c_11(log2^#(x, 0()))
  , log2^#(x, y) ->
    c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
         le^#(x, 0()),
         le^#(x, s(0())),
         inc^#(y))
  , if^#(false(), b, x, y) -> c_14(if2^#(b, x, y))
  , if2^#(false(), x, y) ->
    c_16(log2^#(quot(x, s(s(0()))), y), quot^#(x, s(s(0())))) }
Weak DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , inc^#(0()) -> c_4()
  , minus^#(x, 0()) -> c_6()
  , minus^#(0(), y) -> c_7()
  , quot^#(0(), s(y)) -> c_9()
  , if^#(true(), b, x, y) -> c_13()
  , if2^#(true(), x, s(y)) -> c_15() }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> 0()
  , inc(s(x)) -> s(inc(x))
  , minus(x, 0()) -> x
  , minus(0(), y) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , log(x) -> log2(x, 0())
  , log2(x, y) -> if(le(x, 0()), le(x, s(0())), x, inc(y))
  , if(true(), b, x, y) -> log_undefined()
  , if(false(), b, x, y) -> if2(b, x, y)
  , if2(true(), x, s(y)) -> y
  , if2(false(), x, y) -> log2(quot(x, s(s(0()))), y) }
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()
, inc^#(0()) -> c_4()
, minus^#(x, 0()) -> c_6()
, minus^#(0(), y) -> c_7()
, quot^#(0(), s(y)) -> c_9()
, if^#(true(), b, x, y) -> c_13()
, if2^#(true(), x, s(y)) -> c_15() }

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))
  , inc^#(s(x)) -> c_5(inc^#(x))
  , minus^#(s(x), s(y)) -> c_8(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_10(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(x) -> c_11(log2^#(x, 0()))
  , log2^#(x, y) ->
    c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
         le^#(x, 0()),
         le^#(x, s(0())),
         inc^#(y))
  , if^#(false(), b, x, y) -> c_14(if2^#(b, x, y))
  , if2^#(false(), x, y) ->
    c_16(log2^#(quot(x, s(s(0()))), y), quot^#(x, s(s(0())))) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> 0()
  , inc(s(x)) -> s(inc(x))
  , minus(x, 0()) -> x
  , minus(0(), y) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , log(x) -> log2(x, 0())
  , log2(x, y) -> if(le(x, 0()), le(x, s(0())), x, inc(y))
  , if(true(), b, x, y) -> log_undefined()
  , if(false(), b, x, y) -> if2(b, x, y)
  , if2(true(), x, s(y)) -> y
  , if2(false(), x, y) -> log2(quot(x, s(s(0()))), y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { log2^#(x, y) ->
    c_12(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
         le^#(x, 0()),
         le^#(x, s(0())),
         inc^#(y)) }

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))
  , inc^#(s(x)) -> c_2(inc^#(x))
  , minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(x) -> c_5(log2^#(x, 0()))
  , log2^#(x, y) ->
    c_6(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
        le^#(x, s(0())),
        inc^#(y))
  , if^#(false(), b, x, y) -> c_7(if2^#(b, x, y))
  , if2^#(false(), x, y) ->
    c_8(log2^#(quot(x, s(s(0()))), y), quot^#(x, s(s(0())))) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> 0()
  , inc(s(x)) -> s(inc(x))
  , minus(x, 0()) -> x
  , minus(0(), y) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
  , log(x) -> log2(x, 0())
  , log2(x, y) -> if(le(x, 0()), le(x, s(0())), x, inc(y))
  , if(true(), b, x, y) -> log_undefined()
  , if(false(), b, x, y) -> if2(b, x, y)
  , if2(true(), x, s(y)) -> y
  , if2(false(), x, y) -> log2(quot(x, s(s(0()))), y) }
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)
    , inc(0()) -> 0()
    , inc(s(x)) -> s(inc(x))
    , minus(x, 0()) -> x
    , minus(0(), y) -> 0()
    , minus(s(x), s(y)) -> 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_1(le^#(x, y))
  , inc^#(s(x)) -> c_2(inc^#(x))
  , minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log^#(x) -> c_5(log2^#(x, 0()))
  , log2^#(x, y) ->
    c_6(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
        le^#(x, s(0())),
        inc^#(y))
  , if^#(false(), b, x, y) -> c_7(if2^#(b, x, y))
  , if2^#(false(), x, y) ->
    c_8(log2^#(quot(x, s(s(0()))), y), quot^#(x, s(s(0())))) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> 0()
  , inc(s(x)) -> s(inc(x))
  , minus(x, 0()) -> x
  , minus(0(), y) -> 0()
  , minus(s(x), s(y)) -> 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

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: inc^#(s(x)) -> c_2(inc^#(x))
     -->_1 inc^#(s(x)) -> c_2(inc^#(x)) :2
  
  3: minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
     -->_1 minus^#(s(x), s(y)) -> c_3(minus^#(x, y)) :3
  
  4: quot^#(s(x), s(y)) ->
     c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
     -->_1 quot^#(s(x), s(y)) ->
           c_4(quot^#(minus(x, y), s(y)), minus^#(x, y)) :4
     -->_2 minus^#(s(x), s(y)) -> c_3(minus^#(x, y)) :3
  
  5: log^#(x) -> c_5(log2^#(x, 0()))
     -->_1 log2^#(x, y) ->
           c_6(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
               le^#(x, s(0())),
               inc^#(y)) :6
  
  6: log2^#(x, y) ->
     c_6(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
         le^#(x, s(0())),
         inc^#(y))
     -->_1 if^#(false(), b, x, y) -> c_7(if2^#(b, x, y)) :7
     -->_3 inc^#(s(x)) -> c_2(inc^#(x)) :2
     -->_2 le^#(s(x), s(y)) -> c_1(le^#(x, y)) :1
  
  7: if^#(false(), b, x, y) -> c_7(if2^#(b, x, y))
     -->_1 if2^#(false(), x, y) ->
           c_8(log2^#(quot(x, s(s(0()))), y), quot^#(x, s(s(0())))) :8
  
  8: if2^#(false(), x, y) ->
     c_8(log2^#(quot(x, s(s(0()))), y), quot^#(x, s(s(0()))))
     -->_1 log2^#(x, y) ->
           c_6(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
               le^#(x, s(0())),
               inc^#(y)) :6
     -->_2 quot^#(s(x), s(y)) ->
           c_4(quot^#(minus(x, y), s(y)), minus^#(x, y)) :4
  

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^#(x) -> c_5(log2^#(x, 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))
  , inc^#(s(x)) -> c_2(inc^#(x))
  , minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
  , quot^#(s(x), s(y)) ->
    c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
  , log2^#(x, y) ->
    c_6(if^#(le(x, 0()), le(x, s(0())), x, inc(y)),
        le^#(x, s(0())),
        inc^#(y))
  , if^#(false(), b, x, y) -> c_7(if2^#(b, x, y))
  , if2^#(false(), x, y) ->
    c_8(log2^#(quot(x, s(s(0()))), y), quot^#(x, s(s(0())))) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , inc(0()) -> 0()
  , inc(s(x)) -> s(inc(x))
  , minus(x, 0()) -> x
  , minus(0(), y) -> 0()
  , minus(s(x), s(y)) -> 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..