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)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , id(0()) -> 0()
  , id(s(x)) -> s(id(x))
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , mod(x, y) -> if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
  , if_mod(true(), b1, b2, x, y) -> 0()
  , if_mod(false(), b1, b2, x, y) -> if2(b1, b2, x, y)
  , if2(true(), b2, x, y) -> 0()
  , if2(false(), b2, x, y) -> if3(b2, x, y)
  , if3(true(), x, y) -> mod(minus(x, y), s(y))
  , if3(false(), x, y) -> x }
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))
  , zero^#(0()) -> c_4()
  , zero^#(s(x)) -> c_5()
  , id^#(0()) -> c_6()
  , id^#(s(x)) -> c_7(id^#(x))
  , minus^#(x, 0()) -> c_8()
  , minus^#(s(x), s(y)) -> c_9(minus^#(x, y))
  , mod^#(x, y) ->
    c_10(if_mod^#(zero(x), zero(y), le(y, x), id(x), id(y)),
         zero^#(x),
         zero^#(y),
         le^#(y, x),
         id^#(x),
         id^#(y))
  , if_mod^#(true(), b1, b2, x, y) -> c_11()
  , if_mod^#(false(), b1, b2, x, y) -> c_12(if2^#(b1, b2, x, y))
  , if2^#(true(), b2, x, y) -> c_13()
  , if2^#(false(), b2, x, y) -> c_14(if3^#(b2, x, y))
  , if3^#(true(), x, y) ->
    c_15(mod^#(minus(x, y), s(y)), minus^#(x, y))
  , if3^#(false(), x, y) -> c_16() }

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))
  , zero^#(0()) -> c_4()
  , zero^#(s(x)) -> c_5()
  , id^#(0()) -> c_6()
  , id^#(s(x)) -> c_7(id^#(x))
  , minus^#(x, 0()) -> c_8()
  , minus^#(s(x), s(y)) -> c_9(minus^#(x, y))
  , mod^#(x, y) ->
    c_10(if_mod^#(zero(x), zero(y), le(y, x), id(x), id(y)),
         zero^#(x),
         zero^#(y),
         le^#(y, x),
         id^#(x),
         id^#(y))
  , if_mod^#(true(), b1, b2, x, y) -> c_11()
  , if_mod^#(false(), b1, b2, x, y) -> c_12(if2^#(b1, b2, x, y))
  , if2^#(true(), b2, x, y) -> c_13()
  , if2^#(false(), b2, x, y) -> c_14(if3^#(b2, x, y))
  , if3^#(true(), x, y) ->
    c_15(mod^#(minus(x, y), s(y)), minus^#(x, y))
  , if3^#(false(), x, y) -> c_16() }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , id(0()) -> 0()
  , id(s(x)) -> s(id(x))
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , mod(x, y) -> if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
  , if_mod(true(), b1, b2, x, y) -> 0()
  , if_mod(false(), b1, b2, x, y) -> if2(b1, b2, x, y)
  , if2(true(), b2, x, y) -> 0()
  , if2(false(), b2, x, y) -> if3(b2, x, y)
  , if3(true(), x, y) -> mod(minus(x, y), s(y))
  , if3(false(), x, y) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,4,5,6,8,11,13,16} by
applications of Pre({1,2,4,5,6,8,11,13,16}) = {3,7,9,10,12,14,15}.
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: zero^#(0()) -> c_4()
    , 5: zero^#(s(x)) -> c_5()
    , 6: id^#(0()) -> c_6()
    , 7: id^#(s(x)) -> c_7(id^#(x))
    , 8: minus^#(x, 0()) -> c_8()
    , 9: minus^#(s(x), s(y)) -> c_9(minus^#(x, y))
    , 10: mod^#(x, y) ->
          c_10(if_mod^#(zero(x), zero(y), le(y, x), id(x), id(y)),
               zero^#(x),
               zero^#(y),
               le^#(y, x),
               id^#(x),
               id^#(y))
    , 11: if_mod^#(true(), b1, b2, x, y) -> c_11()
    , 12: if_mod^#(false(), b1, b2, x, y) -> c_12(if2^#(b1, b2, x, y))
    , 13: if2^#(true(), b2, x, y) -> c_13()
    , 14: if2^#(false(), b2, x, y) -> c_14(if3^#(b2, x, y))
    , 15: if3^#(true(), x, y) ->
          c_15(mod^#(minus(x, y), s(y)), minus^#(x, y))
    , 16: if3^#(false(), x, y) -> c_16() }

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))
  , id^#(s(x)) -> c_7(id^#(x))
  , minus^#(s(x), s(y)) -> c_9(minus^#(x, y))
  , mod^#(x, y) ->
    c_10(if_mod^#(zero(x), zero(y), le(y, x), id(x), id(y)),
         zero^#(x),
         zero^#(y),
         le^#(y, x),
         id^#(x),
         id^#(y))
  , if_mod^#(false(), b1, b2, x, y) -> c_12(if2^#(b1, b2, x, y))
  , if2^#(false(), b2, x, y) -> c_14(if3^#(b2, x, y))
  , if3^#(true(), x, y) ->
    c_15(mod^#(minus(x, y), s(y)), minus^#(x, y)) }
Weak DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , zero^#(0()) -> c_4()
  , zero^#(s(x)) -> c_5()
  , id^#(0()) -> c_6()
  , minus^#(x, 0()) -> c_8()
  , if_mod^#(true(), b1, b2, x, y) -> c_11()
  , if2^#(true(), b2, x, y) -> c_13()
  , if3^#(false(), x, y) -> c_16() }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , id(0()) -> 0()
  , id(s(x)) -> s(id(x))
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , mod(x, y) -> if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
  , if_mod(true(), b1, b2, x, y) -> 0()
  , if_mod(false(), b1, b2, x, y) -> if2(b1, b2, x, y)
  , if2(true(), b2, x, y) -> 0()
  , if2(false(), b2, x, y) -> if3(b2, x, y)
  , if3(true(), x, y) -> mod(minus(x, y), s(y))
  , if3(false(), x, y) -> x }
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()
, zero^#(0()) -> c_4()
, zero^#(s(x)) -> c_5()
, id^#(0()) -> c_6()
, minus^#(x, 0()) -> c_8()
, if_mod^#(true(), b1, b2, x, y) -> c_11()
, if2^#(true(), b2, x, y) -> c_13()
, if3^#(false(), x, y) -> c_16() }

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))
  , id^#(s(x)) -> c_7(id^#(x))
  , minus^#(s(x), s(y)) -> c_9(minus^#(x, y))
  , mod^#(x, y) ->
    c_10(if_mod^#(zero(x), zero(y), le(y, x), id(x), id(y)),
         zero^#(x),
         zero^#(y),
         le^#(y, x),
         id^#(x),
         id^#(y))
  , if_mod^#(false(), b1, b2, x, y) -> c_12(if2^#(b1, b2, x, y))
  , if2^#(false(), b2, x, y) -> c_14(if3^#(b2, x, y))
  , if3^#(true(), x, y) ->
    c_15(mod^#(minus(x, y), s(y)), minus^#(x, y)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , id(0()) -> 0()
  , id(s(x)) -> s(id(x))
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , mod(x, y) -> if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
  , if_mod(true(), b1, b2, x, y) -> 0()
  , if_mod(false(), b1, b2, x, y) -> if2(b1, b2, x, y)
  , if2(true(), b2, x, y) -> 0()
  , if2(false(), b2, x, y) -> if3(b2, x, y)
  , if3(true(), x, y) -> mod(minus(x, y), s(y))
  , if3(false(), x, y) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { mod^#(x, y) ->
    c_10(if_mod^#(zero(x), zero(y), le(y, x), id(x), id(y)),
         zero^#(x),
         zero^#(y),
         le^#(y, x),
         id^#(x),
         id^#(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))
  , id^#(s(x)) -> c_2(id^#(x))
  , minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
  , mod^#(x, y) ->
    c_4(if_mod^#(zero(x), zero(y), le(y, x), id(x), id(y)),
        le^#(y, x),
        id^#(x),
        id^#(y))
  , if_mod^#(false(), b1, b2, x, y) -> c_5(if2^#(b1, b2, x, y))
  , if2^#(false(), b2, x, y) -> c_6(if3^#(b2, x, y))
  , if3^#(true(), x, y) ->
    c_7(mod^#(minus(x, y), s(y)), minus^#(x, y)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , id(0()) -> 0()
  , id(s(x)) -> s(id(x))
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , mod(x, y) -> if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
  , if_mod(true(), b1, b2, x, y) -> 0()
  , if_mod(false(), b1, b2, x, y) -> if2(b1, b2, x, y)
  , if2(true(), b2, x, y) -> 0()
  , if2(false(), b2, x, y) -> if3(b2, x, y)
  , if3(true(), x, y) -> mod(minus(x, y), s(y))
  , if3(false(), x, y) -> x }
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)
    , zero(0()) -> true()
    , zero(s(x)) -> false()
    , id(0()) -> 0()
    , id(s(x)) -> s(id(x))
    , minus(x, 0()) -> x
    , minus(s(x), s(y)) -> minus(x, 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))
  , id^#(s(x)) -> c_2(id^#(x))
  , minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
  , mod^#(x, y) ->
    c_4(if_mod^#(zero(x), zero(y), le(y, x), id(x), id(y)),
        le^#(y, x),
        id^#(x),
        id^#(y))
  , if_mod^#(false(), b1, b2, x, y) -> c_5(if2^#(b1, b2, x, y))
  , if2^#(false(), b2, x, y) -> c_6(if3^#(b2, x, y))
  , if3^#(true(), x, y) ->
    c_7(mod^#(minus(x, y), s(y)), minus^#(x, y)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , zero(0()) -> true()
  , zero(s(x)) -> false()
  , id(0()) -> 0()
  , id(s(x)) -> s(id(x))
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..