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(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , mod(x, y) -> if_mod(isZero(y), le(y, x), x, y, minus(x, y))
  , if_mod(true(), b, x, y, z) -> divByZeroError()
  , if_mod(false(), true(), x, y, z) -> mod(z, y)
  , if_mod(false(), false(), x, y, z) -> 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))
  , minus^#(x, x) -> c_4()
  , minus^#(x, 0()) -> c_5()
  , minus^#(0(), x) -> c_6()
  , minus^#(s(x), s(y)) -> c_7(minus^#(x, y))
  , isZero^#(0()) -> c_8()
  , isZero^#(s(x)) -> c_9()
  , mod^#(x, y) ->
    c_10(if_mod^#(isZero(y), le(y, x), x, y, minus(x, y)),
         isZero^#(y),
         le^#(y, x),
         minus^#(x, y))
  , if_mod^#(true(), b, x, y, z) -> c_11()
  , if_mod^#(false(), true(), x, y, z) -> c_12(mod^#(z, y))
  , if_mod^#(false(), false(), x, y, z) -> c_13() }

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^#(x, x) -> c_4()
  , minus^#(x, 0()) -> c_5()
  , minus^#(0(), x) -> c_6()
  , minus^#(s(x), s(y)) -> c_7(minus^#(x, y))
  , isZero^#(0()) -> c_8()
  , isZero^#(s(x)) -> c_9()
  , mod^#(x, y) ->
    c_10(if_mod^#(isZero(y), le(y, x), x, y, minus(x, y)),
         isZero^#(y),
         le^#(y, x),
         minus^#(x, y))
  , if_mod^#(true(), b, x, y, z) -> c_11()
  , if_mod^#(false(), true(), x, y, z) -> c_12(mod^#(z, y))
  , if_mod^#(false(), false(), x, y, z) -> c_13() }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , mod(x, y) -> if_mod(isZero(y), le(y, x), x, y, minus(x, y))
  , if_mod(true(), b, x, y, z) -> divByZeroError()
  , if_mod(false(), true(), x, y, z) -> mod(z, y)
  , if_mod(false(), false(), x, y, z) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,4,5,6,8,9,11,13} by
applications of Pre({1,2,4,5,6,8,9,11,13}) = {3,7,10}. 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^#(x, x) -> c_4()
    , 5: minus^#(x, 0()) -> c_5()
    , 6: minus^#(0(), x) -> c_6()
    , 7: minus^#(s(x), s(y)) -> c_7(minus^#(x, y))
    , 8: isZero^#(0()) -> c_8()
    , 9: isZero^#(s(x)) -> c_9()
    , 10: mod^#(x, y) ->
          c_10(if_mod^#(isZero(y), le(y, x), x, y, minus(x, y)),
               isZero^#(y),
               le^#(y, x),
               minus^#(x, y))
    , 11: if_mod^#(true(), b, x, y, z) -> c_11()
    , 12: if_mod^#(false(), true(), x, y, z) -> c_12(mod^#(z, y))
    , 13: if_mod^#(false(), false(), x, y, z) -> c_13() }

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), s(y)) -> c_7(minus^#(x, y))
  , mod^#(x, y) ->
    c_10(if_mod^#(isZero(y), le(y, x), x, y, minus(x, y)),
         isZero^#(y),
         le^#(y, x),
         minus^#(x, y))
  , if_mod^#(false(), true(), x, y, z) -> c_12(mod^#(z, y)) }
Weak DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , minus^#(x, x) -> c_4()
  , minus^#(x, 0()) -> c_5()
  , minus^#(0(), x) -> c_6()
  , isZero^#(0()) -> c_8()
  , isZero^#(s(x)) -> c_9()
  , if_mod^#(true(), b, x, y, z) -> c_11()
  , if_mod^#(false(), false(), x, y, z) -> c_13() }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , mod(x, y) -> if_mod(isZero(y), le(y, x), x, y, minus(x, y))
  , if_mod(true(), b, x, y, z) -> divByZeroError()
  , if_mod(false(), true(), x, y, z) -> mod(z, y)
  , if_mod(false(), false(), x, y, z) -> 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()
, minus^#(x, x) -> c_4()
, minus^#(x, 0()) -> c_5()
, minus^#(0(), x) -> c_6()
, isZero^#(0()) -> c_8()
, isZero^#(s(x)) -> c_9()
, if_mod^#(true(), b, x, y, z) -> c_11()
, if_mod^#(false(), false(), x, y, z) -> c_13() }

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), s(y)) -> c_7(minus^#(x, y))
  , mod^#(x, y) ->
    c_10(if_mod^#(isZero(y), le(y, x), x, y, minus(x, y)),
         isZero^#(y),
         le^#(y, x),
         minus^#(x, y))
  , if_mod^#(false(), true(), x, y, z) -> c_12(mod^#(z, y)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , mod(x, y) -> if_mod(isZero(y), le(y, x), x, y, minus(x, y))
  , if_mod(true(), b, x, y, z) -> divByZeroError()
  , if_mod(false(), true(), x, y, z) -> mod(z, y)
  , if_mod(false(), false(), x, y, z) -> 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^#(isZero(y), le(y, x), x, y, minus(x, y)),
         isZero^#(y),
         le^#(y, x),
         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))
  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , mod^#(x, y) ->
    c_3(if_mod^#(isZero(y), le(y, x), x, y, minus(x, y)),
        le^#(y, x),
        minus^#(x, y))
  , if_mod^#(false(), true(), x, y, z) -> c_4(mod^#(z, y)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , mod(x, y) -> if_mod(isZero(y), le(y, x), x, y, minus(x, y))
  , if_mod(true(), b, x, y, z) -> divByZeroError()
  , if_mod(false(), true(), x, y, z) -> mod(z, y)
  , if_mod(false(), false(), x, y, z) -> 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)
    , minus(x, x) -> 0()
    , minus(x, 0()) -> x
    , minus(0(), x) -> 0()
    , minus(s(x), s(y)) -> minus(x, y)
    , isZero(0()) -> true()
    , isZero(s(x)) -> false() }

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))
  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , mod^#(x, y) ->
    c_3(if_mod^#(isZero(y), le(y, x), x, y, minus(x, y)),
        le^#(y, x),
        minus^#(x, y))
  , if_mod^#(false(), true(), x, y, z) -> c_4(mod^#(z, y)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , minus(s(x), s(y)) -> minus(x, y)
  , isZero(0()) -> true()
  , isZero(s(x)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..