MAYBE

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

Strict Trs:
  { minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(s(x), s(y)) -> gt(x, y)
  , gt(s(x), 0()) -> true()
  , gt(0(), y) -> false()
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(x)) -> false()
  , div(x, y) -> if1(ge(x, y), x, y)
  , if1(true(), x, y) -> if2(gt(y, 0()), x, y)
  , if1(false(), x, y) -> 0()
  , if2(true(), x, y) -> s(div(minus(x, y), y))
  , if2(false(), x, y) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { minus^#(s(x), y) -> c_1(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_2(minus^#(x, y))
  , if^#(false(), x, y) -> c_3()
  , gt^#(s(x), s(y)) -> c_4(gt^#(x, y))
  , gt^#(s(x), 0()) -> c_5()
  , gt^#(0(), y) -> c_6()
  , ge^#(x, 0()) -> c_7()
  , ge^#(s(x), s(y)) -> c_8(ge^#(x, y))
  , ge^#(0(), s(x)) -> c_9()
  , div^#(x, y) -> c_10(if1^#(ge(x, y), x, y), ge^#(x, y))
  , if1^#(true(), x, y) ->
    c_11(if2^#(gt(y, 0()), x, y), gt^#(y, 0()))
  , if1^#(false(), x, y) -> c_12()
  , if2^#(true(), x, y) -> c_13(div^#(minus(x, y), y), minus^#(x, y))
  , if2^#(false(), x, y) -> c_14() }

and mark the set of starting terms.

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

Strict DPs:
  { minus^#(s(x), y) -> c_1(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_2(minus^#(x, y))
  , if^#(false(), x, y) -> c_3()
  , gt^#(s(x), s(y)) -> c_4(gt^#(x, y))
  , gt^#(s(x), 0()) -> c_5()
  , gt^#(0(), y) -> c_6()
  , ge^#(x, 0()) -> c_7()
  , ge^#(s(x), s(y)) -> c_8(ge^#(x, y))
  , ge^#(0(), s(x)) -> c_9()
  , div^#(x, y) -> c_10(if1^#(ge(x, y), x, y), ge^#(x, y))
  , if1^#(true(), x, y) ->
    c_11(if2^#(gt(y, 0()), x, y), gt^#(y, 0()))
  , if1^#(false(), x, y) -> c_12()
  , if2^#(true(), x, y) -> c_13(div^#(minus(x, y), y), minus^#(x, y))
  , if2^#(false(), x, y) -> c_14() }
Weak Trs:
  { minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(s(x), s(y)) -> gt(x, y)
  , gt(s(x), 0()) -> true()
  , gt(0(), y) -> false()
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(x)) -> false()
  , div(x, y) -> if1(ge(x, y), x, y)
  , if1(true(), x, y) -> if2(gt(y, 0()), x, y)
  , if1(false(), x, y) -> 0()
  , if2(true(), x, y) -> s(div(minus(x, y), y))
  , if2(false(), x, y) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {3,5,6,7,9,12,14} by
applications of Pre({3,5,6,7,9,12,14}) = {1,4,8,10,11}. Here rules
are labeled as follows:

  DPs:
    { 1: minus^#(s(x), y) ->
         c_1(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
    , 2: if^#(true(), x, y) -> c_2(minus^#(x, y))
    , 3: if^#(false(), x, y) -> c_3()
    , 4: gt^#(s(x), s(y)) -> c_4(gt^#(x, y))
    , 5: gt^#(s(x), 0()) -> c_5()
    , 6: gt^#(0(), y) -> c_6()
    , 7: ge^#(x, 0()) -> c_7()
    , 8: ge^#(s(x), s(y)) -> c_8(ge^#(x, y))
    , 9: ge^#(0(), s(x)) -> c_9()
    , 10: div^#(x, y) -> c_10(if1^#(ge(x, y), x, y), ge^#(x, y))
    , 11: if1^#(true(), x, y) ->
          c_11(if2^#(gt(y, 0()), x, y), gt^#(y, 0()))
    , 12: if1^#(false(), x, y) -> c_12()
    , 13: if2^#(true(), x, y) ->
          c_13(div^#(minus(x, y), y), minus^#(x, y))
    , 14: if2^#(false(), x, y) -> c_14() }

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

Strict DPs:
  { minus^#(s(x), y) -> c_1(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_2(minus^#(x, y))
  , gt^#(s(x), s(y)) -> c_4(gt^#(x, y))
  , ge^#(s(x), s(y)) -> c_8(ge^#(x, y))
  , div^#(x, y) -> c_10(if1^#(ge(x, y), x, y), ge^#(x, y))
  , if1^#(true(), x, y) ->
    c_11(if2^#(gt(y, 0()), x, y), gt^#(y, 0()))
  , if2^#(true(), x, y) ->
    c_13(div^#(minus(x, y), y), minus^#(x, y)) }
Weak DPs:
  { if^#(false(), x, y) -> c_3()
  , gt^#(s(x), 0()) -> c_5()
  , gt^#(0(), y) -> c_6()
  , ge^#(x, 0()) -> c_7()
  , ge^#(0(), s(x)) -> c_9()
  , if1^#(false(), x, y) -> c_12()
  , if2^#(false(), x, y) -> c_14() }
Weak Trs:
  { minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(s(x), s(y)) -> gt(x, y)
  , gt(s(x), 0()) -> true()
  , gt(0(), y) -> false()
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(x)) -> false()
  , div(x, y) -> if1(ge(x, y), x, y)
  , if1(true(), x, y) -> if2(gt(y, 0()), x, y)
  , if1(false(), x, y) -> 0()
  , if2(true(), x, y) -> s(div(minus(x, y), y))
  , if2(false(), x, y) -> 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.

{ if^#(false(), x, y) -> c_3()
, gt^#(s(x), 0()) -> c_5()
, gt^#(0(), y) -> c_6()
, ge^#(x, 0()) -> c_7()
, ge^#(0(), s(x)) -> c_9()
, if1^#(false(), x, y) -> c_12()
, if2^#(false(), x, y) -> c_14() }

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

Strict DPs:
  { minus^#(s(x), y) -> c_1(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_2(minus^#(x, y))
  , gt^#(s(x), s(y)) -> c_4(gt^#(x, y))
  , ge^#(s(x), s(y)) -> c_8(ge^#(x, y))
  , div^#(x, y) -> c_10(if1^#(ge(x, y), x, y), ge^#(x, y))
  , if1^#(true(), x, y) ->
    c_11(if2^#(gt(y, 0()), x, y), gt^#(y, 0()))
  , if2^#(true(), x, y) ->
    c_13(div^#(minus(x, y), y), minus^#(x, y)) }
Weak Trs:
  { minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(s(x), s(y)) -> gt(x, y)
  , gt(s(x), 0()) -> true()
  , gt(0(), y) -> false()
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(x)) -> false()
  , div(x, y) -> if1(ge(x, y), x, y)
  , if1(true(), x, y) -> if2(gt(y, 0()), x, y)
  , if1(false(), x, y) -> 0()
  , if2(true(), x, y) -> s(div(minus(x, y), y))
  , if2(false(), x, y) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { if1^#(true(), x, y) ->
    c_11(if2^#(gt(y, 0()), x, y), gt^#(y, 0())) }

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

Strict DPs:
  { minus^#(s(x), y) -> c_1(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_2(minus^#(x, y))
  , gt^#(s(x), s(y)) -> c_3(gt^#(x, y))
  , ge^#(s(x), s(y)) -> c_4(ge^#(x, y))
  , div^#(x, y) -> c_5(if1^#(ge(x, y), x, y), ge^#(x, y))
  , if1^#(true(), x, y) -> c_6(if2^#(gt(y, 0()), x, y))
  , if2^#(true(), x, y) ->
    c_7(div^#(minus(x, y), y), minus^#(x, y)) }
Weak Trs:
  { minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(s(x), s(y)) -> gt(x, y)
  , gt(s(x), 0()) -> true()
  , gt(0(), y) -> false()
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(x)) -> false()
  , div(x, y) -> if1(ge(x, y), x, y)
  , if1(true(), x, y) -> if2(gt(y, 0()), x, y)
  , if1(false(), x, y) -> 0()
  , if2(true(), x, y) -> s(div(minus(x, y), y))
  , if2(false(), x, y) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { minus(s(x), y) -> if(gt(s(x), y), x, y)
    , if(true(), x, y) -> s(minus(x, y))
    , if(false(), x, y) -> 0()
    , gt(s(x), s(y)) -> gt(x, y)
    , gt(s(x), 0()) -> true()
    , gt(0(), y) -> false()
    , ge(x, 0()) -> true()
    , ge(s(x), s(y)) -> ge(x, y)
    , ge(0(), s(x)) -> false() }

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

Strict DPs:
  { minus^#(s(x), y) -> c_1(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_2(minus^#(x, y))
  , gt^#(s(x), s(y)) -> c_3(gt^#(x, y))
  , ge^#(s(x), s(y)) -> c_4(ge^#(x, y))
  , div^#(x, y) -> c_5(if1^#(ge(x, y), x, y), ge^#(x, y))
  , if1^#(true(), x, y) -> c_6(if2^#(gt(y, 0()), x, y))
  , if2^#(true(), x, y) ->
    c_7(div^#(minus(x, y), y), minus^#(x, y)) }
Weak Trs:
  { minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(s(x), s(y)) -> gt(x, y)
  , gt(s(x), 0()) -> true()
  , gt(0(), y) -> false()
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(x)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..