MAYBE

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

Strict Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , plus(x, 0()) -> x
  , plus(x, s(y)) -> s(plus(x, y))
  , quot(x, s(y)) -> help(x, s(y), 0())
  , help(x, s(y), c) -> if(lt(c, x), x, s(y), c)
  , if(false(), x, s(y), c) -> 0()
  , if(true(), x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { lt^#(x, 0()) -> c_1()
  , lt^#(0(), s(y)) -> c_2()
  , lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , plus^#(x, 0()) -> c_4()
  , plus^#(x, s(y)) -> c_5(plus^#(x, y))
  , quot^#(x, s(y)) -> c_6(help^#(x, s(y), 0()))
  , help^#(x, s(y), c) -> c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x))
  , if^#(false(), x, s(y), c) -> c_8()
  , if^#(true(), x, s(y), c) ->
    c_9(help^#(x, s(y), plus(c, s(y))), plus^#(c, s(y))) }

and mark the set of starting terms.

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

Strict DPs:
  { lt^#(x, 0()) -> c_1()
  , lt^#(0(), s(y)) -> c_2()
  , lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , plus^#(x, 0()) -> c_4()
  , plus^#(x, s(y)) -> c_5(plus^#(x, y))
  , quot^#(x, s(y)) -> c_6(help^#(x, s(y), 0()))
  , help^#(x, s(y), c) -> c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x))
  , if^#(false(), x, s(y), c) -> c_8()
  , if^#(true(), x, s(y), c) ->
    c_9(help^#(x, s(y), plus(c, s(y))), plus^#(c, s(y))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , plus(x, 0()) -> x
  , plus(x, s(y)) -> s(plus(x, y))
  , quot(x, s(y)) -> help(x, s(y), 0())
  , help(x, s(y), c) -> if(lt(c, x), x, s(y), c)
  , if(false(), x, s(y), c) -> 0()
  , if(true(), x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: lt^#(x, 0()) -> c_1()
    , 2: lt^#(0(), s(y)) -> c_2()
    , 3: lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
    , 4: plus^#(x, 0()) -> c_4()
    , 5: plus^#(x, s(y)) -> c_5(plus^#(x, y))
    , 6: quot^#(x, s(y)) -> c_6(help^#(x, s(y), 0()))
    , 7: help^#(x, s(y), c) ->
         c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x))
    , 8: if^#(false(), x, s(y), c) -> c_8()
    , 9: if^#(true(), x, s(y), c) ->
         c_9(help^#(x, s(y), plus(c, s(y))), plus^#(c, s(y))) }

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

Strict DPs:
  { lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , plus^#(x, s(y)) -> c_5(plus^#(x, y))
  , quot^#(x, s(y)) -> c_6(help^#(x, s(y), 0()))
  , help^#(x, s(y), c) -> c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x))
  , if^#(true(), x, s(y), c) ->
    c_9(help^#(x, s(y), plus(c, s(y))), plus^#(c, s(y))) }
Weak DPs:
  { lt^#(x, 0()) -> c_1()
  , lt^#(0(), s(y)) -> c_2()
  , plus^#(x, 0()) -> c_4()
  , if^#(false(), x, s(y), c) -> c_8() }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , plus(x, 0()) -> x
  , plus(x, s(y)) -> s(plus(x, y))
  , quot(x, s(y)) -> help(x, s(y), 0())
  , help(x, s(y), c) -> if(lt(c, x), x, s(y), c)
  , if(false(), x, s(y), c) -> 0()
  , if(true(), x, s(y), c) -> s(help(x, s(y), plus(c, s(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.

{ lt^#(x, 0()) -> c_1()
, lt^#(0(), s(y)) -> c_2()
, plus^#(x, 0()) -> c_4()
, if^#(false(), x, s(y), c) -> c_8() }

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

Strict DPs:
  { lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , plus^#(x, s(y)) -> c_5(plus^#(x, y))
  , quot^#(x, s(y)) -> c_6(help^#(x, s(y), 0()))
  , help^#(x, s(y), c) -> c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x))
  , if^#(true(), x, s(y), c) ->
    c_9(help^#(x, s(y), plus(c, s(y))), plus^#(c, s(y))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , plus(x, 0()) -> x
  , plus(x, s(y)) -> s(plus(x, y))
  , quot(x, s(y)) -> help(x, s(y), 0())
  , help(x, s(y), c) -> if(lt(c, x), x, s(y), c)
  , if(false(), x, s(y), c) -> 0()
  , if(true(), x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
     -->_1 lt^#(s(x), s(y)) -> c_3(lt^#(x, y)) :1
  
  2: plus^#(x, s(y)) -> c_5(plus^#(x, y))
     -->_1 plus^#(x, s(y)) -> c_5(plus^#(x, y)) :2
  
  3: quot^#(x, s(y)) -> c_6(help^#(x, s(y), 0()))
     -->_1 help^#(x, s(y), c) ->
           c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x)) :4
  
  4: help^#(x, s(y), c) ->
     c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x))
     -->_1 if^#(true(), x, s(y), c) ->
           c_9(help^#(x, s(y), plus(c, s(y))), plus^#(c, s(y))) :5
     -->_2 lt^#(s(x), s(y)) -> c_3(lt^#(x, y)) :1
  
  5: if^#(true(), x, s(y), c) ->
     c_9(help^#(x, s(y), plus(c, s(y))), plus^#(c, s(y)))
     -->_1 help^#(x, s(y), c) ->
           c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x)) :4
     -->_2 plus^#(x, s(y)) -> c_5(plus^#(x, y)) :2
  

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).

  { quot^#(x, s(y)) -> c_6(help^#(x, s(y), 0())) }


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

Strict DPs:
  { lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , plus^#(x, s(y)) -> c_5(plus^#(x, y))
  , help^#(x, s(y), c) -> c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x))
  , if^#(true(), x, s(y), c) ->
    c_9(help^#(x, s(y), plus(c, s(y))), plus^#(c, s(y))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , plus(x, 0()) -> x
  , plus(x, s(y)) -> s(plus(x, y))
  , quot(x, s(y)) -> help(x, s(y), 0())
  , help(x, s(y), c) -> if(lt(c, x), x, s(y), c)
  , if(false(), x, s(y), c) -> 0()
  , if(true(), x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { lt(x, 0()) -> false()
    , lt(0(), s(y)) -> true()
    , lt(s(x), s(y)) -> lt(x, y)
    , plus(x, 0()) -> x
    , plus(x, s(y)) -> s(plus(x, y)) }

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

Strict DPs:
  { lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , plus^#(x, s(y)) -> c_5(plus^#(x, y))
  , help^#(x, s(y), c) -> c_7(if^#(lt(c, x), x, s(y), c), lt^#(c, x))
  , if^#(true(), x, s(y), c) ->
    c_9(help^#(x, s(y), plus(c, s(y))), plus^#(c, s(y))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , plus(x, 0()) -> x
  , plus(x, s(y)) -> s(plus(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..