YES(?,O(n^3))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , +(x, 0()) -> x
  , +(x, s(y)) -> s(+(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

We add following dependency tuples:

Strict DPs:
  { sum^#(0()) -> c_1()
  , sum^#(s(x)) -> c_2(+^#(sum(x), s(x)), sum^#(x))
  , +^#(x, 0()) -> c_3()
  , +^#(x, s(y)) -> c_4(+^#(x, y)) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict DPs:
  { sum^#(0()) -> c_1()
  , sum^#(s(x)) -> c_2(+^#(sum(x), s(x)), sum^#(x))
  , +^#(x, 0()) -> c_3()
  , +^#(x, s(y)) -> c_4(+^#(x, y)) }
Weak Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , +(x, 0()) -> x
  , +(x, s(y)) -> s(+(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

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

  DPs:
    { 1: sum^#(0()) -> c_1()
    , 2: sum^#(s(x)) -> c_2(+^#(sum(x), s(x)), sum^#(x))
    , 3: +^#(x, 0()) -> c_3()
    , 4: +^#(x, s(y)) -> c_4(+^#(x, y)) }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict DPs:
  { sum^#(s(x)) -> c_2(+^#(sum(x), s(x)), sum^#(x))
  , +^#(x, s(y)) -> c_4(+^#(x, y)) }
Weak DPs:
  { sum^#(0()) -> c_1()
  , +^#(x, 0()) -> c_3() }
Weak Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , +(x, 0()) -> x
  , +(x, s(y)) -> s(+(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ sum^#(0()) -> c_1()
, +^#(x, 0()) -> c_3() }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict DPs:
  { sum^#(s(x)) -> c_2(+^#(sum(x), s(x)), sum^#(x))
  , +^#(x, s(y)) -> c_4(+^#(x, y)) }
Weak Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , +(x, 0()) -> x
  , +(x, s(y)) -> s(+(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS)' as induced by the safe mapping

 safe(sum) = {1}, safe(0) = {}, safe(s) = {1}, safe(+) = {2},
 safe(sum^#) = {}, safe(c_2) = {}, safe(+^#) = {1}, safe(c_4) = {}

and precedence

 sum^# > sum, sum^# > +, sum^# > +^#, +^# > sum, +^# > +, sum ~ + .

Following symbols are considered recursive:

 {sum, +, sum^#, +^#}

The recursion depth is 3.

Further, following argument filtering is employed:

 pi(sum) = [], pi(0) = [], pi(s) = [1], pi(+) = [2],
 pi(sum^#) = [1], pi(c_2) = [1, 2], pi(+^#) = [2], pi(c_4) = [1]

Usable defined function symbols are a subset of:

 {sum^#, +^#}

For your convenience, here are the satisfied ordering constraints:

   pi(sum^#(s(x))) = sum^#(s(; x);)                      
                   > c_2(+^#(s(; x);),  sum^#(x;);)      
                   = pi(c_2(+^#(sum(x), s(x)), sum^#(x)))
                                                         
  pi(+^#(x, s(y))) = +^#(s(; y);)                        
                   > c_4(+^#(y;);)                       
                   = pi(c_4(+^#(x, y)))                  
                                                         

Hurray, we answered YES(?,O(n^3))