YES(O(1),O(n^2))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , sum1(0()) -> 0()
  , sum1(s(x)) -> s(+(sum1(x), +(x, x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We add following weak dependency pairs:

Strict DPs:
  { sum^#(0()) -> c_1()
  , sum^#(s(x)) -> c_2(sum^#(x))
  , sum1^#(0()) -> c_3()
  , sum1^#(s(x)) -> c_4(sum1^#(x)) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs:
  { sum^#(0()) -> c_1()
  , sum^#(s(x)) -> c_2(sum^#(x))
  , sum1^#(0()) -> c_3()
  , sum1^#(s(x)) -> c_4(sum1^#(x)) }
Strict Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , sum1(0()) -> 0()
  , sum1(s(x)) -> s(+(sum1(x), +(x, x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

No rule is usable, rules are removed from the input problem.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs:
  { sum^#(0()) -> c_1()
  , sum^#(s(x)) -> c_2(sum^#(x))
  , sum1^#(0()) -> c_3()
  , sum1^#(s(x)) -> c_4(sum1^#(x)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

The weightgap principle applies (using the following constant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(c_2) = {1}, Uargs(c_4) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

           [0] = [0]         
                             
       [s](x1) = [1] x1 + [0]
                             
   [sum^#](x1) = [1]         
                             
         [c_1] = [0]         
                             
     [c_2](x1) = [1] x1 + [1]
                             
  [sum1^#](x1) = [0]         
                             
         [c_3] = [1]         
                             
     [c_4](x1) = [1] x1 + [2]

This order satisfies following ordering constraints:


Further, it can be verified that all rules not oriented are covered by the weightgap condition.

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^2)).

Strict DPs:
  { sum^#(s(x)) -> c_2(sum^#(x))
  , sum1^#(0()) -> c_3()
  , sum1^#(s(x)) -> c_4(sum1^#(x)) }
Weak DPs: { sum^#(0()) -> c_1() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

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

  DPs:
    { 1: sum^#(s(x)) -> c_2(sum^#(x))
    , 2: sum1^#(0()) -> c_3()
    , 3: sum1^#(s(x)) -> c_4(sum1^#(x))
    , 4: sum^#(0()) -> c_1() }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^2)).

Strict DPs:
  { sum^#(s(x)) -> c_2(sum^#(x))
  , sum1^#(s(x)) -> c_4(sum1^#(x)) }
Weak DPs:
  { sum^#(0()) -> c_1()
  , sum1^#(0()) -> c_3() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

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

{ sum^#(0()) -> c_1()
, sum1^#(0()) -> c_3() }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^2)).

Strict DPs:
  { sum^#(s(x)) -> c_2(sum^#(x))
  , sum1^#(s(x)) -> c_4(sum1^#(x)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

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

 safe(s) = {1}, safe(sum^#) = {}, safe(c_2) = {}, safe(sum1^#) = {},
 safe(c_4) = {}

and precedence

 sum1^# > sum^# .

Following symbols are considered recursive:

 {sum^#, sum1^#}

The recursion depth is 2.

Further, following argument filtering is employed:

 pi(s) = [1], pi(sum^#) = [1], pi(c_2) = [1], pi(sum1^#) = [1],
 pi(c_4) = [1]

Usable defined function symbols are a subset of:

 {sum^#, sum1^#}

For your convenience, here are the satisfied ordering constraints:

   pi(sum^#(s(x))) = sum^#(s(; x);)    
                   > c_2(sum^#(x;);)   
                   = pi(c_2(sum^#(x))) 
                                       
  pi(sum1^#(s(x))) = sum1^#(s(; x);)   
                   > c_4(sum1^#(x;);)  
                   = pi(c_4(sum1^#(x)))
                                       

Hurray, we answered YES(O(1),O(n^2))