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))
  , +(x, 0()) -> x
  , +(x, s(y)) -> s(+(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'Small Polynomial Path Order (PS,2-bounded)'
to orient following rules strictly.

Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , +(x, 0()) -> x
  , +(x, s(y)) -> s(+(x, y)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,2-bounded)' as induced by the safe mapping
  
   safe(sum) = {}, safe(0) = {}, safe(s) = {1}, safe(+) = {1}
  
  and precedence
  
   sum > + .
  
  Following symbols are considered recursive:
  
   {sum, +}
  
  The recursion depth is 2.
  
  For your convenience, here are the satisfied ordering constraints:
  
       sum(0();) > 0()               
                                     
    sum(s(; x);) > +(s(; x); sum(x;))
                                     
       +(0(); x) > x                 
                                     
    +(s(; y); x) > s(; +(y; x))      
                                     

We return to the main proof.

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

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(1),O(1))

Empty rules are trivially bounded

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