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

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

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^1))

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , sum1(0()) -> 0()
  , sum1(s(x)) -> s(+(sum1(x), +(x, x))) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . 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,1-bounded)' as induced by the safe mapping
  
   safe(sum) = {}, safe(0) = {}, safe(s) = {1}, safe(+) = {1, 2},
   safe(sum1) = {}
  
  and precedence
  
   empty .
  
  Following symbols are considered recursive:
  
   {sum, sum1}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
        sum(0();) > 0()                            
                                                   
     sum(s(; x);) > +(; sum(x;),  s(; x))          
                                                   
       sum1(0();) > 0()                            
                                                   
    sum1(s(; x);) > s(; +(; sum1(x;),  +(; x,  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))
  , sum1(0()) -> 0()
  , sum1(s(x)) -> s(+(sum1(x), +(x, x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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