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)) -> +(sqr(s(x)), sum(x))
  , sum(s(x)) -> +(*(s(x), s(x)), sum(x))
  , sqr(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)) -> +(sqr(s(x)), sum(x))
  , sum(s(x)) -> +(*(s(x), s(x)), sum(x))
  , sqr(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(sqr) = {}, safe(*) = {1, 2}
  
  and precedence
  
   sum > sqr .
  
  Following symbols are considered recursive:
  
   {sum}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
       sum(0();) > 0()                                
                                                      
    sum(s(; x);) > +(; sqr(s(; x);),  sum(x;))        
                                                      
    sum(s(; x);) > +(; *(; s(; x),  s(; x)),  sum(x;))
                                                      
         sqr(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)) -> +(sqr(s(x)), sum(x))
  , sum(s(x)) -> +(*(s(x), s(x)), sum(x))
  , sqr(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))