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 'custom shape polynomial interpretation' to
orient following rules strictly.

Trs:
  { sum(0()) -> 0()
  , +(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:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
    [sum](x1) = 2 + x1 + 2*x1^2
                               
        [0]() = 1              
                               
      [s](x1) = 1 + x1         
                               
  [+](x1, x2) = x1 + 3*x2      
                               
  
  This order satisfies the following ordering constraints.
  
      [sum(0())] =  5                
                 >  1                
                 =  [0()]            
                                     
     [sum(s(x))] =  5 + 5*x + 2*x^2  
                 >= 5 + 4*x + 2*x^2  
                 =  [+(sum(x), s(x))]
                                     
     [+(x, 0())] =  x + 3            
                 >  x                
                 =  [x]              
                                     
    [+(x, s(y))] =  x + 3 + 3*y      
                 >  1 + x + 3*y      
                 =  [s(+(x, y))]     
                                     

We return to the main proof.

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

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

We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.

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

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:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
    [sum](x1) = 2 + 2*x1 + x1^2
                               
        [0]() = 1              
                               
      [s](x1) = 1 + x1         
                               
  [+](x1, x2) = x1 + 2*x2      
                               
  
  This order satisfies the following ordering constraints.
  
      [sum(0())] = 5                
                 > 1                
                 = [0()]            
                                    
     [sum(s(x))] = 5 + 4*x + x^2    
                 > 4 + 4*x + x^2    
                 = [+(sum(x), s(x))]
                                    
     [+(x, 0())] = x + 2            
                 > x                
                 = [x]              
                                    
    [+(x, s(y))] = x + 2 + 2*y      
                 > 1 + x + 2*y      
                 = [s(+(x, y))]     
                                    

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))