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

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

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

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [+](x1, x2) = [1] x1 + [2] x2 + [0]
                                       
            [0] = [0]                  
                                       
        [s](x1) = [1] x1 + [1]         
  
  This order satisfies the following ordering constraints:
  
        [+(0(), y)] =  [2] y + [0]            
                    >= [1] y + [0]            
                    =  [y]                    
                                              
     [+(s(x), 0())] =  [1] x + [1]            
                    >= [1] x + [1]            
                    =  [s(x)]                 
                                              
    [+(s(x), s(y))] =  [2] y + [1] x + [3]    
                    >  [2] y + [1] x + [2]    
                    =  [s(+(s(x), +(y, 0())))]
                                              

We return to the main proof.

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

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

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { +(0(), y) -> y
  , +(s(x), 0()) -> s(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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [+](x1, x2) = [1] x1 + [2] x2 + [1]
                                       
            [0] = [0]                  
                                       
        [s](x1) = [1] x1 + [2]         
  
  This order satisfies the following ordering constraints:
  
        [+(0(), y)] =  [2] y + [1]            
                    >  [1] y + [0]            
                    =  [y]                    
                                              
     [+(s(x), 0())] =  [1] x + [3]            
                    >  [1] x + [2]            
                    =  [s(x)]                 
                                              
    [+(s(x), s(y))] =  [2] y + [1] x + [7]    
                    >= [2] y + [1] x + [7]    
                    =  [s(+(s(x), +(y, 0())))]
                                              

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

Empty rules are trivially bounded

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