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), y) -> +(x, s(y))
  , +(s(x), y) -> s(+(x, y)) }
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 }

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) = [3] x1 + [3] x2 + [0]
                                       
            [0] = [1]                  
                                       
        [s](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
     [+(0(), y)] =  [3] y + [3]        
                 >  [1] y + [0]        
                 =  [y]                
                                       
    [+(s(x), y)] =  [3] y + [3] x + [0]
                 >= [3] y + [3] x + [0]
                 =  [+(x, s(y))]       
                                       
    [+(s(x), y)] =  [3] y + [3] x + [0]
                 >= [3] y + [3] x + [0]
                 =  [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^1)).

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

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) = [2] x1 + [2] x2 + [0]
                                       
            [0] = [3]                  
                                       
        [s](x1) = [1] x1 + [2]         
  
  This order satisfies the following ordering constraints:
  
     [+(0(), y)] =  [2] y + [6]        
                 >  [1] y + [0]        
                 =  [y]                
                                       
    [+(s(x), y)] =  [2] y + [2] x + [4]
                 >= [2] y + [2] x + [4]
                 =  [+(x, s(y))]       
                                       
    [+(s(x), y)] =  [2] y + [2] x + [4]
                 >  [2] y + [2] x + [2]
                 =  [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^1)).

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

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) = [3] x1 + [2] x2 + [0]
                                       
            [0] = [1]                  
                                       
        [s](x1) = [1] x1 + [2]         
  
  This order satisfies the following ordering constraints:
  
     [+(0(), y)] = [2] y + [3]        
                 > [1] y + [0]        
                 = [y]                
                                      
    [+(s(x), y)] = [2] y + [3] x + [6]
                 > [2] y + [3] x + [4]
                 = [+(x, s(y))]       
                                      
    [+(s(x), y)] = [2] y + [3] x + [6]
                 > [2] y + [3] x + [2]
                 = [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:
  { +(0(), y) -> y
  , +(s(x), y) -> +(x, s(y))
  , +(s(x), 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^1))