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