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