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:
  { rev(a()) -> a()
  , rev(b()) -> b()
  , rev(++(x, x)) -> rev(x)
  , rev(++(x, y)) -> ++(rev(y), rev(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:
  { rev(++(x, x)) -> rev(x)
  , rev(++(x, y)) -> ++(rev(y), rev(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).
  
       [rev](x1) = [2] x1 + [0]         
                                        
             [a] = [0]                  
                                        
             [b] = [0]                  
                                        
    [++](x1, x2) = [1] x1 + [1] x2 + [2]
  
  This order satisfies the following ordering constraints:
  
         [rev(a())] =  [0]                 
                    >= [0]                 
                    =  [a()]               
                                           
         [rev(b())] =  [0]                 
                    >= [0]                 
                    =  [b()]               
                                           
    [rev(++(x, x))] =  [4] x + [4]         
                    >  [2] x + [0]         
                    =  [rev(x)]            
                                           
    [rev(++(x, y))] =  [2] x + [2] y + [4] 
                    >  [2] x + [2] y + [2] 
                    =  [++(rev(y), rev(x))]
                                           

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:
  { rev(a()) -> a()
  , rev(b()) -> b() }
Weak Trs:
  { rev(++(x, x)) -> rev(x)
  , rev(++(x, y)) -> ++(rev(y), rev(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:
  { rev(a()) -> a()
  , rev(b()) -> b() }

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

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:
  { rev(a()) -> a()
  , rev(b()) -> b()
  , rev(++(x, x)) -> rev(x)
  , rev(++(x, y)) -> ++(rev(y), rev(x)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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