MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { f(n__a(), X, X) -> f(activate(X), b(), n__b())
  , activate(X) -> X
  , activate(n__a()) -> a()
  , activate(n__b()) -> b()
  , b() -> n__b()
  , b() -> a()
  , a() -> n__a() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs: { activate(n__a()) -> a() }

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) and not(IDA(1)).
  
    [f](x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 1] x3 + [1]
                      [0 0]      [0 0]      [0 0]      [1]
                                                          
             [n__a] = [0]                                 
                      [1]                                 
                                                          
     [activate](x1) = [2 1] x1 + [0]                      
                      [0 3]      [1]                      
                                                          
                [b] = [0]                                 
                      [1]                                 
                                                          
             [n__b] = [0]                                 
                      [0]                                 
                                                          
                [a] = [0]                                 
                      [1]                                 
  
  This order satisfies the following ordering constraints:
  
     [f(n__a(), X, X)] =  [2 1] X + [1]                
                          [0 0]     [1]                
                       >= [2 1] X + [1]                
                          [0 0]     [1]                
                       =  [f(activate(X), b(), n__b())]
                                                       
         [activate(X)] =  [2 1] X + [0]                
                          [0 3]     [1]                
                       >= [1 0] X + [0]                
                          [0 1]     [0]                
                       =  [X]                          
                                                       
    [activate(n__a())] =  [1]                          
                          [4]                          
                       >  [0]                          
                          [1]                          
                       =  [a()]                        
                                                       
    [activate(n__b())] =  [0]                          
                          [1]                          
                       >= [0]                          
                          [1]                          
                       =  [b()]                        
                                                       
                 [b()] =  [0]                          
                          [1]                          
                       >= [0]                          
                          [0]                          
                       =  [n__b()]                     
                                                       
                 [b()] =  [0]                          
                          [1]                          
                       >= [0]                          
                          [1]                          
                       =  [a()]                        
                                                       
                 [a()] =  [0]                          
                          [1]                          
                       >= [0]                          
                          [1]                          
                       =  [n__a()]                     
                                                       

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { f(n__a(), X, X) -> f(activate(X), b(), n__b())
  , activate(X) -> X
  , activate(n__b()) -> b()
  , b() -> n__b()
  , b() -> a()
  , a() -> n__a() }
Weak Trs: { activate(n__a()) -> a() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Empty strict component of the problem is NOT empty.

Arrrr..