MAYBE

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

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

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { activate(X) -> X
  , activate(n__b()) -> b()
  , activate(n__c()) -> c() }

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).
  
    [f](x1, x2, x3) = [1] x1 + [3] x2 + [2] x3 + [1]
                                                    
             [n__b] = [0]                           
                                                    
             [n__c] = [0]                           
                                                    
                [c] = [0]                           
                                                    
                [b] = [0]                           
                                                    
     [activate](x1) = [2] x1 + [1]                  
  
  This order satisfies the following ordering constraints:
  
    [f(n__b(), X, n__c())] =  [3] X + [1]   
                           >= [3] X + [1]   
                           =  [f(X, c(), X)]
                                            
                     [c()] =  [0]           
                           >= [0]           
                           =  [n__c()]      
                                            
                     [c()] =  [0]           
                           >= [0]           
                           =  [b()]         
                                            
                     [b()] =  [0]           
                           >= [0]           
                           =  [n__b()]      
                                            
             [activate(X)] =  [2] X + [1]   
                           >  [1] X + [0]   
                           =  [X]           
                                            
        [activate(n__b())] =  [1]           
                           >  [0]           
                           =  [b()]         
                                            
        [activate(n__c())] =  [1]           
                           >  [0]           
                           =  [c()]         
                                            

We return to the main proof.

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

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

Empty strict component of the problem is NOT empty.

Arrrr..