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:
  { zeros() -> cons(0(), n__zeros())
  , zeros() -> n__zeros()
  , tail(cons(X, XS)) -> activate(XS)
  , activate(X) -> X
  , activate(n__zeros()) -> zeros() }
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:
  { zeros() -> cons(0(), n__zeros())
  , zeros() -> n__zeros()
  , activate(X) -> X
  , activate(n__zeros()) -> zeros() }

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).
  
           [zeros] = [3]                  
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [0]
                                          
               [0] = [1]                  
                                          
        [n__zeros] = [1]                  
                                          
        [tail](x1) = [3] x1 + [3]         
                                          
    [activate](x1) = [3] x1 + [3]         
  
  This order satisfies the following ordering constraints:
  
                 [zeros()] =  [3]                    
                           >  [2]                    
                           =  [cons(0(), n__zeros())]
                                                     
                 [zeros()] =  [3]                    
                           >  [1]                    
                           =  [n__zeros()]           
                                                     
       [tail(cons(X, XS))] =  [3] X + [3] XS + [3]   
                           >= [3] XS + [3]           
                           =  [activate(XS)]         
                                                     
             [activate(X)] =  [3] X + [3]            
                           >  [1] X + [0]            
                           =  [X]                    
                                                     
    [activate(n__zeros())] =  [6]                    
                           >  [3]                    
                           =  [zeros()]              
                                                     

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: { tail(cons(X, XS)) -> activate(XS) }
Weak Trs:
  { zeros() -> cons(0(), n__zeros())
  , zeros() -> n__zeros()
  , activate(X) -> X
  , activate(n__zeros()) -> zeros() }
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: { tail(cons(X, XS)) -> activate(XS) }

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).
  
           [zeros] = [3]                  
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [3]
                                          
               [0] = [0]                  
                                          
        [n__zeros] = [0]                  
                                          
        [tail](x1) = [2] x1 + [1]         
                                          
    [activate](x1) = [2] x1 + [3]         
  
  This order satisfies the following ordering constraints:
  
                 [zeros()] =  [3]                    
                           >= [3]                    
                           =  [cons(0(), n__zeros())]
                                                     
                 [zeros()] =  [3]                    
                           >  [0]                    
                           =  [n__zeros()]           
                                                     
       [tail(cons(X, XS))] =  [2] X + [2] XS + [7]   
                           >  [2] XS + [3]           
                           =  [activate(XS)]         
                                                     
             [activate(X)] =  [2] X + [3]            
                           >  [1] X + [0]            
                           =  [X]                    
                                                     
    [activate(n__zeros())] =  [3]                    
                           >= [3]                    
                           =  [zeros()]              
                                                     

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:
  { zeros() -> cons(0(), n__zeros())
  , zeros() -> n__zeros()
  , tail(cons(X, XS)) -> activate(XS)
  , activate(X) -> X
  , activate(n__zeros()) -> zeros() }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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