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:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(0())) -> 0()
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(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: { mark(0()) -> 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).
  
        [a__f](x1) = [1] x1 + [0]         
                                          
               [0] = [0]                  
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [0]
                                          
           [f](x1) = [1] x1 + [0]         
                                          
           [s](x1) = [1] x1 + [0]         
                                          
        [a__p](x1) = [1] x1 + [0]         
                                          
        [mark](x1) = [2] x1 + [1]         
                                          
           [p](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
               [a__f(X)] =  [1] X + [0]           
                         >= [1] X + [0]           
                         =  [f(X)]                
                                                  
             [a__f(0())] =  [0]                   
                         >= [0]                   
                         =  [cons(0(), f(s(0())))]
                                                  
          [a__f(s(0()))] =  [0]                   
                         >= [0]                   
                         =  [a__f(a__p(s(0())))]  
                                                  
               [a__p(X)] =  [1] X + [0]           
                         >= [1] X + [0]           
                         =  [p(X)]                
                                                  
          [a__p(s(0()))] =  [0]                   
                         >= [0]                   
                         =  [0()]                 
                                                  
             [mark(0())] =  [1]                   
                         >  [0]                   
                         =  [0()]                 
                                                  
    [mark(cons(X1, X2))] =  [2] X1 + [2] X2 + [1] 
                         >= [2] X1 + [1] X2 + [1] 
                         =  [cons(mark(X1), X2)]  
                                                  
            [mark(f(X))] =  [2] X + [1]           
                         >= [2] X + [1]           
                         =  [a__f(mark(X))]       
                                                  
            [mark(s(X))] =  [2] X + [1]           
                         >= [2] X + [1]           
                         =  [s(mark(X))]          
                                                  
            [mark(p(X))] =  [2] X + [1]           
                         >= [2] X + [1]           
                         =  [a__p(mark(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:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(0())) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(X)) }
Weak Trs: { mark(0()) -> 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:
  { a__f(X) -> f(X)
  , mark(f(X)) -> a__f(mark(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).
  
        [a__f](x1) = [1] x1 + [3]         
                                          
               [0] = [1]                  
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [0]
                                          
           [f](x1) = [1] x1 + [2]         
                                          
           [s](x1) = [1] x1 + [0]         
                                          
        [a__p](x1) = [1] x1 + [0]         
                                          
        [mark](x1) = [2] x1 + [0]         
                                          
           [p](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
               [a__f(X)] =  [1] X + [3]           
                         >  [1] X + [2]           
                         =  [f(X)]                
                                                  
             [a__f(0())] =  [4]                   
                         >= [4]                   
                         =  [cons(0(), f(s(0())))]
                                                  
          [a__f(s(0()))] =  [4]                   
                         >= [4]                   
                         =  [a__f(a__p(s(0())))]  
                                                  
               [a__p(X)] =  [1] X + [0]           
                         >= [1] X + [0]           
                         =  [p(X)]                
                                                  
          [a__p(s(0()))] =  [1]                   
                         >= [1]                   
                         =  [0()]                 
                                                  
             [mark(0())] =  [2]                   
                         >  [1]                   
                         =  [0()]                 
                                                  
    [mark(cons(X1, X2))] =  [2] X1 + [2] X2 + [0] 
                         >= [2] X1 + [1] X2 + [0] 
                         =  [cons(mark(X1), X2)]  
                                                  
            [mark(f(X))] =  [2] X + [4]           
                         >  [2] X + [3]           
                         =  [a__f(mark(X))]       
                                                  
            [mark(s(X))] =  [2] X + [0]           
                         >= [2] X + [0]           
                         =  [s(mark(X))]          
                                                  
            [mark(p(X))] =  [2] X + [0]           
                         >= [2] X + [0]           
                         =  [a__p(mark(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:
  { a__f(0()) -> cons(0(), f(s(0())))
  , a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(0())) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(X)) }
Weak Trs:
  { a__f(X) -> f(X)
  , mark(0()) -> 0()
  , mark(f(X)) -> a__f(mark(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: { a__f(0()) -> cons(0(), f(s(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).
  
        [a__f](x1) = [1] x1 + [3]         
                                          
               [0] = [0]                  
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [0]
                                          
           [f](x1) = [1] x1 + [2]         
                                          
           [s](x1) = [1] x1 + [0]         
                                          
        [a__p](x1) = [1] x1 + [0]         
                                          
        [mark](x1) = [2] x1 + [0]         
                                          
           [p](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
               [a__f(X)] =  [1] X + [3]           
                         >  [1] X + [2]           
                         =  [f(X)]                
                                                  
             [a__f(0())] =  [3]                   
                         >  [2]                   
                         =  [cons(0(), f(s(0())))]
                                                  
          [a__f(s(0()))] =  [3]                   
                         >= [3]                   
                         =  [a__f(a__p(s(0())))]  
                                                  
               [a__p(X)] =  [1] X + [0]           
                         >= [1] X + [0]           
                         =  [p(X)]                
                                                  
          [a__p(s(0()))] =  [0]                   
                         >= [0]                   
                         =  [0()]                 
                                                  
             [mark(0())] =  [0]                   
                         >= [0]                   
                         =  [0()]                 
                                                  
    [mark(cons(X1, X2))] =  [2] X1 + [2] X2 + [0] 
                         >= [2] X1 + [1] X2 + [0] 
                         =  [cons(mark(X1), X2)]  
                                                  
            [mark(f(X))] =  [2] X + [4]           
                         >  [2] X + [3]           
                         =  [a__f(mark(X))]       
                                                  
            [mark(s(X))] =  [2] X + [0]           
                         >= [2] X + [0]           
                         =  [s(mark(X))]          
                                                  
            [mark(p(X))] =  [2] X + [0]           
                         >= [2] X + [0]           
                         =  [a__p(mark(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:
  { a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(0())) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(X)) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , mark(0()) -> 0()
  , mark(f(X)) -> a__f(mark(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: { mark(cons(X1, X2)) -> cons(mark(X1), X2) }

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).
  
        [a__f](x1) = [1] x1 + [2]         
                                          
               [0] = [0]                  
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [1]
                                          
           [f](x1) = [1] x1 + [1]         
                                          
           [s](x1) = [1] x1 + [0]         
                                          
        [a__p](x1) = [1] x1 + [0]         
                                          
        [mark](x1) = [2] x1 + [0]         
                                          
           [p](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
               [a__f(X)] =  [1] X + [2]           
                         >  [1] X + [1]           
                         =  [f(X)]                
                                                  
             [a__f(0())] =  [2]                   
                         >= [2]                   
                         =  [cons(0(), f(s(0())))]
                                                  
          [a__f(s(0()))] =  [2]                   
                         >= [2]                   
                         =  [a__f(a__p(s(0())))]  
                                                  
               [a__p(X)] =  [1] X + [0]           
                         >= [1] X + [0]           
                         =  [p(X)]                
                                                  
          [a__p(s(0()))] =  [0]                   
                         >= [0]                   
                         =  [0()]                 
                                                  
             [mark(0())] =  [0]                   
                         >= [0]                   
                         =  [0()]                 
                                                  
    [mark(cons(X1, X2))] =  [2] X1 + [2] X2 + [2] 
                         >  [2] X1 + [1] X2 + [1] 
                         =  [cons(mark(X1), X2)]  
                                                  
            [mark(f(X))] =  [2] X + [2]           
                         >= [2] X + [2]           
                         =  [a__f(mark(X))]       
                                                  
            [mark(s(X))] =  [2] X + [0]           
                         >= [2] X + [0]           
                         =  [s(mark(X))]          
                                                  
            [mark(p(X))] =  [2] X + [0]           
                         >= [2] X + [0]           
                         =  [a__p(mark(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:
  { a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(0())) -> 0()
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(X)) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(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:
  { a__p(s(0())) -> 0()
  , mark(s(X)) -> s(mark(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).
  
        [a__f](x1) = [1] x1 + [2]         
                                          
               [0] = [0]                  
                                          
    [cons](x1, x2) = [1] x1 + [1] x2 + [0]
                                          
           [f](x1) = [1] x1 + [1]         
                                          
           [s](x1) = [1] x1 + [1]         
                                          
        [a__p](x1) = [1] x1 + [0]         
                                          
        [mark](x1) = [2] x1 + [0]         
                                          
           [p](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
               [a__f(X)] =  [1] X + [2]           
                         >  [1] X + [1]           
                         =  [f(X)]                
                                                  
             [a__f(0())] =  [2]                   
                         >= [2]                   
                         =  [cons(0(), f(s(0())))]
                                                  
          [a__f(s(0()))] =  [3]                   
                         >= [3]                   
                         =  [a__f(a__p(s(0())))]  
                                                  
               [a__p(X)] =  [1] X + [0]           
                         >= [1] X + [0]           
                         =  [p(X)]                
                                                  
          [a__p(s(0()))] =  [1]                   
                         >  [0]                   
                         =  [0()]                 
                                                  
             [mark(0())] =  [0]                   
                         >= [0]                   
                         =  [0()]                 
                                                  
    [mark(cons(X1, X2))] =  [2] X1 + [2] X2 + [0] 
                         >= [2] X1 + [1] X2 + [0] 
                         =  [cons(mark(X1), X2)]  
                                                  
            [mark(f(X))] =  [2] X + [2]           
                         >= [2] X + [2]           
                         =  [a__f(mark(X))]       
                                                  
            [mark(s(X))] =  [2] X + [2]           
                         >  [2] X + [1]           
                         =  [s(mark(X))]          
                                                  
            [mark(p(X))] =  [2] X + [0]           
                         >= [2] X + [0]           
                         =  [a__p(mark(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:
  { a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , mark(p(X)) -> a__p(mark(X)) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__p(s(0())) -> 0()
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

Trs: { mark(p(X)) -> a__p(mark(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) and not(IDA(1)).
  
        [a__f](x1) = [1 2] x1 + [0]           
                     [0 0]      [2]           
                                              
               [0] = [0]                      
                     [1]                      
                                              
    [cons](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                     [0 1]      [0 0]      [0]
                                              
           [f](x1) = [1 1] x1 + [0]           
                     [0 0]      [2]           
                                              
           [s](x1) = [1 0] x1 + [0]           
                     [0 1]      [1]           
                                              
        [a__p](x1) = [1 1] x1 + [0]           
                     [0 0]      [1]           
                                              
        [mark](x1) = [3 1] x1 + [0]           
                     [0 1]      [0]           
                                              
           [p](x1) = [1 1] x1 + [0]           
                     [0 0]      [1]           
  
  This order satisfies the following ordering constraints:
  
               [a__f(X)] =  [1 2] X + [0]            
                            [0 0]     [2]            
                         >= [1 1] X + [0]            
                            [0 0]     [2]            
                         =  [f(X)]                   
                                                     
             [a__f(0())] =  [2]                      
                            [2]                      
                         >= [2]                      
                            [1]                      
                         =  [cons(0(), f(s(0())))]   
                                                     
          [a__f(s(0()))] =  [4]                      
                            [2]                      
                         >= [4]                      
                            [2]                      
                         =  [a__f(a__p(s(0())))]     
                                                     
               [a__p(X)] =  [1 1] X + [0]            
                            [0 0]     [1]            
                         >= [1 1] X + [0]            
                            [0 0]     [1]            
                         =  [p(X)]                   
                                                     
          [a__p(s(0()))] =  [2]                      
                            [1]                      
                         >  [0]                      
                            [1]                      
                         =  [0()]                    
                                                     
             [mark(0())] =  [1]                      
                            [1]                      
                         >  [0]                      
                            [1]                      
                         =  [0()]                    
                                                     
    [mark(cons(X1, X2))] =  [3 1] X1 + [3 0] X2 + [0]
                            [0 1]      [0 0]      [0]
                         >= [3 1] X1 + [1 0] X2 + [0]
                            [0 1]      [0 0]      [0]
                         =  [cons(mark(X1), X2)]     
                                                     
            [mark(f(X))] =  [3 3] X + [2]            
                            [0 0]     [2]            
                         >  [3 3] X + [0]            
                            [0 0]     [2]            
                         =  [a__f(mark(X))]          
                                                     
            [mark(s(X))] =  [3 1] X + [1]            
                            [0 1]     [1]            
                         >  [3 1] X + [0]            
                            [0 1]     [1]            
                         =  [s(mark(X))]             
                                                     
            [mark(p(X))] =  [3 3] X + [1]            
                            [0 0]     [1]            
                         >  [3 2] X + [0]            
                            [0 0]     [1]            
                         =  [a__p(mark(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:
  { a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__p(s(0())) -> 0()
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

Trs: { a__p(X) -> p(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) and not(IDA(1)).
  
        [a__f](x1) = [1 2] x1 + [1]           
                     [0 0]      [1]           
                                              
               [0] = [0]                      
                     [1]                      
                                              
    [cons](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                     [0 1]      [0 0]      [0]
                                              
           [f](x1) = [1 1] x1 + [0]           
                     [0 0]      [1]           
                                              
           [s](x1) = [1 0] x1 + [0]           
                     [0 1]      [2]           
                                              
        [a__p](x1) = [1 1] x1 + [1]           
                     [0 0]      [1]           
                                              
        [mark](x1) = [3 1] x1 + [0]           
                     [0 1]      [0]           
                                              
           [p](x1) = [1 1] x1 + [0]           
                     [0 0]      [1]           
  
  This order satisfies the following ordering constraints:
  
               [a__f(X)] =  [1 2] X + [1]            
                            [0 0]     [1]            
                         >  [1 1] X + [0]            
                            [0 0]     [1]            
                         =  [f(X)]                   
                                                     
             [a__f(0())] =  [3]                      
                            [1]                      
                         >= [3]                      
                            [1]                      
                         =  [cons(0(), f(s(0())))]   
                                                     
          [a__f(s(0()))] =  [7]                      
                            [1]                      
                         >= [7]                      
                            [1]                      
                         =  [a__f(a__p(s(0())))]     
                                                     
               [a__p(X)] =  [1 1] X + [1]            
                            [0 0]     [1]            
                         >  [1 1] X + [0]            
                            [0 0]     [1]            
                         =  [p(X)]                   
                                                     
          [a__p(s(0()))] =  [4]                      
                            [1]                      
                         >  [0]                      
                            [1]                      
                         =  [0()]                    
                                                     
             [mark(0())] =  [1]                      
                            [1]                      
                         >  [0]                      
                            [1]                      
                         =  [0()]                    
                                                     
    [mark(cons(X1, X2))] =  [3 1] X1 + [3 0] X2 + [0]
                            [0 1]      [0 0]      [0]
                         >= [3 1] X1 + [1 0] X2 + [0]
                            [0 1]      [0 0]      [0]
                         =  [cons(mark(X1), X2)]     
                                                     
            [mark(f(X))] =  [3 3] X + [1]            
                            [0 0]     [1]            
                         >= [3 3] X + [1]            
                            [0 0]     [1]            
                         =  [a__f(mark(X))]          
                                                     
            [mark(s(X))] =  [3 1] X + [2]            
                            [0 1]     [2]            
                         >  [3 1] X + [0]            
                            [0 1]     [2]            
                         =  [s(mark(X))]             
                                                     
            [mark(p(X))] =  [3 3] X + [1]            
                            [0 0]     [1]            
                         >= [3 2] X + [1]            
                            [0 0]     [1]            
                         =  [a__p(mark(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: { a__f(s(0())) -> a__f(a__p(s(0()))) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(0())) -> 0()
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

Trs: { a__f(s(0())) -> a__f(a__p(s(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) and not(IDA(1)).
  
        [a__f](x1) = [1 3] x1 + [0]           
                     [0 0]      [0]           
                                              
               [0] = [0]                      
                     [1]                      
                                              
    [cons](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                     [0 0]      [0 0]      [0]
                                              
           [f](x1) = [1 1] x1 + [0]           
                     [0 0]      [0]           
                                              
           [s](x1) = [1 0] x1 + [0]           
                     [0 1]      [1]           
                                              
        [a__p](x1) = [1 0] x1 + [0]           
                     [0 0]      [1]           
                                              
        [mark](x1) = [3 0] x1 + [0]           
                     [0 1]      [0]           
                                              
           [p](x1) = [1 0] x1 + [0]           
                     [0 0]      [1]           
  
  This order satisfies the following ordering constraints:
  
               [a__f(X)] =  [1 3] X + [0]            
                            [0 0]     [0]            
                         >= [1 1] X + [0]            
                            [0 0]     [0]            
                         =  [f(X)]                   
                                                     
             [a__f(0())] =  [3]                      
                            [0]                      
                         >  [2]                      
                            [0]                      
                         =  [cons(0(), f(s(0())))]   
                                                     
          [a__f(s(0()))] =  [6]                      
                            [0]                      
                         >  [3]                      
                            [0]                      
                         =  [a__f(a__p(s(0())))]     
                                                     
               [a__p(X)] =  [1 0] X + [0]            
                            [0 0]     [1]            
                         >= [1 0] X + [0]            
                            [0 0]     [1]            
                         =  [p(X)]                   
                                                     
          [a__p(s(0()))] =  [0]                      
                            [1]                      
                         >= [0]                      
                            [1]                      
                         =  [0()]                    
                                                     
             [mark(0())] =  [0]                      
                            [1]                      
                         >= [0]                      
                            [1]                      
                         =  [0()]                    
                                                     
    [mark(cons(X1, X2))] =  [3 0] X1 + [3 0] X2 + [0]
                            [0 0]      [0 0]      [0]
                         >= [3 0] X1 + [1 0] X2 + [0]
                            [0 0]      [0 0]      [0]
                         =  [cons(mark(X1), X2)]     
                                                     
            [mark(f(X))] =  [3 3] X + [0]            
                            [0 0]     [0]            
                         >= [3 3] X + [0]            
                            [0 0]     [0]            
                         =  [a__f(mark(X))]          
                                                     
            [mark(s(X))] =  [3 0] X + [0]            
                            [0 1]     [1]            
                         >= [3 0] X + [0]            
                            [0 1]     [1]            
                         =  [s(mark(X))]             
                                                     
            [mark(p(X))] =  [3 0] X + [0]            
                            [0 0]     [1]            
                         >= [3 0] X + [0]            
                            [0 0]     [1]            
                         =  [a__p(mark(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:
  { a__f(X) -> f(X)
  , a__f(0()) -> cons(0(), f(s(0())))
  , a__f(s(0())) -> a__f(a__p(s(0())))
  , a__p(X) -> p(X)
  , a__p(s(0())) -> 0()
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(f(X)) -> a__f(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(p(X)) -> a__p(mark(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))