We consider the following Problem:

  Strict Trs:
    {  a__f(0()) -> cons(0(), f(s(0())))
     , a__f(s(0())) -> a__f(a__p(s(0())))
     , a__p(s(0())) -> 0()
     , mark(f(X)) -> a__f(mark(X))
     , mark(p(X)) -> a__p(mark(X))
     , mark(0()) -> 0()
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(s(X)) -> s(mark(X))
     , a__f(X) -> f(X)
     , a__p(X) -> p(X)}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  a__f(0()) -> cons(0(), f(s(0())))
       , a__f(s(0())) -> a__f(a__p(s(0())))
       , a__p(s(0())) -> 0()
       , mark(f(X)) -> a__f(mark(X))
       , mark(p(X)) -> a__p(mark(X))
       , mark(0()) -> 0()
       , mark(cons(X1, X2)) -> cons(mark(X1), X2)
       , mark(s(X)) -> s(mark(X))
       , a__f(X) -> f(X)
       , a__p(X) -> p(X)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  mark(0()) -> 0()
       , a__f(X) -> f(X)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(a__f) = {1}, Uargs(cons) = {1}, Uargs(f) = {},
        Uargs(s) = {1}, Uargs(a__p) = {1}, Uargs(mark) = {}, Uargs(p) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       a__f(x1) = [1 0] x1 + [1]
                  [0 0]      [1]
       0() = [0]
             [0]
       cons(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
                      [0 0]      [0 0]      [1]
       f(x1) = [0 0] x1 + [0]
               [0 0]      [0]
       s(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       a__p(x1) = [1 0] x1 + [0]
                  [0 0]      [1]
       mark(x1) = [0 0] x1 + [1]
                  [0 0]      [1]
       p(x1) = [0 0] x1 + [0]
               [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  a__f(0()) -> cons(0(), f(s(0())))
         , a__f(s(0())) -> a__f(a__p(s(0())))
         , a__p(s(0())) -> 0()
         , mark(f(X)) -> a__f(mark(X))
         , mark(p(X)) -> a__p(mark(X))
         , mark(cons(X1, X2)) -> cons(mark(X1), X2)
         , mark(s(X)) -> s(mark(X))
         , a__p(X) -> p(X)}
      Weak Trs:
        {  mark(0()) -> 0()
         , a__f(X) -> f(X)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {a__p(s(0())) -> 0()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(a__f) = {1}, Uargs(cons) = {1}, Uargs(f) = {},
          Uargs(s) = {1}, Uargs(a__p) = {1}, Uargs(mark) = {}, Uargs(p) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         a__f(x1) = [1 0] x1 + [1]
                    [0 0]      [1]
         0() = [0]
               [0]
         cons(x1, x2) = [1 1] x1 + [0 0] x2 + [1]
                        [0 0]      [0 0]      [1]
         f(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
         s(x1) = [1 0] x1 + [0]
                 [0 0]      [2]
         a__p(x1) = [1 2] x1 + [0]
                    [0 0]      [0]
         mark(x1) = [0 0] x1 + [0]
                    [0 0]      [3]
         p(x1) = [0 0] x1 + [0]
                 [0 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  a__f(0()) -> cons(0(), f(s(0())))
           , a__f(s(0())) -> a__f(a__p(s(0())))
           , mark(f(X)) -> a__f(mark(X))
           , mark(p(X)) -> a__p(mark(X))
           , mark(cons(X1, X2)) -> cons(mark(X1), X2)
           , mark(s(X)) -> s(mark(X))
           , a__p(X) -> p(X)}
        Weak Trs:
          {  a__p(s(0())) -> 0()
           , mark(0()) -> 0()
           , a__f(X) -> f(X)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {a__p(X) -> p(X)}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(a__f) = {1}, Uargs(cons) = {1}, Uargs(f) = {},
            Uargs(s) = {1}, Uargs(a__p) = {1}, Uargs(mark) = {}, Uargs(p) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           a__f(x1) = [1 1] x1 + [1]
                      [0 0]      [1]
           0() = [0]
                 [0]
           cons(x1, x2) = [1 2] x1 + [0 0] x2 + [1]
                          [0 0]      [0 0]      [1]
           f(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
           s(x1) = [1 0] x1 + [0]
                   [0 1]      [0]
           a__p(x1) = [1 0] x1 + [2]
                      [0 1]      [2]
           mark(x1) = [0 0] x1 + [2]
                      [0 0]      [3]
           p(x1) = [0 0] x1 + [0]
                   [0 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  a__f(0()) -> cons(0(), f(s(0())))
             , a__f(s(0())) -> a__f(a__p(s(0())))
             , mark(f(X)) -> a__f(mark(X))
             , mark(p(X)) -> a__p(mark(X))
             , mark(cons(X1, X2)) -> cons(mark(X1), X2)
             , mark(s(X)) -> s(mark(X))}
          Weak Trs:
            {  a__p(X) -> p(X)
             , a__p(s(0())) -> 0()
             , mark(0()) -> 0()
             , a__f(X) -> f(X)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {a__f(0()) -> cons(0(), f(s(0())))}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(a__f) = {1}, Uargs(cons) = {1}, Uargs(f) = {},
              Uargs(s) = {1}, Uargs(a__p) = {1}, Uargs(mark) = {}, Uargs(p) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             a__f(x1) = [1 0] x1 + [1]
                        [0 0]      [1]
             0() = [0]
                   [0]
             cons(x1, x2) = [1 3] x1 + [0 0] x2 + [0]
                            [0 0]      [0 0]      [1]
             f(x1) = [0 0] x1 + [0]
                     [0 0]      [0]
             s(x1) = [1 0] x1 + [2]
                     [0 0]      [0]
             a__p(x1) = [1 0] x1 + [2]
                        [0 0]      [1]
             mark(x1) = [0 0] x1 + [3]
                        [0 0]      [3]
             p(x1) = [0 0] x1 + [0]
                     [0 0]      [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Strict Trs:
              {  a__f(s(0())) -> a__f(a__p(s(0())))
               , mark(f(X)) -> a__f(mark(X))
               , mark(p(X)) -> a__p(mark(X))
               , mark(cons(X1, X2)) -> cons(mark(X1), X2)
               , mark(s(X)) -> s(mark(X))}
            Weak Trs:
              {  a__f(0()) -> cons(0(), f(s(0())))
               , a__p(X) -> p(X)
               , a__p(s(0())) -> 0()
               , mark(0()) -> 0()
               , a__f(X) -> f(X)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component: {a__f(s(0())) -> a__f(a__p(s(0())))}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(a__f) = {1}, Uargs(cons) = {1}, Uargs(f) = {},
                Uargs(s) = {1}, Uargs(a__p) = {1}, Uargs(mark) = {}, Uargs(p) = {}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               a__f(x1) = [1 2] x1 + [0]
                          [0 0]      [0]
               0() = [1]
                     [1]
               cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                              [0 0]      [0 0]      [0]
               f(x1) = [1 0] x1 + [0]
                       [0 0]      [0]
               s(x1) = [1 2] x1 + [0]
                       [0 0]      [3]
               a__p(x1) = [1 0] x1 + [1]
                          [0 0]      [2]
               mark(x1) = [1 0] x1 + [0]
                          [0 0]      [1]
               p(x1) = [1 0] x1 + [0]
                       [0 0]      [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  mark(f(X)) -> a__f(mark(X))
                 , mark(p(X)) -> a__p(mark(X))
                 , mark(cons(X1, X2)) -> cons(mark(X1), X2)
                 , mark(s(X)) -> s(mark(X))}
              Weak Trs:
                {  a__f(s(0())) -> a__f(a__p(s(0())))
                 , a__f(0()) -> cons(0(), f(s(0())))
                 , a__p(X) -> p(X)
                 , a__p(s(0())) -> 0()
                 , mark(0()) -> 0()
                 , a__f(X) -> f(X)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              The weightgap principle applies, where following rules are oriented strictly:
              
              TRS Component: {mark(f(X)) -> a__f(mark(X))}
              
              Interpretation of nonconstant growth:
              -------------------------------------
                The following argument positions are usable:
                  Uargs(a__f) = {1}, Uargs(cons) = {1}, Uargs(f) = {},
                  Uargs(s) = {1}, Uargs(a__p) = {1}, Uargs(mark) = {}, Uargs(p) = {}
                We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                Interpretation Functions:
                 a__f(x1) = [1 0] x1 + [0]
                            [0 1]      [1]
                 0() = [0]
                       [1]
                 cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                                [0 1]      [0 0]      [0]
                 f(x1) = [0 0] x1 + [0]
                         [0 1]      [1]
                 s(x1) = [1 0] x1 + [0]
                         [0 1]      [0]
                 a__p(x1) = [1 0] x1 + [0]
                            [0 1]      [0]
                 mark(x1) = [0 1] x1 + [0]
                            [0 1]      [0]
                 p(x1) = [0 0] x1 + [0]
                         [0 1]      [0]
              
              The strictly oriented rules are moved into the weak component.
              
              We consider the following Problem:
              
                Strict Trs:
                  {  mark(p(X)) -> a__p(mark(X))
                   , mark(cons(X1, X2)) -> cons(mark(X1), X2)
                   , mark(s(X)) -> s(mark(X))}
                Weak Trs:
                  {  mark(f(X)) -> a__f(mark(X))
                   , a__f(s(0())) -> a__f(a__p(s(0())))
                   , a__f(0()) -> cons(0(), f(s(0())))
                   , a__p(X) -> p(X)
                   , a__p(s(0())) -> 0()
                   , mark(0()) -> 0()
                   , a__f(X) -> f(X)}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                The weightgap principle applies, where following rules are oriented strictly:
                
                TRS Component: {mark(cons(X1, X2)) -> cons(mark(X1), X2)}
                
                Interpretation of nonconstant growth:
                -------------------------------------
                  The following argument positions are usable:
                    Uargs(a__f) = {1}, Uargs(cons) = {1}, Uargs(f) = {},
                    Uargs(s) = {1}, Uargs(a__p) = {1}, Uargs(mark) = {}, Uargs(p) = {}
                  We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                  Interpretation Functions:
                   a__f(x1) = [1 0] x1 + [0]
                              [0 1]      [1]
                   0() = [0]
                         [0]
                   cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                                  [0 1]      [0 0]      [1]
                   f(x1) = [0 0] x1 + [0]
                           [0 1]      [1]
                   s(x1) = [1 0] x1 + [0]
                           [0 1]      [0]
                   a__p(x1) = [1 0] x1 + [0]
                              [0 1]      [0]
                   mark(x1) = [0 2] x1 + [0]
                              [0 1]      [0]
                   p(x1) = [0 0] x1 + [0]
                           [0 1]      [0]
                
                The strictly oriented rules are moved into the weak component.
                
                We consider the following Problem:
                
                  Strict Trs:
                    {  mark(p(X)) -> a__p(mark(X))
                     , mark(s(X)) -> s(mark(X))}
                  Weak Trs:
                    {  mark(cons(X1, X2)) -> cons(mark(X1), X2)
                     , mark(f(X)) -> a__f(mark(X))
                     , a__f(s(0())) -> a__f(a__p(s(0())))
                     , a__f(0()) -> cons(0(), f(s(0())))
                     , a__p(X) -> p(X)
                     , a__p(s(0())) -> 0()
                     , mark(0()) -> 0()
                     , a__f(X) -> f(X)}
                  StartTerms: basic terms
                  Strategy: innermost
                
                Certificate: YES(?,O(n^1))
                
                Proof:
                  The weightgap principle applies, where following rules are oriented strictly:
                  
                  TRS Component: {mark(s(X)) -> s(mark(X))}
                  
                  Interpretation of nonconstant growth:
                  -------------------------------------
                    The following argument positions are usable:
                      Uargs(a__f) = {1}, Uargs(cons) = {1}, Uargs(f) = {},
                      Uargs(s) = {1}, Uargs(a__p) = {1}, Uargs(mark) = {}, Uargs(p) = {}
                    We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
                    Interpretation Functions:
                     a__f(x1) = [1 0] x1 + [0]
                                [0 1]      [0]
                     0() = [0]
                           [0]
                     cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                                    [0 1]      [0 0]      [0]
                     f(x1) = [0 0] x1 + [0]
                             [0 1]      [0]
                     s(x1) = [1 0] x1 + [0]
                             [0 1]      [1]
                     a__p(x1) = [1 0] x1 + [0]
                                [0 1]      [0]
                     mark(x1) = [0 1] x1 + [0]
                                [0 1]      [0]
                     p(x1) = [0 0] x1 + [0]
                             [0 1]      [0]
                  
                  The strictly oriented rules are moved into the weak component.
                  
                  We consider the following Problem:
                  
                    Strict Trs: {mark(p(X)) -> a__p(mark(X))}
                    Weak Trs:
                      {  mark(s(X)) -> s(mark(X))
                       , mark(cons(X1, X2)) -> cons(mark(X1), X2)
                       , mark(f(X)) -> a__f(mark(X))
                       , a__f(s(0())) -> a__f(a__p(s(0())))
                       , a__f(0()) -> cons(0(), f(s(0())))
                       , a__p(X) -> p(X)
                       , a__p(s(0())) -> 0()
                       , mark(0()) -> 0()
                       , a__f(X) -> f(X)}
                    StartTerms: basic terms
                    Strategy: innermost
                  
                  Certificate: YES(?,O(n^1))
                  
                  Proof:
                    We consider the following Problem:
                    
                      Strict Trs: {mark(p(X)) -> a__p(mark(X))}
                      Weak Trs:
                        {  mark(s(X)) -> s(mark(X))
                         , mark(cons(X1, X2)) -> cons(mark(X1), X2)
                         , mark(f(X)) -> a__f(mark(X))
                         , a__f(s(0())) -> a__f(a__p(s(0())))
                         , a__f(0()) -> cons(0(), f(s(0())))
                         , a__p(X) -> p(X)
                         , a__p(s(0())) -> 0()
                         , mark(0()) -> 0()
                         , a__f(X) -> f(X)}
                      StartTerms: basic terms
                      Strategy: innermost
                    
                    Certificate: YES(?,O(n^1))
                    
                    Proof:
                      The problem is match-bounded by 1.
                      The enriched problem is compatible with the following automaton:
                      {  a__f_0(1) -> 1
                       , a__f_0(2) -> 1
                       , a__f_1(3) -> 3
                       , a__f_1(4) -> 1
                       , 0_0() -> 1
                       , 0_0() -> 2
                       , 0_1() -> 1
                       , 0_1() -> 3
                       , 0_1() -> 4
                       , 0_1() -> 6
                       , cons_0(1, 2) -> 1
                       , cons_0(2, 2) -> 1
                       , cons_0(2, 2) -> 2
                       , cons_1(3, 2) -> 3
                       , cons_1(6, 7) -> 1
                       , cons_1(6, 7) -> 3
                       , f_0(1) -> 1
                       , f_0(2) -> 1
                       , f_0(2) -> 2
                       , f_1(3) -> 3
                       , f_1(4) -> 1
                       , f_1(5) -> 7
                       , s_0(1) -> 1
                       , s_0(2) -> 2
                       , s_1(3) -> 3
                       , s_1(6) -> 5
                       , a__p_0(2) -> 1
                       , a__p_1(3) -> 1
                       , a__p_1(3) -> 3
                       , a__p_1(5) -> 4
                       , mark_0(2) -> 1
                       , mark_1(2) -> 3
                       , p_0(2) -> 1
                       , p_0(2) -> 2
                       , p_1(3) -> 1
                       , p_1(3) -> 3
                       , p_1(5) -> 4}

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