We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  active(f(0())) -> mark(cons(0(), f(s(0()))))
       , active(f(s(0()))) -> mark(f(p(s(0()))))
       , active(p(s(X))) -> mark(X)
       , mark(f(X)) -> active(f(mark(X)))
       , mark(0()) -> active(0())
       , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
       , mark(s(X)) -> active(s(mark(X)))
       , mark(p(X)) -> active(p(mark(X)))
       , f(mark(X)) -> f(X)
       , f(active(X)) -> f(X)
       , cons(mark(X1), X2) -> cons(X1, X2)
       , cons(X1, mark(X2)) -> cons(X1, X2)
       , cons(active(X1), X2) -> cons(X1, X2)
       , cons(X1, active(X2)) -> cons(X1, X2)
       , s(mark(X)) -> s(X)
       , s(active(X)) -> s(X)
       , p(mark(X)) -> p(X)
       , p(active(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:
      {  f(mark(X)) -> f(X)
       , f(active(X)) -> f(X)
       , cons(mark(X1), X2) -> cons(X1, X2)
       , cons(active(X1), X2) -> cons(X1, X2)
       , s(mark(X)) -> s(X)
       , s(active(X)) -> s(X)
       , p(mark(X)) -> p(X)
       , p(active(X)) -> p(X)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {},
        Uargs(cons) = {1}, Uargs(s) = {1}, Uargs(p) = {1}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       active(x1) = [1 0] x1 + [1]
                    [0 0]      [1]
       f(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       0() = [0]
             [0]
       mark(x1) = [1 0] x1 + [1]
                  [1 0]      [1]
       cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                      [0 0]      [0 0]      [1]
       s(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       p(x1) = [1 0] x1 + [0]
               [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  active(f(0())) -> mark(cons(0(), f(s(0()))))
         , active(f(s(0()))) -> mark(f(p(s(0()))))
         , active(p(s(X))) -> mark(X)
         , mark(f(X)) -> active(f(mark(X)))
         , mark(0()) -> active(0())
         , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
         , mark(s(X)) -> active(s(mark(X)))
         , mark(p(X)) -> active(p(mark(X)))
         , cons(X1, mark(X2)) -> cons(X1, X2)
         , cons(X1, active(X2)) -> cons(X1, X2)}
      Weak Trs:
        {  f(mark(X)) -> f(X)
         , f(active(X)) -> f(X)
         , cons(mark(X1), X2) -> cons(X1, X2)
         , cons(active(X1), X2) -> cons(X1, X2)
         , s(mark(X)) -> s(X)
         , s(active(X)) -> s(X)
         , p(mark(X)) -> p(X)
         , p(active(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: {active(p(s(X))) -> mark(X)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {},
          Uargs(cons) = {1}, Uargs(s) = {1}, Uargs(p) = {1}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         active(x1) = [1 0] x1 + [1]
                      [0 1]      [1]
         f(x1) = [1 0] x1 + [0]
                 [1 0]      [0]
         0() = [0]
               [0]
         mark(x1) = [1 0] x1 + [1]
                    [0 0]      [1]
         cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                        [1 0]      [0 0]      [3]
         s(x1) = [1 0] x1 + [1]
                 [0 0]      [0]
         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:
          {  active(f(0())) -> mark(cons(0(), f(s(0()))))
           , active(f(s(0()))) -> mark(f(p(s(0()))))
           , mark(f(X)) -> active(f(mark(X)))
           , mark(0()) -> active(0())
           , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
           , mark(s(X)) -> active(s(mark(X)))
           , mark(p(X)) -> active(p(mark(X)))
           , cons(X1, mark(X2)) -> cons(X1, X2)
           , cons(X1, active(X2)) -> cons(X1, X2)}
        Weak Trs:
          {  active(p(s(X))) -> mark(X)
           , f(mark(X)) -> f(X)
           , f(active(X)) -> f(X)
           , cons(mark(X1), X2) -> cons(X1, X2)
           , cons(active(X1), X2) -> cons(X1, X2)
           , s(mark(X)) -> s(X)
           , s(active(X)) -> s(X)
           , p(mark(X)) -> p(X)
           , p(active(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()) -> active(0())}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {},
            Uargs(cons) = {1}, Uargs(s) = {1}, Uargs(p) = {1}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           active(x1) = [1 0] x1 + [1]
                        [0 1]      [1]
           f(x1) = [1 0] x1 + [0]
                   [0 1]      [0]
           0() = [0]
                 [1]
           mark(x1) = [1 0] x1 + [3]
                      [0 1]      [1]
           cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                          [0 0]      [0 1]      [3]
           s(x1) = [1 0] x1 + [0]
                   [0 1]      [0]
           p(x1) = [1 0] x1 + [2]
                   [0 1]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  active(f(0())) -> mark(cons(0(), f(s(0()))))
             , active(f(s(0()))) -> mark(f(p(s(0()))))
             , mark(f(X)) -> active(f(mark(X)))
             , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
             , mark(s(X)) -> active(s(mark(X)))
             , mark(p(X)) -> active(p(mark(X)))
             , cons(X1, mark(X2)) -> cons(X1, X2)
             , cons(X1, active(X2)) -> cons(X1, X2)}
          Weak Trs:
            {  mark(0()) -> active(0())
             , active(p(s(X))) -> mark(X)
             , f(mark(X)) -> f(X)
             , f(active(X)) -> f(X)
             , cons(mark(X1), X2) -> cons(X1, X2)
             , cons(active(X1), X2) -> cons(X1, X2)
             , s(mark(X)) -> s(X)
             , s(active(X)) -> s(X)
             , p(mark(X)) -> p(X)
             , p(active(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:
            {  active(f(0())) -> mark(cons(0(), f(s(0()))))
             , active(f(s(0()))) -> mark(f(p(s(0()))))}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {},
              Uargs(cons) = {1}, Uargs(s) = {1}, Uargs(p) = {1}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             active(x1) = [1 1] x1 + [0]
                          [0 0]      [0]
             f(x1) = [1 0] x1 + [0]
                     [0 0]      [1]
             0() = [0]
                   [0]
             mark(x1) = [1 0] x1 + [0]
                        [0 0]      [0]
             cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                            [0 0]      [0 0]      [0]
             s(x1) = [1 0] x1 + [0]
                     [0 0]      [0]
             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)) -> active(f(mark(X)))
               , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
               , mark(s(X)) -> active(s(mark(X)))
               , mark(p(X)) -> active(p(mark(X)))
               , cons(X1, mark(X2)) -> cons(X1, X2)
               , cons(X1, active(X2)) -> cons(X1, X2)}
            Weak Trs:
              {  active(f(0())) -> mark(cons(0(), f(s(0()))))
               , active(f(s(0()))) -> mark(f(p(s(0()))))
               , mark(0()) -> active(0())
               , active(p(s(X))) -> mark(X)
               , f(mark(X)) -> f(X)
               , f(active(X)) -> f(X)
               , cons(mark(X1), X2) -> cons(X1, X2)
               , cons(active(X1), X2) -> cons(X1, X2)
               , s(mark(X)) -> s(X)
               , s(active(X)) -> s(X)
               , p(mark(X)) -> p(X)
               , p(active(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:
              {  cons(X1, mark(X2)) -> cons(X1, X2)
               , cons(X1, active(X2)) -> cons(X1, X2)}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {},
                Uargs(cons) = {1}, Uargs(s) = {1}, Uargs(p) = {1}
              We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
              Interpretation Functions:
               active(x1) = [1 0] x1 + [0]
                            [0 1]      [1]
               f(x1) = [1 0] x1 + [0]
                       [0 0]      [0]
               0() = [0]
                     [0]
               mark(x1) = [1 0] x1 + [0]
                          [0 1]      [1]
               cons(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
                              [0 0]      [0 0]      [0]
               s(x1) = [1 0] x1 + [0]
                       [0 1]      [0]
               p(x1) = [1 0] x1 + [0]
                       [0 1]      [0]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  mark(f(X)) -> active(f(mark(X)))
                 , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
                 , mark(s(X)) -> active(s(mark(X)))
                 , mark(p(X)) -> active(p(mark(X)))}
              Weak Trs:
                {  cons(X1, mark(X2)) -> cons(X1, X2)
                 , cons(X1, active(X2)) -> cons(X1, X2)
                 , active(f(0())) -> mark(cons(0(), f(s(0()))))
                 , active(f(s(0()))) -> mark(f(p(s(0()))))
                 , mark(0()) -> active(0())
                 , active(p(s(X))) -> mark(X)
                 , f(mark(X)) -> f(X)
                 , f(active(X)) -> f(X)
                 , cons(mark(X1), X2) -> cons(X1, X2)
                 , cons(active(X1), X2) -> cons(X1, X2)
                 , s(mark(X)) -> s(X)
                 , s(active(X)) -> s(X)
                 , p(mark(X)) -> p(X)
                 , p(active(X)) -> p(X)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              We consider the following Problem:
              
                Strict Trs:
                  {  mark(f(X)) -> active(f(mark(X)))
                   , mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
                   , mark(s(X)) -> active(s(mark(X)))
                   , mark(p(X)) -> active(p(mark(X)))}
                Weak Trs:
                  {  cons(X1, mark(X2)) -> cons(X1, X2)
                   , cons(X1, active(X2)) -> cons(X1, X2)
                   , active(f(0())) -> mark(cons(0(), f(s(0()))))
                   , active(f(s(0()))) -> mark(f(p(s(0()))))
                   , mark(0()) -> active(0())
                   , active(p(s(X))) -> mark(X)
                   , f(mark(X)) -> f(X)
                   , f(active(X)) -> f(X)
                   , cons(mark(X1), X2) -> cons(X1, X2)
                   , cons(active(X1), X2) -> cons(X1, X2)
                   , s(mark(X)) -> s(X)
                   , s(active(X)) -> s(X)
                   , p(mark(X)) -> p(X)
                   , p(active(X)) -> p(X)}
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(?,O(n^1))
              
              Proof:
                The problem is match-bounded by 0.
                The enriched problem is compatible with the following automaton:
                {  active_0(2) -> 1
                 , f_0(2) -> 1
                 , 0_0() -> 2
                 , mark_0(2) -> 1
                 , cons_0(2, 2) -> 1
                 , s_0(2) -> 1
                 , p_0(2) -> 1}

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