We consider the following Problem:

  Strict Trs:
    {  active(f(X, X)) -> mark(f(a(), b()))
     , active(b()) -> mark(a())
     , active(f(X1, X2)) -> f(active(X1), X2)
     , f(mark(X1), X2) -> mark(f(X1, X2))
     , proper(f(X1, X2)) -> f(proper(X1), proper(X2))
     , proper(a()) -> ok(a())
     , proper(b()) -> ok(b())
     , f(ok(X1), ok(X2)) -> ok(f(X1, X2))
     , top(mark(X)) -> top(proper(X))
     , top(ok(X)) -> top(active(X))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  active(f(X, X)) -> mark(f(a(), b()))
       , active(b()) -> mark(a())
       , active(f(X1, X2)) -> f(active(X1), X2)
       , f(mark(X1), X2) -> mark(f(X1, X2))
       , proper(f(X1, X2)) -> f(proper(X1), proper(X2))
       , proper(a()) -> ok(a())
       , proper(b()) -> ok(b())
       , f(ok(X1), ok(X2)) -> ok(f(X1, X2))
       , top(mark(X)) -> top(proper(X))
       , top(ok(X)) -> top(active(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(ok(X1), ok(X2)) -> ok(f(X1, X2))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(active) = {}, Uargs(f) = {1, 2}, Uargs(mark) = {1},
        Uargs(proper) = {}, Uargs(ok) = {1}, Uargs(top) = {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, x2) = [1 0] x1 + [1 0] x2 + [0]
                   [0 0]      [0 0]      [1]
       mark(x1) = [1 0] x1 + [1]
                  [0 0]      [1]
       a() = [0]
             [0]
       b() = [0]
             [0]
       proper(x1) = [0 0] x1 + [1]
                    [0 0]      [1]
       ok(x1) = [1 0] x1 + [1]
                [0 0]      [1]
       top(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(X, X)) -> mark(f(a(), b()))
         , active(b()) -> mark(a())
         , active(f(X1, X2)) -> f(active(X1), X2)
         , f(mark(X1), X2) -> mark(f(X1, X2))
         , proper(f(X1, X2)) -> f(proper(X1), proper(X2))
         , proper(a()) -> ok(a())
         , proper(b()) -> ok(b())
         , top(mark(X)) -> top(proper(X))
         , top(ok(X)) -> top(active(X))}
      Weak Trs: {f(ok(X1), ok(X2)) -> ok(f(X1, X2))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component:
        {  active(b()) -> mark(a())
         , top(ok(X)) -> top(active(X))}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(active) = {}, Uargs(f) = {1, 2}, Uargs(mark) = {1},
          Uargs(proper) = {}, Uargs(ok) = {1}, Uargs(top) = {1}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         active(x1) = [1 0] x1 + [0]
                      [0 0]      [1]
         f(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                     [0 0]      [0 0]      [1]
         mark(x1) = [1 0] x1 + [0]
                    [0 0]      [1]
         a() = [0]
               [0]
         b() = [1]
               [0]
         proper(x1) = [0 0] x1 + [0]
                      [0 0]      [1]
         ok(x1) = [1 0] x1 + [1]
                  [0 0]      [1]
         top(x1) = [1 0] x1 + [1]
                   [0 1]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  active(f(X, X)) -> mark(f(a(), b()))
           , active(f(X1, X2)) -> f(active(X1), X2)
           , f(mark(X1), X2) -> mark(f(X1, X2))
           , proper(f(X1, X2)) -> f(proper(X1), proper(X2))
           , proper(a()) -> ok(a())
           , proper(b()) -> ok(b())
           , top(mark(X)) -> top(proper(X))}
        Weak Trs:
          {  active(b()) -> mark(a())
           , top(ok(X)) -> top(active(X))
           , f(ok(X1), ok(X2)) -> ok(f(X1, X2))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component:
          {  proper(a()) -> ok(a())
           , proper(b()) -> ok(b())}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(active) = {}, Uargs(f) = {1, 2}, Uargs(mark) = {1},
            Uargs(proper) = {}, Uargs(ok) = {1}, Uargs(top) = {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, x2) = [1 0] x1 + [1 0] x2 + [0]
                       [0 0]      [0 0]      [1]
           mark(x1) = [1 0] x1 + [1]
                      [0 0]      [1]
           a() = [0]
                 [0]
           b() = [0]
                 [0]
           proper(x1) = [0 0] x1 + [3]
                        [0 0]      [0]
           ok(x1) = [1 0] x1 + [1]
                    [0 0]      [0]
           top(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(X, X)) -> mark(f(a(), b()))
             , active(f(X1, X2)) -> f(active(X1), X2)
             , f(mark(X1), X2) -> mark(f(X1, X2))
             , proper(f(X1, X2)) -> f(proper(X1), proper(X2))
             , top(mark(X)) -> top(proper(X))}
          Weak Trs:
            {  proper(a()) -> ok(a())
             , proper(b()) -> ok(b())
             , active(b()) -> mark(a())
             , top(ok(X)) -> top(active(X))
             , f(ok(X1), ok(X2)) -> ok(f(X1, X2))}
          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(X, X)) -> mark(f(a(), b()))}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(active) = {}, Uargs(f) = {1, 2}, Uargs(mark) = {1},
              Uargs(proper) = {}, Uargs(ok) = {1}, Uargs(top) = {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, x2) = [1 0] x1 + [1 0] x2 + [0]
                         [0 0]      [0 0]      [1]
             mark(x1) = [1 0] x1 + [0]
                        [0 0]      [1]
             a() = [0]
                   [0]
             b() = [0]
                   [0]
             proper(x1) = [0 0] x1 + [1]
                          [0 0]      [1]
             ok(x1) = [1 0] x1 + [1]
                      [0 0]      [1]
             top(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(X1, X2)) -> f(active(X1), X2)
               , f(mark(X1), X2) -> mark(f(X1, X2))
               , proper(f(X1, X2)) -> f(proper(X1), proper(X2))
               , top(mark(X)) -> top(proper(X))}
            Weak Trs:
              {  active(f(X, X)) -> mark(f(a(), b()))
               , proper(a()) -> ok(a())
               , proper(b()) -> ok(b())
               , active(b()) -> mark(a())
               , top(ok(X)) -> top(active(X))
               , f(ok(X1), ok(X2)) -> ok(f(X1, X2))}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The weightgap principle applies, where following rules are oriented strictly:
            
            TRS Component: {top(mark(X)) -> top(proper(X))}
            
            Interpretation of nonconstant growth:
            -------------------------------------
              The following argument positions are usable:
                Uargs(active) = {}, Uargs(f) = {1, 2}, Uargs(mark) = {1},
                Uargs(proper) = {}, Uargs(ok) = {1}, Uargs(top) = {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, x2) = [1 0] x1 + [1 0] x2 + [0]
                           [0 1]      [0 0]      [2]
               mark(x1) = [1 0] x1 + [1]
                          [0 0]      [0]
               a() = [0]
                     [0]
               b() = [1]
                     [0]
               proper(x1) = [1 0] x1 + [0]
                            [0 0]      [0]
               ok(x1) = [1 0] x1 + [0]
                        [0 1]      [0]
               top(x1) = [1 1] x1 + [0]
                         [0 0]      [1]
            
            The strictly oriented rules are moved into the weak component.
            
            We consider the following Problem:
            
              Strict Trs:
                {  active(f(X1, X2)) -> f(active(X1), X2)
                 , f(mark(X1), X2) -> mark(f(X1, X2))
                 , proper(f(X1, X2)) -> f(proper(X1), proper(X2))}
              Weak Trs:
                {  top(mark(X)) -> top(proper(X))
                 , active(f(X, X)) -> mark(f(a(), b()))
                 , proper(a()) -> ok(a())
                 , proper(b()) -> ok(b())
                 , active(b()) -> mark(a())
                 , top(ok(X)) -> top(active(X))
                 , f(ok(X1), ok(X2)) -> ok(f(X1, X2))}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(?,O(n^1))
            
            Proof:
              We consider the following Problem:
              
                Strict Trs:
                  {  active(f(X1, X2)) -> f(active(X1), X2)
                   , f(mark(X1), X2) -> mark(f(X1, X2))
                   , proper(f(X1, X2)) -> f(proper(X1), proper(X2))}
                Weak Trs:
                  {  top(mark(X)) -> top(proper(X))
                   , active(f(X, X)) -> mark(f(a(), b()))
                   , proper(a()) -> ok(a())
                   , proper(b()) -> ok(b())
                   , active(b()) -> mark(a())
                   , top(ok(X)) -> top(active(X))
                   , f(ok(X1), ok(X2)) -> ok(f(X1, X2))}
                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:
                {  active_0(3) -> 1
                 , active_0(4) -> 1
                 , active_0(5) -> 1
                 , active_0(7) -> 1
                 , f_0(3, 3) -> 2
                 , f_0(3, 4) -> 2
                 , f_0(3, 5) -> 2
                 , f_0(3, 7) -> 2
                 , f_0(4, 3) -> 2
                 , f_0(4, 4) -> 2
                 , f_0(4, 5) -> 2
                 , f_0(4, 7) -> 2
                 , f_0(5, 3) -> 2
                 , f_0(5, 4) -> 2
                 , f_0(5, 5) -> 2
                 , f_0(5, 7) -> 2
                 , f_0(7, 3) -> 2
                 , f_0(7, 4) -> 2
                 , f_0(7, 5) -> 2
                 , f_0(7, 7) -> 2
                 , f_1(3, 3) -> 9
                 , f_1(3, 4) -> 9
                 , f_1(3, 5) -> 9
                 , f_1(3, 7) -> 9
                 , f_1(4, 3) -> 9
                 , f_1(4, 4) -> 9
                 , f_1(4, 5) -> 9
                 , f_1(4, 7) -> 9
                 , f_1(5, 3) -> 9
                 , f_1(5, 4) -> 9
                 , f_1(5, 5) -> 9
                 , f_1(5, 7) -> 9
                 , f_1(7, 3) -> 9
                 , f_1(7, 4) -> 9
                 , f_1(7, 5) -> 9
                 , f_1(7, 7) -> 9
                 , mark_0(3) -> 3
                 , mark_0(4) -> 1
                 , mark_0(4) -> 3
                 , mark_0(5) -> 3
                 , mark_0(7) -> 3
                 , mark_1(9) -> 2
                 , mark_1(9) -> 9
                 , a_0() -> 4
                 , b_0() -> 5
                 , proper_0(3) -> 6
                 , proper_0(4) -> 6
                 , proper_0(5) -> 6
                 , proper_0(7) -> 6
                 , ok_0(2) -> 2
                 , ok_0(3) -> 7
                 , ok_0(4) -> 6
                 , ok_0(4) -> 7
                 , ok_0(5) -> 6
                 , ok_0(5) -> 7
                 , ok_0(7) -> 7
                 , ok_1(9) -> 9
                 , top_0(1) -> 8
                 , top_0(3) -> 8
                 , top_0(4) -> 8
                 , top_0(5) -> 8
                 , top_0(6) -> 8
                 , top_0(7) -> 8}

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