We consider the following Problem:

  Strict Trs:
    {  rev(a()) -> a()
     , rev(b()) -> b()
     , rev(++(x, y)) -> ++(rev(y), rev(x))
     , rev(++(x, x)) -> rev(x)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  rev(a()) -> a()
       , rev(b()) -> b()
       , rev(++(x, y)) -> ++(rev(y), rev(x))
       , rev(++(x, x)) -> rev(x)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  rev(a()) -> a()
       , rev(b()) -> b()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(rev) = {}, Uargs(++) = {1, 2}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       rev(x1) = [0 0] x1 + [1]
                 [1 0]      [1]
       a() = [0]
             [0]
       b() = [0]
             [0]
       ++(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                    [0 0]      [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  rev(++(x, y)) -> ++(rev(y), rev(x))
         , rev(++(x, x)) -> rev(x)}
      Weak Trs:
        {  rev(a()) -> a()
         , rev(b()) -> b()}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {rev(++(x, x)) -> rev(x)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(rev) = {}, Uargs(++) = {1, 2}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         rev(x1) = [0 1] x1 + [1]
                   [0 0]      [1]
         a() = [0]
               [0]
         b() = [0]
               [0]
         ++(x1, x2) = [1 0] x1 + [1 0] x2 + [3]
                      [0 1]      [0 1]      [1]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {rev(++(x, y)) -> ++(rev(y), rev(x))}
        Weak Trs:
          {  rev(++(x, x)) -> rev(x)
           , rev(a()) -> a()
           , rev(b()) -> b()}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {rev(++(x, y)) -> ++(rev(y), rev(x))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(rev) = {}, Uargs(++) = {1, 2}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           rev(x1) = [0 1] x1 + [0]
                     [0 1]      [0]
           a() = [0]
                 [0]
           b() = [0]
                 [0]
           ++(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                        [0 1]      [0 1]      [1]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Weak Trs:
            {  rev(++(x, y)) -> ++(rev(y), rev(x))
             , rev(++(x, x)) -> rev(x)
             , rev(a()) -> a()
             , rev(b()) -> b()}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          We consider the following Problem:
          
            Weak Trs:
              {  rev(++(x, y)) -> ++(rev(y), rev(x))
               , rev(++(x, x)) -> rev(x)
               , rev(a()) -> a()
               , rev(b()) -> b()}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            Empty rules are trivially bounded

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