We consider the following Problem:

  Strict Trs:
    {  flatten(nil()) -> nil()
     , flatten(unit(x)) -> flatten(x)
     , flatten(++(x, y)) -> ++(flatten(x), flatten(y))
     , flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
     , flatten(flatten(x)) -> flatten(x)
     , rev(nil()) -> nil()
     , rev(unit(x)) -> unit(x)
     , rev(++(x, y)) -> ++(rev(y), rev(x))
     , rev(rev(x)) -> x
     , ++(x, nil()) -> x
     , ++(nil(), y) -> y
     , ++(++(x, y), z) -> ++(x, ++(y, z))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  flatten(nil()) -> nil()
       , flatten(unit(x)) -> flatten(x)
       , flatten(++(x, y)) -> ++(flatten(x), flatten(y))
       , flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
       , flatten(flatten(x)) -> flatten(x)
       , rev(nil()) -> nil()
       , rev(unit(x)) -> unit(x)
       , rev(++(x, y)) -> ++(rev(y), rev(x))
       , rev(rev(x)) -> x
       , ++(x, nil()) -> x
       , ++(nil(), y) -> y
       , ++(++(x, y), z) -> ++(x, ++(y, z))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  flatten(nil()) -> nil()
       , rev(nil()) -> nil()
       , rev(unit(x)) -> unit(x)
       , rev(rev(x)) -> x
       , ++(x, nil()) -> x
       , ++(nil(), y) -> y}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(flatten) = {}, Uargs(unit) = {}, Uargs(++) = {1, 2},
        Uargs(rev) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       flatten(x1) = [0 0] x1 + [1]
                     [0 0]      [1]
       nil() = [0]
               [0]
       unit(x1) = [0 0] x1 + [0]
                  [0 0]      [0]
       ++(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                    [0 1]      [0 1]      [1]
       rev(x1) = [0 1] x1 + [1]
                 [1 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  flatten(unit(x)) -> flatten(x)
         , flatten(++(x, y)) -> ++(flatten(x), flatten(y))
         , flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
         , flatten(flatten(x)) -> flatten(x)
         , rev(++(x, y)) -> ++(rev(y), rev(x))
         , ++(++(x, y), z) -> ++(x, ++(y, z))}
      Weak Trs:
        {  flatten(nil()) -> nil()
         , rev(nil()) -> nil()
         , rev(unit(x)) -> unit(x)
         , rev(rev(x)) -> x
         , ++(x, nil()) -> x
         , ++(nil(), y) -> y}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component:
        {  flatten(unit(x)) -> flatten(x)
         , flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
         , flatten(flatten(x)) -> flatten(x)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(flatten) = {}, Uargs(unit) = {}, Uargs(++) = {1, 2},
          Uargs(rev) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         flatten(x1) = [0 2] x1 + [0]
                       [0 1]      [1]
         nil() = [0]
                 [3]
         unit(x1) = [0 0] x1 + [0]
                    [0 1]      [2]
         ++(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                      [0 1]      [0 1]      [0]
         rev(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:
          {  flatten(++(x, y)) -> ++(flatten(x), flatten(y))
           , rev(++(x, y)) -> ++(rev(y), rev(x))
           , ++(++(x, y), z) -> ++(x, ++(y, z))}
        Weak Trs:
          {  flatten(unit(x)) -> flatten(x)
           , flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
           , flatten(flatten(x)) -> flatten(x)
           , flatten(nil()) -> nil()
           , rev(nil()) -> nil()
           , rev(unit(x)) -> unit(x)
           , rev(rev(x)) -> x
           , ++(x, nil()) -> x
           , ++(nil(), y) -> y}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {flatten(++(x, y)) -> ++(flatten(x), flatten(y))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(flatten) = {}, Uargs(unit) = {}, Uargs(++) = {1, 2},
            Uargs(rev) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           flatten(x1) = [0 1] x1 + [0]
                         [0 1]      [0]
           nil() = [0]
                   [3]
           unit(x1) = [1 0] x1 + [0]
                      [0 1]      [3]
           ++(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                        [0 1]      [0 1]      [1]
           rev(x1) = [1 0] x1 + [0]
                     [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))
             , ++(++(x, y), z) -> ++(x, ++(y, z))}
          Weak Trs:
            {  flatten(++(x, y)) -> ++(flatten(x), flatten(y))
             , flatten(unit(x)) -> flatten(x)
             , flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
             , flatten(flatten(x)) -> flatten(x)
             , flatten(nil()) -> nil()
             , rev(nil()) -> nil()
             , rev(unit(x)) -> unit(x)
             , rev(rev(x)) -> x
             , ++(x, nil()) -> x
             , ++(nil(), y) -> y}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          We consider the following Problem:
          
            Strict Trs:
              {  rev(++(x, y)) -> ++(rev(y), rev(x))
               , ++(++(x, y), z) -> ++(x, ++(y, z))}
            Weak Trs:
              {  flatten(++(x, y)) -> ++(flatten(x), flatten(y))
               , flatten(unit(x)) -> flatten(x)
               , flatten(++(unit(x), y)) -> ++(flatten(x), flatten(y))
               , flatten(flatten(x)) -> flatten(x)
               , flatten(nil()) -> nil()
               , rev(nil()) -> nil()
               , rev(unit(x)) -> unit(x)
               , rev(rev(x)) -> x
               , ++(x, nil()) -> x
               , ++(nil(), y) -> y}
            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:
            {  flatten_0(2) -> 1
             , nil_0() -> 1
             , nil_0() -> 2
             , unit_0(2) -> 1
             , unit_0(2) -> 2
             , ++_0(2, 2) -> 1
             , rev_0(2) -> 1}

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