We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  rev(nil()) -> nil()
       , rev(rev(x)) -> x
       , rev(++(x, y)) -> ++(rev(y), rev(x))
       , ++(nil(), y) -> y
       , ++(x, nil()) -> x
       , ++(.(x, y), z) -> .(x, ++(y, z))
       , ++(x, ++(y, z)) -> ++(++(x, y), z)
       , make(x) -> .(x, nil())}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  rev(nil()) -> nil()
       , rev(rev(x)) -> x}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(rev) = {}, Uargs(++) = {1, 2}, Uargs(.) = {2},
        Uargs(make) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       rev(x1) = [1 0] x1 + [1]
                 [0 1]      [1]
       nil() = [0]
               [0]
       ++(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                    [0 0]      [0 0]      [0]
       .(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                   [0 0]      [0 0]      [1]
       make(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:
        {  rev(++(x, y)) -> ++(rev(y), rev(x))
         , ++(nil(), y) -> y
         , ++(x, nil()) -> x
         , ++(.(x, y), z) -> .(x, ++(y, z))
         , ++(x, ++(y, z)) -> ++(++(x, y), z)
         , make(x) -> .(x, nil())}
      Weak Trs:
        {  rev(nil()) -> nil()
         , rev(rev(x)) -> x}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component:
        {  ++(nil(), y) -> y
         , ++(x, nil()) -> x}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(rev) = {}, Uargs(++) = {1, 2}, Uargs(.) = {2},
          Uargs(make) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         rev(x1) = [0 1] x1 + [1]
                   [1 0]      [0]
         nil() = [0]
                 [0]
         ++(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                      [0 1]      [0 1]      [0]
         .(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                     [0 0]      [0 0]      [0]
         make(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:
          {  rev(++(x, y)) -> ++(rev(y), rev(x))
           , ++(.(x, y), z) -> .(x, ++(y, z))
           , ++(x, ++(y, z)) -> ++(++(x, y), z)
           , make(x) -> .(x, nil())}
        Weak Trs:
          {  ++(nil(), y) -> y
           , ++(x, nil()) -> x
           , rev(nil()) -> nil()
           , rev(rev(x)) -> x}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {make(x) -> .(x, nil())}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(rev) = {}, Uargs(++) = {1, 2}, Uargs(.) = {2},
            Uargs(make) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           rev(x1) = [0 1] x1 + [1]
                     [1 0]      [0]
           nil() = [0]
                   [0]
           ++(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                        [0 1]      [0 1]      [0]
           .(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                       [0 0]      [0 0]      [0]
           make(x1) = [0 0] x1 + [2]
                      [0 0]      [0]
        
        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))
             , ++(x, ++(y, z)) -> ++(++(x, y), z)}
          Weak Trs:
            {  make(x) -> .(x, nil())
             , ++(nil(), y) -> y
             , ++(x, nil()) -> x
             , rev(nil()) -> nil()
             , rev(rev(x)) -> x}
          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))
               , ++(x, ++(y, z)) -> ++(++(x, y), z)}
            Weak Trs:
              {  make(x) -> .(x, nil())
               , ++(nil(), y) -> y
               , ++(x, nil()) -> x
               , rev(nil()) -> nil()
               , rev(rev(x)) -> 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:
            {  rev_0(2) -> 1
             , nil_0() -> 1
             , nil_0() -> 2
             , nil_0() -> 3
             , nil_1() -> 3
             , ++_0(2, 2) -> 1
             , ++_1(2, 2) -> 3
             , ._0(2, 2) -> 1
             , ._0(2, 2) -> 2
             , ._0(2, 2) -> 3
             , ._1(2, 3) -> 1
             , ._1(2, 3) -> 3
             , make_0(2) -> 1}

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