We consider the following Problem:

  Strict Trs:
    {  merge(x, nil()) -> x
     , merge(nil(), y) -> y
     , merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v())))
     , merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v()))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  merge(x, nil()) -> x
       , merge(nil(), y) -> y
       , merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v())))
       , merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v()))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {merge(nil(), y) -> y}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(merge) = {}, Uargs(++) = {2}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       merge(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                       [0 0]      [0 1]      [1]
       nil() = [0]
               [0]
       ++(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                    [0 0]      [0 0]      [0]
       u() = [0]
             [0]
       v() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  merge(x, nil()) -> x
         , merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v())))
         , merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v()))}
      Weak Trs: {merge(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: {merge(x, nil()) -> x}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(merge) = {}, Uargs(++) = {2}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         merge(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                         [0 1]      [1 1]      [0]
         nil() = [1]
                 [0]
         ++(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
                      [0 0]      [0 0]      [0]
         u() = [0]
               [0]
         v() = [2]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v())))
           , merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v()))}
        Weak Trs:
          {  merge(x, nil()) -> x
           , merge(nil(), y) -> y}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs:
            {  merge(++(x, y), ++(u(), v())) -> ++(x, merge(y, ++(u(), v())))
             , merge(++(x, y), ++(u(), v())) -> ++(u(), merge(++(x, y), v()))}
          Weak Trs:
            {  merge(x, nil()) -> x
             , merge(nil(), y) -> y}
          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:
          {  merge_0(2, 2) -> 1
           , merge_1(2, 4) -> 3
           , merge_1(8, 6) -> 7
           , nil_0() -> 1
           , nil_0() -> 2
           , ++_0(2, 2) -> 1
           , ++_0(2, 2) -> 2
           , ++_1(2, 2) -> 8
           , ++_1(2, 3) -> 1
           , ++_1(2, 3) -> 3
           , ++_1(5, 6) -> 3
           , ++_1(5, 6) -> 4
           , ++_1(5, 7) -> 1
           , ++_1(5, 7) -> 3
           , u_0() -> 1
           , u_0() -> 2
           , u_1() -> 5
           , v_0() -> 1
           , v_0() -> 2
           , v_1() -> 6}

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