We consider the following Problem:

  Strict Trs:
    {  g(0(), f(x, x)) -> x
     , g(x, s(y)) -> g(f(x, y), 0())
     , g(s(x), y) -> g(f(x, y), 0())
     , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0()))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  g(0(), f(x, x)) -> x
       , g(x, s(y)) -> g(f(x, y), 0())
       , g(s(x), y) -> g(f(x, y), 0())
       , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0()))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {g(x, s(y)) -> g(f(x, y), 0())}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(g) = {}, Uargs(f) = {1, 2}, Uargs(s) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       g(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
                   [0 0]      [0 0]      [1]
       0() = [0]
             [0]
       f(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                   [0 0]      [0 0]      [1]
       s(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:
        {  g(0(), f(x, x)) -> x
         , g(s(x), y) -> g(f(x, y), 0())
         , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0()))}
      Weak Trs: {g(x, s(y)) -> g(f(x, y), 0())}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {g(0(), f(x, x)) -> x}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(g) = {}, Uargs(f) = {1, 2}, Uargs(s) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         g(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                     [0 0]      [0 1]      [1]
         0() = [0]
               [2]
         f(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                     [0 1]      [0 1]      [1]
         s(x1) = [0 0] x1 + [0]
                 [0 0]      [2]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  g(s(x), y) -> g(f(x, y), 0())
           , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0()))}
        Weak Trs:
          {  g(0(), f(x, x)) -> x
           , g(x, s(y)) -> g(f(x, y), 0())}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {g(s(x), y) -> g(f(x, y), 0())}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(g) = {}, Uargs(f) = {1, 2}, Uargs(s) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           g(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                       [0 0]      [0 1]      [0]
           0() = [0]
                 [0]
           f(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                       [0 1]      [0 1]      [1]
           s(x1) = [1 0] x1 + [1]
                   [0 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs: {g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0()))}
          Weak Trs:
            {  g(s(x), y) -> g(f(x, y), 0())
             , g(0(), f(x, x)) -> x
             , g(x, s(y)) -> g(f(x, y), 0())}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          We consider the following Problem:
          
            Strict Trs: {g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0()))}
            Weak Trs:
              {  g(s(x), y) -> g(f(x, y), 0())
               , g(0(), f(x, x)) -> x
               , g(x, s(y)) -> g(f(x, y), 0())}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The problem is match-bounded by 2.
            The enriched problem is compatible with the following automaton:
            {  g_0(2, 2) -> 1
             , g_1(2, 5) -> 3
             , g_1(2, 6) -> 4
             , g_1(5, 6) -> 4
             , g_1(6, 6) -> 4
             , g_1(7, 6) -> 1
             , g_1(13, 6) -> 8
             , g_1(14, 6) -> 9
             , g_2(2, 10) -> 8
             , g_2(2, 11) -> 9
             , g_2(5, 11) -> 12
             , g_2(6, 11) -> 12
             , g_2(10, 11) -> 12
             , g_2(11, 11) -> 12
             , 0_0() -> 1
             , 0_0() -> 2
             , 0_1() -> 5
             , 0_1() -> 6
             , 0_2() -> 10
             , 0_2() -> 11
             , f_0(2, 2) -> 1
             , f_0(2, 2) -> 2
             , f_1(2, 2) -> 7
             , f_1(2, 5) -> 1
             , f_1(2, 5) -> 2
             , f_1(2, 6) -> 1
             , f_1(2, 6) -> 2
             , f_1(2, 10) -> 13
             , f_1(2, 11) -> 14
             , f_1(3, 4) -> 1
             , f_1(4, 4) -> 3
             , f_1(4, 4) -> 4
             , f_1(4, 4) -> 8
             , f_1(4, 4) -> 9
             , f_2(8, 9) -> 1
             , f_2(9, 12) -> 3
             , f_2(9, 12) -> 4
             , f_2(9, 12) -> 8
             , f_2(9, 12) -> 9
             , s_0(2) -> 1
             , s_0(2) -> 2}

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