We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  s(a()) -> a()
       , s(s(x)) -> x
       , s(f(x, y)) -> f(s(y), s(x))
       , s(g(x, y)) -> g(s(x), s(y))
       , f(x, a()) -> x
       , f(a(), y) -> y
       , f(g(x, y), g(u, v)) -> g(f(x, u), f(y, v))
       , g(a(), a()) -> a()}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  s(a()) -> a()
       , s(s(x)) -> x}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(s) = {}, Uargs(f) = {1, 2}, Uargs(g) = {1, 2}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       s(x1) = [1 0] x1 + [1]
               [0 1]      [1]
       a() = [0]
             [0]
       f(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                   [0 0]      [0 0]      [0]
       g(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                   [0 0]      [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  s(f(x, y)) -> f(s(y), s(x))
         , s(g(x, y)) -> g(s(x), s(y))
         , f(x, a()) -> x
         , f(a(), y) -> y
         , f(g(x, y), g(u, v)) -> g(f(x, u), f(y, v))
         , g(a(), a()) -> a()}
      Weak Trs:
        {  s(a()) -> a()
         , s(s(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:
        {  f(x, a()) -> x
         , f(a(), y) -> y
         , g(a(), a()) -> a()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(s) = {}, Uargs(f) = {1, 2}, Uargs(g) = {1, 2}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         s(x1) = [0 1] x1 + [1]
                 [1 0]      [0]
         a() = [0]
               [0]
         f(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                     [0 1]      [0 1]      [0]
         g(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                     [0 1]      [0 1]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  s(f(x, y)) -> f(s(y), s(x))
           , s(g(x, y)) -> g(s(x), s(y))
           , f(g(x, y), g(u, v)) -> g(f(x, u), f(y, v))}
        Weak Trs:
          {  f(x, a()) -> x
           , f(a(), y) -> y
           , g(a(), a()) -> a()
           , s(a()) -> a()
           , s(s(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: {f(g(x, y), g(u, v)) -> g(f(x, u), f(y, v))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(s) = {}, Uargs(f) = {1, 2}, Uargs(g) = {1, 2}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           s(x1) = [0 1] x1 + [1]
                   [1 0]      [0]
           a() = [0]
                 [0]
           f(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                       [0 1]      [0 1]      [0]
           g(x1, x2) = [1 0] x1 + [1 0] x2 + [3]
                       [0 1]      [0 1]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs:
            {  s(f(x, y)) -> f(s(y), s(x))
             , s(g(x, y)) -> g(s(x), s(y))}
          Weak Trs:
            {  f(g(x, y), g(u, v)) -> g(f(x, u), f(y, v))
             , f(x, a()) -> x
             , f(a(), y) -> y
             , g(a(), a()) -> a()
             , s(a()) -> a()
             , s(s(x)) -> x}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          We consider the following Problem:
          
            Strict Trs:
              {  s(f(x, y)) -> f(s(y), s(x))
               , s(g(x, y)) -> g(s(x), s(y))}
            Weak Trs:
              {  f(g(x, y), g(u, v)) -> g(f(x, u), f(y, v))
               , f(x, a()) -> x
               , f(a(), y) -> y
               , g(a(), a()) -> a()
               , s(a()) -> a()
               , s(s(x)) -> x}
            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:
            {  s_0(2) -> 1
             , a_0() -> 1
             , a_0() -> 2
             , f_0(2, 2) -> 1
             , g_0(2, 2) -> 1}

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