We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(g(x)) -> g(g(f(x)))
       , f(g(x)) -> g(g(g(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)) -> g(g(g(x)))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {}, Uargs(g) = {1}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1) = [1 0] x1 + [1]
               [0 0]      [1]
       g(x1) = [1 0] x1 + [0]
               [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {f(g(x)) -> g(g(f(x)))}
      Weak Trs: {f(g(x)) -> g(g(g(x)))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      We consider the following Problem:
      
        Strict Trs: {f(g(x)) -> g(g(f(x)))}
        Weak Trs: {f(g(x)) -> g(g(g(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:
        {  f_0(2) -> 1
         , f_1(2) -> 4
         , g_0(2) -> 1
         , g_0(2) -> 2
         , g_1(2) -> 4
         , g_1(3) -> 1
         , g_1(3) -> 4
         , g_1(4) -> 3}

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