We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  a(a(x)) -> b(b(x))
       , b(b(a(x))) -> a(b(b(x)))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {a(a(x)) -> b(b(x))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       a(x1) = [1 0] x1 + [2]
               [0 1]      [2]
       b(x1) = [1 0] x1 + [0]
               [0 1]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {b(b(a(x))) -> a(b(b(x)))}
      Weak Trs: {a(a(x)) -> b(b(x))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      We consider the following Problem:
      
        Strict Trs: {b(b(a(x))) -> a(b(b(x)))}
        Weak Trs: {a(a(x)) -> b(b(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:
        {  a_0(2) -> 1
         , b_0(2) -> 1}

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