We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs: {a(b(a(x))) -> b(a(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(b(a(x))) -> b(a(b(x)))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(a) = {}, Uargs(b) = {1}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       a(x1) = [0 3] x1 + [0]
               [0 1]      [3]
       b(x1) = [1 0] x1 + [1]
               [0 1]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Weak Trs: {a(b(a(x))) -> b(a(b(x)))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(O(1),O(1))
    
    Proof:
      We consider the following Problem:
      
        Weak Trs: {a(b(a(x))) -> b(a(b(x)))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(O(1),O(1))
      
      Proof:
        Empty rules are trivially bounded

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