We consider the following Problem:

  Strict Trs:
    {  +(x, +(y, z)) -> +(+(x, y), z)
     , +(*(x, y), +(x, z)) -> *(x, +(y, z))
     , +(*(x, y), +(*(x, z), u)) -> +(*(x, +(y, z)), u)}
  StartTerms: basic terms
  Strategy: innermost

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

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

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