We consider the following Problem:

  Strict Trs:
    {  *(i(x), x) -> 1()
     , *(1(), y) -> y
     , *(x, 0()) -> 0()
     , *(*(x, y), z) -> *(x, *(y, z))}
  StartTerms: basic terms
  Strategy: innermost

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

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

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