We consider the following Problem:

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

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

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

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