We consider the following Problem:

  Strict Trs:
    {  div(X, e()) -> i(X)
     , i(div(X, Y)) -> div(Y, X)
     , div(div(X, Y), Z) -> div(Y, div(i(X), Z))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  div(X, e()) -> i(X)
       , i(div(X, Y)) -> div(Y, X)
       , div(div(X, Y), Z) -> div(Y, div(i(X), 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(div(X, Y)) -> div(Y, X)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(div) = {1, 2}, Uargs(i) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       div(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                     [0 0]      [0 0]      [1]
       e() = [0]
             [0]
       i(x1) = [1 0] x1 + [2]
               [1 0]      [3]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  div(X, e()) -> i(X)
         , div(div(X, Y), Z) -> div(Y, div(i(X), Z))}
      Weak Trs: {i(div(X, Y)) -> div(Y, X)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {div(X, e()) -> i(X)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(div) = {1, 2}, Uargs(i) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         div(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                       [0 0]      [0 0]      [1]
         e() = [0]
               [0]
         i(x1) = [1 0] x1 + [0]
                 [0 0]      [1]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {div(div(X, Y), Z) -> div(Y, div(i(X), Z))}
        Weak Trs:
          {  div(X, e()) -> i(X)
           , i(div(X, Y)) -> div(Y, X)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs: {div(div(X, Y), Z) -> div(Y, div(i(X), Z))}
          Weak Trs:
            {  div(X, e()) -> i(X)
             , i(div(X, Y)) -> div(Y, 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:
          {  div_0(2, 2) -> 1
           , e_0() -> 2
           , i_0(2) -> 1}

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