YES

Problem:
 div(X,e()) -> i(X)
 i(div(X,Y)) -> div(Y,X)
 div(div(X,Y),Z) -> div(Y,div(i(X),Z))

Proof:
 DP Processor:
  DPs:
   div#(X,e()) -> i#(X)
   i#(div(X,Y)) -> div#(Y,X)
   div#(div(X,Y),Z) -> i#(X)
   div#(div(X,Y),Z) -> div#(i(X),Z)
   div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))
  TRS:
   div(X,e()) -> i(X)
   i(div(X,Y)) -> div(Y,X)
   div(div(X,Y),Z) -> div(Y,div(i(X),Z))
  Matrix Interpretation Processor: dim=1
   
   usable rules:
    div(X,e()) -> i(X)
    i(div(X,Y)) -> div(Y,X)
    div(div(X,Y),Z) -> div(Y,div(i(X),Z))
   interpretation:
    [i#](x0) = 3/2x0,
    
    [div#](x0, x1) = 3/2x0 + 1/2x1,
    
    [i](x0) = 2x0,
    
    [div](x0, x1) = 2x0 + x1 + 1,
    
    [e] = 1
   orientation:
    div#(X,e()) = 3/2X + 1/2 >= 3/2X = i#(X)
    
    i#(div(X,Y)) = 3X + 3/2Y + 3/2 >= 1/2X + 3/2Y = div#(Y,X)
    
    div#(div(X,Y),Z) = 3X + 3/2Y + 1/2Z + 3/2 >= 3/2X = i#(X)
    
    div#(div(X,Y),Z) = 3X + 3/2Y + 1/2Z + 3/2 >= 3X + 1/2Z = div#(i(X),Z)
    
    div#(div(X,Y),Z) = 3X + 3/2Y + 1/2Z + 3/2 >= 2X + 3/2Y + 1/2Z + 1/2 = div#(Y,div(i(X),Z))
    
    div(X,e()) = 2X + 2 >= 2X = i(X)
    
    i(div(X,Y)) = 4X + 2Y + 2 >= X + 2Y + 1 = div(Y,X)
    
    div(div(X,Y),Z) = 4X + 2Y + Z + 3 >= 4X + 2Y + Z + 2 = div(Y,div(i(X),Z))
   problem:
    DPs:
     
    TRS:
     div(X,e()) -> i(X)
     i(div(X,Y)) -> div(Y,X)
     div(div(X,Y),Z) -> div(Y,div(i(X),Z))
   Qed