YES

We show the termination of the TRS R:

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

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(X,e()) -> i#(X)
p2: i#(div(X,Y)) -> div#(Y,X)
p3: div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))
p4: div#(div(X,Y),Z) -> div#(i(X),Z)
p5: div#(div(X,Y),Z) -> i#(X)

and R consists of:

r1: div(X,e()) -> i(X)
r2: i(div(X,Y)) -> div(Y,X)
r3: div(div(X,Y),Z) -> div(Y,div(i(X),Z))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(X,e()) -> i#(X)
p2: i#(div(X,Y)) -> div#(Y,X)
p3: div#(div(X,Y),Z) -> i#(X)
p4: div#(div(X,Y),Z) -> div#(i(X),Z)
p5: div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))

and R consists of:

r1: div(X,e()) -> i(X)
r2: i(div(X,Y)) -> div(Y,X)
r3: div(div(X,Y),Z) -> div(Y,div(i(X),Z))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          div#_A(x1,x2) = max{x1 + 12, x2 + 6}
          e_A = 3
          i#_A(x1) = x1 + 12
          div_A(x1,x2) = max{3, x1 + 2, x2}
          i_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        div# = e = i# = div = i
      
      partial status:
    
        pi(div#) = [1]
        pi(e) = []
        pi(i#) = [1]
        pi(div) = [1, 2]
        pi(i) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          div#_A(x1,x2) = max{6, x1 + 3}
          e_A = 0
          i#_A(x1) = max{2, x1}
          div_A(x1,x2) = max{x1 + 7, x2 + 9}
          i_A(x1) = 2
    
      precedence:
      
        div# = e = i# = div = i
      
      partial status:
    
        pi(div#) = []
        pi(e) = []
        pi(i#) = [1]
        pi(div) = [2]
        pi(i) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

We remove them from the problem.  Then no dependency pair remains.