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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        div#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x2 + (0,0,6,0)
        e_A() = (1,1,1,1)
        i#_A(x1) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,0,0)) x1 + (0,6,0,5)
        div_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,1,1,0),(0,1,0,1)) x1 + ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (4,1,1,1)
        i_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,1,0,1)) x1 + (0,3,0,6)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        div# > div > i > e > i#
      
      argument filter:
    
        pi(div#) = [1, 2]
        pi(e) = []
        pi(i#) = []
        pi(div) = []
        pi(i) = 1
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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