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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      div#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,0,0,0)) x2 + (1,0,0,1)
      e_A() = (1,0,1,1)
      i#_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (0,3,4,0)
      div_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,0,1,1)) x2 + (2,0,1,0)
      i_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,0,0)) x1 + (0,0,4,4)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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