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

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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