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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          div#_A(x1,x2) = x1 + x2 + 1
          e_A() = 2
          i#_A(x1) = x1 + 1
          div_A(x1,x2) = x1 + x2 + 2
          i_A(x1) = x1
  
    precedence:
    
      i > e = div > div# = i#
    
    partial status:
  
      pi(div#) = []
      pi(e) = []
      pi(i#) = []
      pi(div) = [1]
      pi(i) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- 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) -> i#(X)
p4: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- 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) -> div#(Y,div(i(X),Z))
p4: 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 set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          div#_A(x1,x2) = x1 + x2 + 2
          e_A() = 1
          i#_A(x1) = x1 + 2
          div_A(x1,x2) = x1 + x2 + 3
          i_A(x1) = x1
  
    precedence:
    
      div# = i# > div > e = i
    
    partial status:
  
      pi(div#) = [1]
      pi(e) = []
      pi(i#) = []
      pi(div) = []
      pi(i) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: i#(div(X,Y)) -> div#(Y,X)
p2: div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))
p3: 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}


-- Reduction pair.

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

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          i#_A(x1) = x1 + 1
          div_A(x1,x2) = x1 + x2 + 2
          div#_A(x1,x2) = x1 + x2 + 1
          i_A(x1) = x1
          e_A() = 1
  
    precedence:
    
      i# > div = div# = i = e
    
    partial status:
  
      pi(i#) = []
      pi(div) = []
      pi(div#) = [1]
      pi(i) = [1]
      pi(e) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: div#(div(X,Y),Z) -> i#(X)
p2: 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 estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          div#_A(x1,x2) = x1 + x2 + 1
          div_A(x1,x2) = x1 + x2 + 1
          i_A(x1) = x1
          e_A() = 1
  
    precedence:
    
      div# > div = i = e
    
    partial status:
  
      pi(div#) = [1]
      pi(div) = []
      pi(i) = []
      pi(e) = []

The next rules are strictly ordered:

  p1

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