YES

We show the termination of the TRS R:

  minus(X,|0|()) -> X
  minus(s(X),s(Y)) -> p(minus(X,Y))
  p(s(X)) -> X
  div(|0|(),s(Y)) -> |0|()
  div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))

-- SCC decomposition.

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

p1: minus#(s(X),s(Y)) -> p#(minus(X,Y))
p2: minus#(s(X),s(Y)) -> minus#(X,Y)
p3: div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))
p4: div#(s(X),s(Y)) -> minus#(X,Y)

and R consists of:

r1: minus(X,|0|()) -> X
r2: minus(s(X),s(Y)) -> p(minus(X,Y))
r3: p(s(X)) -> X
r4: div(|0|(),s(Y)) -> |0|()
r5: div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))

The estimated dependency graph contains the following SCCs:

  {p3}
  {p2}


-- Reduction pair.

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

p1: div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))

and R consists of:

r1: minus(X,|0|()) -> X
r2: minus(s(X),s(Y)) -> p(minus(X,Y))
r3: p(s(X)) -> X
r4: div(|0|(),s(Y)) -> |0|()
r5: div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      div#_A(x1,x2) = ((0,0,0),(1,0,0),(0,0,0)) x1
      s_A(x1) = ((1,0,0),(1,1,0),(0,1,0)) x1 + (2,1,1)
      minus_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,0)) x1 + (1,1,1)
      p_A(x1) = ((1,0,0),(1,1,0),(0,1,0)) x1 + (0,3,1)
      |0|_A() = (1,1,1)

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: minus#(s(X),s(Y)) -> minus#(X,Y)

and R consists of:

r1: minus(X,|0|()) -> X
r2: minus(s(X),s(Y)) -> p(minus(X,Y))
r3: p(s(X)) -> X
r4: div(|0|(),s(Y)) -> |0|()
r5: div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      minus#_A(x1,x2) = x1
      s_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1

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