YES

We show the termination of the relative TRS R/S:

  R:
  minus(x,o()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  div(|0|(),s(y)) -> |0|()
  div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
  divL(x,nil()) -> x
  divL(x,cons(y,xs)) -> divL(div(x,y),xs)

  S:
  cons(x,cons(y,xs)) -> cons(y,cons(x,xs))

-- SCC decomposition.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)
p2: div#(s(x),s(y)) -> div#(minus(x,y),s(y))
p3: div#(s(x),s(y)) -> minus#(x,y)
p4: divL#(x,cons(y,xs)) -> divL#(div(x,y),xs)
p5: divL#(x,cons(y,xs)) -> div#(x,y)

and R consists of:

r1: minus(x,o()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: div(|0|(),s(y)) -> |0|()
r4: div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
r5: divL(x,nil()) -> x
r6: divL(x,cons(y,xs)) -> divL(div(x,y),xs)
r7: cons(x,cons(y,xs)) -> cons(y,cons(x,xs))

The estimated dependency graph contains the following SCCs:

  {p4}
  {p2}
  {p1}


-- Reduction pair.

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

p1: divL#(x,cons(y,xs)) -> divL#(div(x,y),xs)

and R consists of:

r1: minus(x,o()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: div(|0|(),s(y)) -> |0|()
r4: div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
r5: divL(x,nil()) -> x
r6: divL(x,cons(y,xs)) -> divL(div(x,y),xs)
r7: cons(x,cons(y,xs)) -> cons(y,cons(x,xs))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      divL#_A(x1,x2) = x2
      cons_A(x1,x2) = x1 + x2 + 1
      div_A(x1,x2) = 1
      minus_A(x1,x2) = x1 + 1
      o_A() = 1
      s_A(x1) = x1
      |0|_A() = 1
      divL_A(x1,x2) = x1 + x2
      nil_A() = 1

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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,o()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: div(|0|(),s(y)) -> |0|()
r4: div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
r5: divL(x,nil()) -> x
r6: divL(x,cons(y,xs)) -> divL(div(x,y),xs)
r7: cons(x,cons(y,xs)) -> cons(y,cons(x,xs))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      div#_A(x1,x2) = x1
      s_A(x1) = x1 + 1
      minus_A(x1,x2) = x1
      o_A() = 0
      div_A(x1,x2) = x1 + x2
      |0|_A() = 1
      divL_A(x1,x2) = x1 + x2
      nil_A() = 1
      cons_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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,o()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: div(|0|(),s(y)) -> |0|()
r4: div(s(x),s(y)) -> s(div(minus(x,y),s(y)))
r5: divL(x,nil()) -> x
r6: divL(x,cons(y,xs)) -> divL(div(x,y),xs)
r7: cons(x,cons(y,xs)) -> cons(y,cons(x,xs))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      minus#_A(x1,x2) = x1
      s_A(x1) = x1 + 1
      minus_A(x1,x2) = x1
      o_A() = 1
      div_A(x1,x2) = x1
      |0|_A() = 0
      divL_A(x1,x2) = x1 + x2
      nil_A() = 1
      cons_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p1

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