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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        divL#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(0,1,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,1,1,1)) x2
        cons_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,1,1,1)) x1 + x2 + (9,1,1,1)
        div_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,0,1,1)) x1 + ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0)) x2 + (8,11,2,1)
        minus_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,0,0,0)) x1 + (0,1,1,1)
        o_A() = (1,1,1,1)
        s_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,0,0,0)) x1 + (2,1,1,1)
        |0|_A() = (1,1,1,1)
        divL_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,1,1,0)) x2 + (0,1,0,0)
        nil_A() = (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        nil > cons > divL > div > minus > o > |0| > s > divL#
      
      argument filter:
    
        pi(divL#) = 2
        pi(cons) = []
        pi(div) = 1
        pi(minus) = []
        pi(o) = []
        pi(s) = []
        pi(|0|) = []
        pi(divL) = [1]
        pi(nil) = []
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        div#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,1,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x2
        s_A(x1) = x1 + (0,1,1,1)
        minus_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,1,1)) x1 + (0,0,3,1)
        o_A() = (1,1,1,0)
        div_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x2 + (1,1,1,6)
        |0|_A() = (1,1,1,9)
        divL_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x2
        nil_A() = (1,1,1,1)
        cons_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + x2 + (2,2,2,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        cons > divL > div > minus > div# > o > nil > |0| > s
      
      argument filter:
    
        pi(div#) = 2
        pi(s) = []
        pi(minus) = []
        pi(o) = []
        pi(div) = []
        pi(|0|) = []
        pi(divL) = 1
        pi(nil) = []
        pi(cons) = []
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        minus#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2
        s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,1,0)) x1 + (1,1,2,1)
        minus_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (0,0,0,1)
        o_A() = (1,1,0,1)
        div_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,1,0,0),(0,0,1,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,0)) x2 + (3,4,1,1)
        |0|_A() = (1,1,6,8)
        divL_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,1,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2
        nil_A() = (1,1,1,1)
        cons_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,0,1,1)) x1 + x2 + (4,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        o > |0| > minus > nil > divL > cons > div > s > minus#
      
      argument filter:
    
        pi(minus#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(o) = []
        pi(div) = []
        pi(|0|) = []
        pi(divL) = []
        pi(nil) = []
        pi(cons) = [2]
    

The next rules are strictly ordered:

  p1

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