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. lexicographic path order with precedence:
    
      precedence:
      
        nil > div > divL > |0| > minus > s > o > cons > divL#
      
      argument filter:
    
        pi(divL#) = [1, 2]
        pi(cons) = 2
        pi(div) = 1
        pi(minus) = 1
        pi(o) = []
        pi(s) = 1
        pi(|0|) = []
        pi(divL) = [1, 2]
        pi(nil) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        nil > |0| > div > divL > s > minus > o > cons > divL#
      
      argument filter:
    
        pi(divL#) = 2
        pi(cons) = [2]
        pi(div) = [1]
        pi(minus) = [1]
        pi(o) = []
        pi(s) = [1]
        pi(|0|) = []
        pi(divL) = [2]
        pi(nil) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        nil > |0| > divL# > div > divL > minus > s > o > cons
      
      argument filter:
    
        pi(divL#) = []
        pi(cons) = [2]
        pi(div) = []
        pi(minus) = []
        pi(o) = []
        pi(s) = 1
        pi(|0|) = []
        pi(divL) = 2
        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. lexicographic path order with precedence:
    
      precedence:
      
        div > divL > cons > nil > |0| > o > minus > s > div#
      
      argument filter:
    
        pi(div#) = [1, 2]
        pi(s) = [1]
        pi(minus) = 1
        pi(o) = []
        pi(div) = 1
        pi(|0|) = []
        pi(divL) = [1, 2]
        pi(nil) = []
        pi(cons) = 2
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > div > divL > cons > nil > o > minus > s > div#
      
      argument filter:
    
        pi(div#) = [1, 2]
        pi(s) = [1]
        pi(minus) = 1
        pi(o) = []
        pi(div) = [1]
        pi(|0|) = []
        pi(divL) = 2
        pi(nil) = []
        pi(cons) = [2]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        div > divL > cons > nil > |0| > o > minus > s > div#
      
      argument filter:
    
        pi(div#) = 2
        pi(s) = [1]
        pi(minus) = 1
        pi(o) = []
        pi(div) = 1
        pi(|0|) = []
        pi(divL) = 2
        pi(nil) = []
        pi(cons) = 2
    

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. lexicographic path order with precedence:
    
      precedence:
      
        divL > minus > cons > o > div > nil > s > |0| > minus#
      
      argument filter:
    
        pi(minus#) = [2]
        pi(s) = 1
        pi(minus) = 1
        pi(o) = []
        pi(div) = 1
        pi(|0|) = []
        pi(divL) = 1
        pi(nil) = []
        pi(cons) = [2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        div > divL > cons > nil > |0| > o > minus > minus# > s
      
      argument filter:
    
        pi(minus#) = [2]
        pi(s) = [1]
        pi(minus) = 1
        pi(o) = []
        pi(div) = 1
        pi(|0|) = []
        pi(divL) = 1
        pi(nil) = []
        pi(cons) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        div > divL > cons > nil > |0| > o > minus > s > minus#
      
      argument filter:
    
        pi(minus#) = [2]
        pi(s) = 1
        pi(minus) = 1
        pi(o) = []
        pi(div) = 1
        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.