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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          divL#_A(x1,x2) = max{x1 + 2, x2}
          cons_A(x1,x2) = x2 + 2
          div_A(x1,x2) = x1
          minus_A(x1,x2) = x1
          o_A = 0
          s_A(x1) = x1 + 3
          |0|_A = 2
          divL_A(x1,x2) = max{x1 + 2, x2 + 9}
          nil_A = 0
    
      precedence:
      
        divL# = cons = div = minus = o = |0| = divL = nil > s
      
      partial status:
    
        pi(divL#) = [2]
        pi(cons) = []
        pi(div) = [1]
        pi(minus) = [1]
        pi(o) = []
        pi(s) = [1]
        pi(|0|) = []
        pi(divL) = [2]
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          divL#_A(x1,x2) = x2
          cons_A(x1,x2) = 2
          div_A(x1,x2) = x1 + 7
          minus_A(x1,x2) = max{24, x1 + 16}
          o_A = 0
          s_A(x1) = x1 + 8
          |0|_A = 7
          divL_A(x1,x2) = max{2, x2}
          nil_A = 0
    
      precedence:
      
        divL# = cons = div = minus = o = s = |0| = divL = nil
      
      partial status:
    
        pi(divL#) = [2]
        pi(cons) = []
        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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          div#_A(x1,x2) = x1 + 8
          s_A(x1) = max{9, x1 + 2}
          minus_A(x1,x2) = max{4, x1}
          o_A = 0
          div_A(x1,x2) = max{7, x1}
          |0|_A = 2
          divL_A(x1,x2) = max{16, x1 + 8, x2}
          nil_A = 0
          cons_A(x1,x2) = x2 + 17
    
      precedence:
      
        o = div > s > |0| > div# > nil = cons > divL > minus
      
      partial status:
    
        pi(div#) = [1]
        pi(s) = [1]
        pi(minus) = [1]
        pi(o) = []
        pi(div) = [1]
        pi(|0|) = []
        pi(divL) = [2]
        pi(nil) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          div#_A(x1,x2) = x1 + 2
          s_A(x1) = x1 + 5
          minus_A(x1,x2) = x1
          o_A = 0
          div_A(x1,x2) = x1 + 9
          |0|_A = 4
          divL_A(x1,x2) = 0
          nil_A = 0
          cons_A(x1,x2) = 2
    
      precedence:
      
        o = |0| > div > minus = divL > div# = s = nil = cons
      
      partial status:
    
        pi(div#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(o) = []
        pi(div) = []
        pi(|0|) = []
        pi(divL) = []
        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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{0, x1 - 1}
          s_A(x1) = max{3, x1 + 2}
          minus_A(x1,x2) = max{1, x1}
          o_A = 0
          div_A(x1,x2) = x1
          |0|_A = 3
          divL_A(x1,x2) = max{x1 + 2, x2 + 1}
          nil_A = 0
          cons_A(x1,x2) = x2 + 1
    
      precedence:
      
        o > nil > cons > minus > div = |0| > minus# > divL > s
      
      partial status:
    
        pi(minus#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(o) = []
        pi(div) = []
        pi(|0|) = []
        pi(divL) = [2]
        pi(nil) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = 0
          s_A(x1) = 2
          minus_A(x1,x2) = 9
          o_A = 0
          div_A(x1,x2) = 8
          |0|_A = 3
          divL_A(x1,x2) = 0
          nil_A = 0
          cons_A(x1,x2) = 0
    
      precedence:
      
        minus# = s = minus = o = div = |0| = divL = nil = cons
      
      partial status:
    
        pi(minus#) = []
        pi(s) = []
        pi(minus) = []
        pi(o) = []
        pi(div) = []
        pi(|0|) = []
        pi(divL) = []
        pi(nil) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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