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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        divL#_A(x1,x2) = max{x1 + 3, x2}
        cons_A(x1,x2) = x2 + 2
        div_A(x1,x2) = x1
        minus_A(x1,x2) = x1
        o_A = 0
        s_A(x1) = x1 + 6
        |0|_A = 1
        divL_A(x1,x2) = max{x1 + 6, x2 + 4}
        nil_A = 0
  
    precedence:
    
      divL# = cons = div = minus = o = s = |0| = divL = nil
    
    partial status:
  
      pi(divL#) = [2]
      pi(cons) = []
      pi(div) = []
      pi(minus) = [1]
      pi(o) = []
      pi(s) = []
      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:

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

-- 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{4, x1 + 1}
        s_A(x1) = x1 + 3
        minus_A(x1,x2) = x1
        o_A = 0
        div_A(x1,x2) = x1
        |0|_A = 6
        divL_A(x1,x2) = max{x1 + 1, x2 + 5}
        nil_A = 0
        cons_A(x1,x2) = max{x1 - 1, x2}
  
    precedence:
    
      cons > minus# = minus = o = div = |0| = divL = nil > s
    
    partial status:
  
      pi(minus#) = [1]
      pi(s) = []
      pi(minus) = [1]
      pi(o) = []
      pi(div) = [1]
      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.