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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          divL#_A(x1,x2) = ((0,1),(1,1)) x1 + ((1,0),(1,0)) x2 + (2,1)
          cons_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (3,2)
          div_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(0,0)) x2 + (0,2)
          minus_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (2,0)
          o_A() = (0,1)
          s_A(x1) = x1 + (0,1)
          |0|_A() = (0,1)
          divL_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (0,1)
          nil_A() = (1,1)
  
    precedence:
    
      divL# = cons = div = s = |0| = divL = nil > minus > o
    
    partial status:
  
      pi(divL#) = []
      pi(cons) = []
      pi(div) = []
      pi(minus) = [1]
      pi(o) = []
      pi(s) = []
      pi(|0|) = []
      pi(divL) = []
      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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          div#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,0)
          s_A(x1) = x1 + (2,1)
          minus_A(x1,x2) = x1
          o_A() = (1,0)
          div_A(x1,x2) = x1 + ((0,1),(0,1)) x2 + (2,1)
          |0|_A() = (1,1)
          divL_A(x1,x2) = x1 + ((1,1),(0,1)) x2 + (0,2)
          nil_A() = (1,0)
          cons_A(x1,x2) = ((0,0),(1,1)) x1 + x2 + (1,2)
  
    precedence:
    
      div# > o = nil = cons > div > s = minus = |0| = divL
    
    partial status:
  
      pi(div#) = [2]
      pi(s) = []
      pi(minus) = [1]
      pi(o) = []
      pi(div) = []
      pi(|0|) = []
      pi(divL) = [2]
      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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          minus#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,3)
          s_A(x1) = x1 + (2,2)
          minus_A(x1,x2) = x1
          o_A() = (1,1)
          div_A(x1,x2) = ((0,1),(1,0)) x1 + ((0,0),(1,0)) x2 + (3,1)
          |0|_A() = (1,1)
          divL_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,0)
          nil_A() = (1,1)
          cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,0)) x2 + (2,3)
  
    precedence:
    
      o = divL > minus# = nil = cons > minus > s = div > |0|
    
    partial status:
  
      pi(minus#) = [1]
      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.