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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            divL#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (2,1)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,5)
            div_A(x1,x2) = x1 + x2 + (1,4)
            minus_A(x1,x2) = x1
            o_A() = (1,1)
            s_A(x1) = x1 + (2,1)
            |0|_A() = (1,2)
            divL_A(x1,x2) = x1 + x2 + (2,2)
            nil_A() = (1,1)
    
      precedence:
      
        divL# = cons = div = minus = o = s = |0| = divL = nil
      
      partial status:
    
        pi(divL#) = []
        pi(cons) = []
        pi(div) = []
        pi(minus) = []
        pi(o) = []
        pi(s) = []
        pi(|0|) = []
        pi(divL) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            divL#_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
            cons_A(x1,x2) = (0,0)
            div_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,2)
            minus_A(x1,x2) = ((1,0),(0,0)) x1 + (1,5)
            o_A() = (1,1)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,2)
            |0|_A() = (0,1)
            divL_A(x1,x2) = x2
            nil_A() = (1,1)
    
      precedence:
      
        minus = |0| > o > div = divL > s > nil > divL# = cons
      
      partial status:
    
        pi(divL#) = [1]
        pi(cons) = []
        pi(div) = []
        pi(minus) = []
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,0)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,2)
            minus_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2
            o_A() = (1,1)
            div_A(x1,x2) = ((1,0),(0,0)) x1 + (2,3)
            |0|_A() = (2,4)
            divL_A(x1,x2) = x1 + x2 + (2,1)
            nil_A() = (1,1)
            cons_A(x1,x2) = ((1,0),(1,1)) x2 + (3,5)
    
      precedence:
      
        minus > o = div = |0| = divL > nil = cons > div# = s
      
      partial status:
    
        pi(div#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(o) = []
        pi(div) = []
        pi(|0|) = []
        pi(divL) = [2]
        pi(nil) = []
        pi(cons) = [2]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = (1,1)
            s_A(x1) = (1,1)
            minus_A(x1,x2) = ((1,0),(0,0)) x1 + (5,2)
            o_A() = (0,1)
            div_A(x1,x2) = (4,1)
            |0|_A() = (5,2)
            divL_A(x1,x2) = x2 + (1,1)
            nil_A() = (1,1)
            cons_A(x1,x2) = ((1,0),(1,1)) x2 + (2,2)
    
      precedence:
      
        div# = minus = o = nil > |0| > s = div = divL = cons
      
      partial status:
    
        pi(div#) = []
        pi(s) = []
        pi(minus) = []
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,2)
            s_A(x1) = ((1,0),(0,0)) x1 + (2,1)
            minus_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (0,4)
            o_A() = (1,1)
            div_A(x1,x2) = ((1,0),(0,0)) x1 + (1,3)
            |0|_A() = (3,2)
            divL_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
            nil_A() = (1,1)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (3,5)
    
      precedence:
      
        minus = div = cons > s > minus# > o = |0| = nil > divL
      
      partial status:
    
        pi(minus#) = []
        pi(s) = []
        pi(minus) = []
        pi(o) = []
        pi(div) = []
        pi(|0|) = []
        pi(divL) = [1]
        pi(nil) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = (1,2)
            s_A(x1) = (0,1)
            minus_A(x1,x2) = (1,2)
            o_A() = (1,1)
            div_A(x1,x2) = (2,0)
            |0|_A() = (1,2)
            divL_A(x1,x2) = (0,0)
            nil_A() = (0,0)
            cons_A(x1,x2) = (3,0)
    
      precedence:
      
        o > minus# > nil > cons > s = minus = div > |0| = divL
      
      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.