YES

We show the termination of the TRS R:

  -(x,|0|()) -> x
  -(|0|(),s(y)) -> |0|()
  -(s(x),s(y)) -> -(x,y)
  lt(x,|0|()) -> false()
  lt(|0|(),s(y)) -> true()
  lt(s(x),s(y)) -> lt(x,y)
  if(true(),x,y) -> x
  if(false(),x,y) -> y
  div(x,|0|()) -> |0|()
  div(|0|(),y) -> |0|()
  div(s(x),s(y)) -> if(lt(x,y),|0|(),s(div(-(x,y),s(y))))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: -#(s(x),s(y)) -> -#(x,y)
p2: lt#(s(x),s(y)) -> lt#(x,y)
p3: div#(s(x),s(y)) -> if#(lt(x,y),|0|(),s(div(-(x,y),s(y))))
p4: div#(s(x),s(y)) -> lt#(x,y)
p5: div#(s(x),s(y)) -> div#(-(x,y),s(y))
p6: div#(s(x),s(y)) -> -#(x,y)

and R consists of:

r1: -(x,|0|()) -> x
r2: -(|0|(),s(y)) -> |0|()
r3: -(s(x),s(y)) -> -(x,y)
r4: lt(x,|0|()) -> false()
r5: lt(|0|(),s(y)) -> true()
r6: lt(s(x),s(y)) -> lt(x,y)
r7: if(true(),x,y) -> x
r8: if(false(),x,y) -> y
r9: div(x,|0|()) -> |0|()
r10: div(|0|(),y) -> |0|()
r11: div(s(x),s(y)) -> if(lt(x,y),|0|(),s(div(-(x,y),s(y))))

The estimated dependency graph contains the following SCCs:

  {p5}
  {p1}
  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(s(x),s(y)) -> div#(-(x,y),s(y))

and R consists of:

r1: -(x,|0|()) -> x
r2: -(|0|(),s(y)) -> |0|()
r3: -(s(x),s(y)) -> -(x,y)
r4: lt(x,|0|()) -> false()
r5: lt(|0|(),s(y)) -> true()
r6: lt(s(x),s(y)) -> lt(x,y)
r7: if(true(),x,y) -> x
r8: if(false(),x,y) -> y
r9: div(x,|0|()) -> |0|()
r10: div(|0|(),y) -> |0|()
r11: div(s(x),s(y)) -> if(lt(x,y),|0|(),s(div(-(x,y),s(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > s > - > div#
      
      argument filter:
    
        pi(div#) = [1]
        pi(s) = [1]
        pi(-) = [1]
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        div#_A(x1,x2) = x1
        s_A(x1) = x1 + 1
        -_A(x1,x2) = 2
        |0|_A() = 3
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: -#(s(x),s(y)) -> -#(x,y)

and R consists of:

r1: -(x,|0|()) -> x
r2: -(|0|(),s(y)) -> |0|()
r3: -(s(x),s(y)) -> -(x,y)
r4: lt(x,|0|()) -> false()
r5: lt(|0|(),s(y)) -> true()
r6: lt(s(x),s(y)) -> lt(x,y)
r7: if(true(),x,y) -> x
r8: if(false(),x,y) -> y
r9: div(x,|0|()) -> |0|()
r10: div(|0|(),y) -> |0|()
r11: div(s(x),s(y)) -> if(lt(x,y),|0|(),s(div(-(x,y),s(y))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > -#
      
      argument filter:
    
        pi(-#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        -#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: lt#(s(x),s(y)) -> lt#(x,y)

and R consists of:

r1: -(x,|0|()) -> x
r2: -(|0|(),s(y)) -> |0|()
r3: -(s(x),s(y)) -> -(x,y)
r4: lt(x,|0|()) -> false()
r5: lt(|0|(),s(y)) -> true()
r6: lt(s(x),s(y)) -> lt(x,y)
r7: if(true(),x,y) -> x
r8: if(false(),x,y) -> y
r9: div(x,|0|()) -> |0|()
r10: div(|0|(),y) -> |0|()
r11: div(s(x),s(y)) -> if(lt(x,y),|0|(),s(div(-(x,y),s(y))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > lt#
      
      argument filter:
    
        pi(lt#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        lt#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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