YES
We show the termination of the TRS R:

  plus(x,|0|()) -> x
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  times(|0|(),y) -> |0|()
  times(s(|0|()),y) -> y
  times(s(x),y) -> plus(y,times(x,y))
  div(|0|(),y) -> |0|()
  div(x,y) -> quot(x,y,y)
  quot(|0|(),s(y),z) -> |0|()
  quot(s(x),s(y),z) -> quot(x,y,z)
  quot(x,|0|(),s(z)) -> s(div(x,s(z)))
  div(div(x,y),z) -> div(x,times(y,z))


-- SCC decomposition.

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

p1: plus#(s(x),y) -> plus#(x,y)
p2: times#(s(x),y) -> plus#(y,times(x,y))
p3: times#(s(x),y) -> times#(x,y)
p4: div#(x,y) -> quot#(x,y,y)
p5: quot#(s(x),s(y),z) -> quot#(x,y,z)
p6: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p7: div#(div(x,y),z) -> div#(x,times(y,z))
p8: div#(div(x,y),z) -> times#(y,z)

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))

The estimated dependency graph contains the following SCCs:

  {p4, p5, p6, p7}
  {p3}
  {p1}


-- Reduction pair.

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

p1: div#(div(x,y),z) -> div#(x,times(y,z))
p2: div#(x,y) -> quot#(x,y,y)
p3: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p4: quot#(s(x),s(y),z) -> quot#(x,y,z)

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))

The set of usable rules consists of

  r1, r2, r3, r5, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          div#_A(x1,x2) = x1 + 1
          div_A(x1,x2) = x1 + x2 + 2
          times_A(x1,x2) = x1 + x2 + 1
          quot#_A(x1,x2,x3) = x1 + 1
          |0|_A() = 8
          s_A(x1) = x1 + 3
          plus_A(x1,x2) = x1 + x2 + 1
  
    base order 2:
  
      precedence:
      
        plus > times > div# = quot# > div = s > |0|
      
  
    partial status:
  
      pi(div#) = [1]
      pi(div) = []
      pi(times) = []
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = [1]
      pi(plus) = [1]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: div#(x,y) -> quot#(x,y,y)
p2: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p3: quot#(s(x),s(y),z) -> quot#(x,y,z)

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: div#(x,y) -> quot#(x,y,y)
p2: quot#(s(x),s(y),z) -> quot#(x,y,z)
p3: quot#(x,|0|(),s(z)) -> div#(x,s(z))

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))

The set of usable rules consists of

  r2

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          div#_A(x1,x2) = x1 + 1
          quot#_A(x1,x2,x3) = x1 + 1
          s_A(x1) = x1 + 2
          |0|_A() = 3
          plus_A(x1,x2) = x2
  
    base order 2:
  
      precedence:
      
        div# = quot# > s = plus > |0|
      
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = []
      pi(s) = [1]
      pi(|0|) = []
      pi(plus) = [2]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: div#(x,y) -> quot#(x,y,y)
p2: quot#(x,|0|(),s(z)) -> div#(x,s(z))

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: div#(x,y) -> quot#(x,y,y)
p2: quot#(x,|0|(),s(z)) -> div#(x,s(z))

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))

The set of usable rules consists of

  r2, r3, r5, r7, r9, r11

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          div#_A(x1,x2) = x2
          quot#_A(x1,x2,x3) = x2
          |0|_A() = 0
          s_A(x1) = 0
          plus_A(x1,x2) = x2 + 1
          times_A(x1,x2) = x1 + x2 + 1
          div_A(x1,x2) = x1 + x2 + 2
          quot_A(x1,x2,x3) = x2 + 1
  
    base order 2:
  
      precedence:
      
        plus > times = quot > |0| > div# = div > s > quot#
      
  
    partial status:
  
      pi(div#) = [2]
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(plus) = [2]
      pi(times) = []
      pi(div) = [1]
      pi(quot) = [2]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: quot#(x,|0|(),s(z)) -> div#(x,s(z))

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

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

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))

The set of usable rules consists of

  r2, r5, r7, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          times#_A(x1,x2) = x1 + x2 + 3
          s_A(x1) = x1 + 2
          plus_A(x1,x2) = x2 + 1
          |0|_A() = 1
          times_A(x1,x2) = x1 + x2 + 3
          div_A(x1,x2) = x1 + x2 + 4
          quot_A(x1,x2,x3) = x1 + x3 + 7
  
    base order 2:
  
      precedence:
      
        times# = times > div > s = plus > quot > |0|
      
  
    partial status:
  
      pi(times#) = [2]
      pi(s) = []
      pi(plus) = [2]
      pi(|0|) = []
      pi(times) = []
      pi(div) = [2]
      pi(quot) = [3]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

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

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))

The set of usable rules consists of

  r2, r5, r7, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1 + x2 + 3
          s_A(x1) = x1 + 2
          plus_A(x1,x2) = x2 + 1
          |0|_A() = 1
          times_A(x1,x2) = x1 + x2 + 3
          div_A(x1,x2) = x1 + x2 + 4
          quot_A(x1,x2,x3) = x1 + x3 + 7
  
    base order 2:
  
      precedence:
      
        plus# = times > div > s = plus > quot > |0|
      
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = []
      pi(plus) = [2]
      pi(|0|) = []
      pi(times) = []
      pi(div) = [2]
      pi(quot) = [3]

The following rules are removed:

  p1

Then no dependency pair remains.