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

  r4, r7, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = x1 + 5
        div_A(x1,x2) = max{15, x1 + 10, x2 + 3}
        times_A(x1,x2) = max{21, x1 - 13, x2 - 6}
        quot#_A(x1,x2,x3) = x1 + 5
        |0|_A = 4
        s_A(x1) = max{5, x1 + 1}
        quot_A(x1,x2,x3) = max{16, x1 - 6, x2 - 6, x3 - 6}
  
    base order 2:
  
      precedence:
      
        div = times = quot > s > |0| > div# = quot#
      
  
    partial status:
  
      pi(div#) = []
      pi(div) = [2]
      pi(times) = []
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []
      pi(quot) = []

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

  r4, r7, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = x1 + 1
        quot#_A(x1,x2,x3) = x1 + 1
        s_A(x1) = max{6, x1 + 2}
        |0|_A = 10
        times_A(x1,x2) = max{13, x1 - 11, x2 - 11}
        div_A(x1,x2) = max{12, x1 - 13, x2 - 27}
        quot_A(x1,x2,x3) = max{11, x1 - 14, x2 - 28}
  
    base order 2:
  
      precedence:
      
        times > div > div# = quot# = s = quot > |0|
      
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = []
      pi(s) = [1]
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []

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

  r4, r7, r9, r11

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = max{4, x2 + 2}
        quot#_A(x1,x2,x3) = max{0, x2 - 1, x3 - 1}
        |0|_A = 6
        s_A(x1) = 2
        times_A(x1,x2) = max{7, x1 - 16, x2 - 16}
        div_A(x1,x2) = max{16, x1 - 7, x2 + 15}
        quot_A(x1,x2,x3) = max{16, x1 + 1, x2 - 7, x3 - 1}
  
    base order 2:
  
      precedence:
      
        times > div = quot > quot# = |0| > div# > s
      
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []
      pi(times) = []
      pi(div) = [2]
      pi(quot) = []

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

  r4, r7, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{9, x1 + 3, x2 + 1}
        s_A(x1) = max{6, x1 + 2}
        times_A(x1,x2) = max{8, x1 - 8, x2 - 2}
        |0|_A = 4
        div_A(x1,x2) = max{7, x1 - 10, x2 - 10}
        quot_A(x1,x2,x3) = max{5, x1 - 7, x2 - 9, x3 - 7}
  
    base order 2:
  
      precedence:
      
        times > times# = div > s = quot > |0|
      
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = [1]
      pi(times) = []
      pi(|0|) = []
      pi(div) = []
      pi(quot) = []

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

  r4, r7, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{7, x1 + 1, x2 + 1}
        s_A(x1) = max{6, x1 + 2}
        times_A(x1,x2) = max{9, x1 - 8, x2 - 2}
        |0|_A = 4
        div_A(x1,x2) = max{8, x1 - 10, x2 - 10}
        quot_A(x1,x2,x3) = max{7, x1 - 11, x2 - 9, x3 - 7}
  
    base order 2:
  
      precedence:
      
        times > div > plus# = s = quot > |0|
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(times) = []
      pi(|0|) = []
      pi(div) = []
      pi(quot) = []

The following rules are removed:

  p1

Then no dependency pair remains.