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, r4, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = 1
        div_A(x1,x2) = 2
        times_A(x1,x2) = max{x1 + 4, x2 + 1}
        quot#_A(x1,x2,x3) = 1
        |0|_A = 4
        s_A(x1) = x1
        plus_A(x1,x2) = max{4, x1 + 1, x2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{0, x1 - 6, x2 - 9}
          div_A(x1,x2) = max{4, x1 + 2, x2 - 15}
          times_A(x1,x2) = max{8, x1 - 12, x2}
          quot#_A(x1,x2,x3) = max{0, x1 - 6, x2 - 9, x3 - 9}
          |0|_A = 5
          s_A(x1) = max{7, x1}
          plus_A(x1,x2) = max{8, x1, x2}
          div#!_A(x1,x2) = max{0, x1 - 1}
          div!_A(x1,x2) = max{1, x1 - 8, x2 - 17}
          times!_A(x1,x2) = max{x1 - 17, x2 + 15}
          quot#!_A(x1,x2,x3) = max{0, x1 - 1}
          |0|!_A = 6
          s!_A(x1) = 5
          plus!_A(x1,x2) = max{x1 + 14, x2 - 26}
  
    partial status:
  
      pi(div#) = []
      pi(div) = []
      pi(times) = [1]
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(plus) = [1, 2]

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = 0
        quot#_A(x1,x2,x3) = 0
        s_A(x1) = max{4, x1 - 18}
        |0|_A = 10
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{0, x1 - 8, x2 - 16}
          quot#_A(x1,x2,x3) = max{0, x1 - 8, x2 - 17, x3 - 16}
          s_A(x1) = x1 + 15
          |0|_A = 9
          div#!_A(x1,x2) = max{7, x1 - 1}
          quot#!_A(x1,x2,x3) = max{7, x1 - 1}
          s!_A(x1) = max{0, x1 - 44}
          |0|!_A = 6
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = []
      pi(s) = []
      pi(|0|) = []

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = x2 + 5
        quot#_A(x1,x2,x3) = max{2, x2 + 1, x3 - 1}
        |0|_A = 12
        s_A(x1) = 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{0, x1 - 2, x2 - 7}
          quot#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 7, x3 - 7}
          |0|_A = 2
          s_A(x1) = max{6, x1 - 2}
          div#!_A(x1,x2) = 0
          quot#!_A(x1,x2,x3) = max{x1 + 3, x2 - 1}
          |0|!_A = 3
          s!_A(x1) = 2
  
    partial status:
  
      pi(div#) = [2]
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- 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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = x1 + 1
        s_A(x1) = x1 + 14
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          times#_A(x1,x2) = 0
          s_A(x1) = max{8, x1 - 3}
          times#!_A(x1,x2) = x2 + 2
          s!_A(x1) = max{3, x1 - 25}
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = [1]

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x1 + 4
        s_A(x1) = x1 + 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = 0
          s_A(x1) = max{8, x1 - 3}
          plus#!_A(x1,x2) = x2 + 5
          s!_A(x1) = max{3, x1 - 24}
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.