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

  r5, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{1, x1 - 17}
            div_A(x1,x2) = max{16, x1 + 11, x2 - 14}
            times_A(x1,x2) = max{2, x1 - 26, x2 + 1}
            quot#_A(x1,x2,x3) = max{1, x1 - 17}
            |0|_A = 8
            s_A(x1) = x1 + 6
            div#!_A(x1,x2) = max{23, x1 + 22}
            div!_A(x1,x2) = max{5, x1 - 10, x2 - 3}
            times!_A(x1,x2) = x2 + 9
            quot#!_A(x1,x2,x3) = max{23, x1 + 22}
            |0|!_A = 2
            s!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{1, x1 - 3}
            div_A(x1,x2) = max{2, x1 + 1, x2 - 13}
            times_A(x1,x2) = max{17, x1 - 10, x2 + 6}
            quot#_A(x1,x2,x3) = max{1, x1 - 3}
            |0|_A = 0
            s_A(x1) = x1 + 6
            div#!_A(x1,x2) = 0
            div!_A(x1,x2) = max{20, x2 + 2}
            times!_A(x1,x2) = max{4, x1 - 1}
            quot#!_A(x1,x2,x3) = 0
            |0|!_A = 3
            s!_A(x1) = 2
        
  
    partial status:
  
      pi(div#) = []
      pi(div) = [1, 2]
      pi(times) = [2]
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []

The following rules are removed:

  p1, p4



-- 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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{0, x1 - 2, x2 - 8}
            quot#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 8, x3 - 8}
            |0|_A = 3
            s_A(x1) = max{7, x1 - 2}
            times_A(x1,x2) = max{9, x1 - 9, x2 - 11}
            div#!_A(x1,x2) = max{3, x2 + 1}
            quot#!_A(x1,x2,x3) = max{2, x3 + 1}
            |0|!_A = 9
            s!_A(x1) = 0
            times!_A(x1,x2) = max{12, x1 - 8, x2 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{0, x1 - 2, x2 - 3}
            quot#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 3, x3 - 3}
            |0|_A = 0
            s_A(x1) = max{2, x1 - 1}
            times_A(x1,x2) = max{1, x1 - 3, x2 - 5}
            div#!_A(x1,x2) = max{1, x1 - 1, x2 - 4}
            quot#!_A(x1,x2,x3) = max{1, x1 - 1, x3 - 4}
            |0|!_A = 2
            s!_A(x1) = 4
            times!_A(x1,x2) = max{5, x1 - 4, x2 - 7}
        
  
    partial status:
  
      pi(div#) = [2]
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(times) = []

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)

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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = 0
            s_A(x1) = max{14, x1 - 7}
            times_A(x1,x2) = max{5, x1 - 16, x2 - 17}
            |0|_A = 5
            times#!_A(x1,x2) = max{1, x1 - 43}
            s!_A(x1) = max{2, x1 - 42}
            times!_A(x1,x2) = max{3, x1 - 33, x2 - 67}
            |0|!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = 0
            s_A(x1) = max{7, x1 - 2}
            times_A(x1,x2) = max{11, x1 - 11, x2 - 15}
            |0|_A = 3
            times#!_A(x1,x2) = max{1, x1 - 8, x2 - 1}
            s!_A(x1) = 0
            times!_A(x1,x2) = max{1, x1 - 8, x2 - 2}
            |0|!_A = 0
        
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = [1]
      pi(times) = []
      pi(|0|) = []

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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{13, x1 - 8}
            times_A(x1,x2) = max{3, x1 - 15, x2 - 16}
            |0|_A = 3
            plus#!_A(x1,x2) = max{1, x1 - 12}
            s!_A(x1) = max{6, x1 + 2}
            times!_A(x1,x2) = max{7, x1 - 21, x2 - 25}
            |0|!_A = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{17, x1 - 12}
            times_A(x1,x2) = max{3, x1 - 19, x2 - 20}
            |0|_A = 3
            plus#!_A(x1,x2) = max{0, x1 - 25}
            s!_A(x1) = max{5, x1 - 24}
            times!_A(x1,x2) = max{4, x1 - 25, x2 - 33}
            |0|!_A = 1
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(times) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.