YES
We show the termination of the TRS R:

  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)))


-- SCC decomposition.

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: div(|0|(),y) -> |0|()
r2: div(x,y) -> quot(x,y,y)
r3: quot(|0|(),s(y),z) -> |0|()
r4: quot(s(x),s(y),z) -> quot(x,y,z)
r5: quot(x,|0|(),s(z)) -> s(div(x,s(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#(x,|0|(),s(z)) -> div#(x,s(z))
p3: quot#(s(x),s(y),z) -> quot#(x,y,z)

and R consists of:

r1: div(|0|(),y) -> |0|()
r2: div(x,y) -> quot(x,y,y)
r3: quot(|0|(),s(y),z) -> |0|()
r4: quot(s(x),s(y),z) -> quot(x,y,z)
r5: quot(x,|0|(),s(z)) -> s(div(x,s(z)))

The set of usable rules consists of

  (no rules)

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 - 9}
            quot#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 9, x3 - 9}
            |0|_A = 7
            s_A(x1) = max{8, x1 + 6}
            div#!_A(x1,x2) = max{3, x1 - 9, x2 - 18}
            quot#!_A(x1,x2,x3) = max{3, x1 - 9, x2 - 18, x3 - 18}
            |0|!_A = 1
            s!_A(x1) = max{2, x1 - 12}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{0, x1 - 2, x2 - 6}
            quot#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 9, x3 - 6}
            |0|_A = 7
            s_A(x1) = max{5, x1 + 4}
            div#!_A(x1,x2) = max{2, x1 - 6, x2 - 7}
            quot#!_A(x1,x2,x3) = max{2, x1 - 6, x2 - 8, x3 - 7}
            |0|!_A = 1
            s!_A(x1) = max{3, x1 - 9}
        
  
    partial status:
  
      pi(div#) = [1]
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p3



-- 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: div(|0|(),y) -> |0|()
r2: div(x,y) -> quot(x,y,y)
r3: quot(|0|(),s(y),z) -> |0|()
r4: quot(s(x),s(y),z) -> quot(x,y,z)
r5: quot(x,|0|(),s(z)) -> s(div(x,s(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: div(|0|(),y) -> |0|()
r2: div(x,y) -> quot(x,y,y)
r3: quot(|0|(),s(y),z) -> |0|()
r4: quot(s(x),s(y),z) -> quot(x,y,z)
r5: quot(x,|0|(),s(z)) -> s(div(x,s(z)))

The set of usable rules consists of

  (no rules)

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 - 1, x2 - 2}
            quot#_A(x1,x2,x3) = max{1, x1 - 1, x2 - 10}
            |0|_A = 9
            s_A(x1) = 3
            div#!_A(x1,x2) = max{6, x2 - 1}
            quot#!_A(x1,x2,x3) = max{3, x2 - 2}
            |0|!_A = 5
            s!_A(x1) = max{8, x1 + 3}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{3, x1 - 1, x2 - 1}
            quot#_A(x1,x2,x3) = max{3, x1 - 1, x2 - 9, x3 - 1}
            |0|_A = 8
            s_A(x1) = 3
            div#!_A(x1,x2) = x2 + 3
            quot#!_A(x1,x2,x3) = max{2, x2 - 1}
            |0|!_A = 7
            s!_A(x1) = x1 + 9
        
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.