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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          div#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,2)
          quot#_A(x1,x2,x3) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x3 + (2,2)
          |0|_A() = (4,1)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,0)
  
    precedence:
    
      s > div# = quot# = |0|
    
    partial status:
  
      pi(div#) = [1]
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          div#_A(x1,x2) = ((0,0),(1,0)) x2 + (1,2)
          quot#_A(x1,x2,x3) = ((0,0),(1,0)) x2 + (1,1)
          |0|_A() = (4,7)
          s_A(x1) = (2,6)
  
    precedence:
    
      quot# > div# = |0| = s
    
    partial status:
  
      pi(div#) = []
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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: 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:

  (no SCCs)