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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = ((1,0),(0,0)) x1
            quot#_A(x1,x2,x3) = ((1,0),(0,0)) x1
            |0|_A() = (2,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (1,0)
    
      precedence:
      
        div# = quot# = |0| = s
      
      partial status:
    
        pi(div#) = []
        pi(quot#) = []
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = (3,3)
            quot#_A(x1,x2,x3) = (3,3)
            |0|_A() = (2,2)
            s_A(x1) = (1,1)
    
      precedence:
      
        div# = quot# = |0| = s
      
      partial status:
    
        pi(div#) = []
        pi(quot#) = []
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (2,2)
            quot#_A(x1,x2,x3) = ((0,0),(1,0)) x1 + x2 + (1,1)
            |0|_A() = (3,4)
            s_A(x1) = (0,1)
    
      precedence:
      
        div# > quot# = |0| = s
      
      partial status:
    
        pi(div#) = [2]
        pi(quot#) = [2]
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = ((1,0),(1,1)) x2 + (1,2)
            quot#_A(x1,x2,x3) = ((1,0),(1,1)) x2 + (0,1)
            |0|_A() = (2,5)
            s_A(x1) = (1,1)
    
      precedence:
      
        |0| > div# > quot# = s
      
      partial status:
    
        pi(div#) = []
        pi(quot#) = [2]
        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)