YES

We show the termination of the TRS R:

  quot(|0|(),s(y),s(z)) -> |0|()
  quot(s(x),s(y),z) -> quot(x,y,z)
  quot(x,|0|(),s(z)) -> s(quot(x,s(z),s(z)))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: quot#(s(x),s(y),z) -> quot#(x,y,z)
p2: quot#(x,|0|(),s(z)) -> quot#(x,s(z),s(z))

and R consists of:

r1: quot(|0|(),s(y),s(z)) -> |0|()
r2: quot(s(x),s(y),z) -> quot(x,y,z)
r3: quot(x,|0|(),s(z)) -> s(quot(x,s(z),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: quot#(s(x),s(y),z) -> quot#(x,y,z)
p2: quot#(x,|0|(),s(z)) -> quot#(x,s(z),s(z))

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > quot# > |0|
      
      argument filter:
    
        pi(quot#) = [1, 3]
        pi(s) = 1
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > |0| > quot#
      
      argument filter:
    
        pi(quot#) = [1, 3]
        pi(s) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: quot#(x,|0|(),s(z)) -> quot#(x,s(z),s(z))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)