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)
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  quot(x,|0|(),s(z)) -> s(quot(x,plus(z,s(|0|())),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: plus#(s(x),y) -> plus#(x,y)
p3: quot#(x,|0|(),s(z)) -> quot#(x,plus(z,s(|0|())),s(z))
p4: quot#(x,|0|(),s(z)) -> plus#(z,s(|0|()))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p3}
  {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,plus(z,s(|0|())),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: plus(|0|(),y) -> y
r4: plus(s(x),y) -> s(plus(x,y))
r5: quot(x,|0|(),s(z)) -> s(quot(x,plus(z,s(|0|())),s(z)))

The set of usable rules consists of

  r3, r4

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      quot# > plus > s > |0|
    
    argument filter:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(|0|) = []
      pi(plus) = [1, 2]

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,plus(z,s(|0|())),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: plus(|0|(),y) -> y
r4: plus(s(x),y) -> s(plus(x,y))
r5: quot(x,|0|(),s(z)) -> s(quot(x,plus(z,s(|0|())),s(z)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: quot#(x,|0|(),s(z)) -> quot#(x,plus(z,s(|0|())),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: plus(|0|(),y) -> y
r4: plus(s(x),y) -> s(plus(x,y))
r5: quot(x,|0|(),s(z)) -> s(quot(x,plus(z,s(|0|())),s(z)))

The set of usable rules consists of

  r3, r4

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      |0| > plus > s > quot#
    
    argument filter:
  
      pi(quot#) = [2, 3]
      pi(|0|) = []
      pi(s) = []
      pi(plus) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.  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: quot(|0|(),s(y),s(z)) -> |0|()
r2: quot(s(x),s(y),z) -> quot(x,y,z)
r3: plus(|0|(),y) -> y
r4: plus(s(x),y) -> s(plus(x,y))
r5: quot(x,|0|(),s(z)) -> s(quot(x,plus(z,s(|0|())),s(z)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      s > plus#
    
    argument filter:
  
      pi(plus#) = [1, 2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5

We remove them from the problem.  Then no dependency pair remains.