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

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

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 reduction pair:

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

The next rules are strictly ordered:

  p1

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