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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      quot#_A(x1,x2,x3) = ((0,1),(0,1)) x1
      s_A(x1) = x1 + (0,1)
      |0|_A() = (0,1)
      plus_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (0,1)

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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      quot#_A(x1,x2,x3) = ((1,0),(1,1)) x2
      |0|_A() = (1,0)
      s_A(x1) = ((0,1),(0,0)) x1
      plus_A(x1,x2) = ((1,1),(1,1)) x2

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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      plus#_A(x1,x2) = ((1,0),(1,0)) x1
      s_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1

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