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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      quot#_A(x1,x2,x3) = ((1,0,0),(1,1,0),(0,0,1)) x1 + ((1,0,0),(0,0,0),(0,1,0)) x3
      s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
      |0|_A() = (2,1,1)

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,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)