YES

We show the termination of R/S, where R is

  minus(x,0) -> x
  minus(s(x),s(y)) -> minus(x,y)
  quot(0,s(y)) -> 0
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
  plus(0,y) -> y
  plus(s(x),y) -> s(plus(x,y))
  minus(minus(x,y),z) -> minus(x,plus(y,z))

and S is:

  rand(x) -> x
  rand(x) -> rand(s(x))

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  minus#(s(x),s(y)) -> minus#(x,y)
  quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
  quot#(s(x),s(y)) -> minus#(x,y)
  plus#(s(x),y) -> plus#(x,y)
  minus#(minus(x,y),z) -> minus#(x,plus(y,z))
  minus#(minus(x,y),z) -> plus#(y,z)

and Q consists of the rules:

  minus(x,0) -> x
  minus(s(x),s(y)) -> minus(x,y)
  quot(0,s(y)) -> 0
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
  plus(0,y) -> y
  plus(s(x),y) -> s(plus(x,y))
  minus(minus(x,y),z) -> minus(x,plus(y,z))
  rand(x) -> x
  rand(x) -> rand(s(x))

The weakly monotone algebra (N^3, >_lex) with

  minus#_A((x1,y1,z1),(x2,y2,z2)) = (x2 + 1, 1, 0)
  s_A((x1,y1,z1)) = (x1, 0, 0)
  quot#_A((x1,y1,z1),(x2,y2,z2)) = (x2 + 2, 0, 1)
  minus_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2 + 1, 1, 1)
  plus#_A((x1,y1,z1),(x2,y2,z2)) = (0, 2, 1)
  plus_A((x1,y1,z1),(x2,y2,z2)) = (x2, y2 + 1, z2)
  0_A = (1, 1, 1)
  quot_A((x1,y1,z1),(x2,y2,z2)) = (2, 0, 0)
  rand_A((x1,y1,z1)) = (x1 + 1, y1 + 1, z1)

strictly orients the following dependency pairs:

  quot#(s(x),s(y)) -> minus#(x,y)
  minus#(minus(x,y),z) -> plus#(y,z)

We remove them from the set of dependency pairs.

The weakly monotone algebra (N^3, >_lex) with

  minus#_A((x1,y1,z1),(x2,y2,z2)) = (x1, 0, 0)
  s_A((x1,y1,z1)) = (x1, 1, 2)
  quot#_A((x1,y1,z1),(x2,y2,z2)) = (0, 0, 0)
  minus_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2 + 2, 2, 3)
  plus#_A((x1,y1,z1),(x2,y2,z2)) = (0, 0, 0)
  plus_A((x1,y1,z1),(x2,y2,z2)) = (x2 + 1, y2 + 2, 1)
  0_A = (0, 1, 1)
  quot_A((x1,y1,z1),(x2,y2,z2)) = (1, 2, 1)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 0)

strictly orients the following dependency pairs:

  minus#(minus(x,y),z) -> minus#(x,plus(y,z))

We remove them from the set of dependency pairs.

The weakly monotone algebra (N^3, >_lex) with

  minus#_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2, y1 + y2, z1)
  s_A((x1,y1,z1)) = (x1, y1 + 1, z1 + 1)
  quot#_A((x1,y1,z1),(x2,y2,z2)) = (0, 0, 0)
  minus_A((x1,y1,z1),(x2,y2,z2)) = (x1, y1, z1)
  plus#_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2, y1 + y2, z1 + z2)
  0_A = (1, 1, 1)
  quot_A((x1,y1,z1),(x2,y2,z2)) = (x1 + 1, y1 + 1, z1 + 1)
  plus_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2 + 1, y1 + y2 + 1, z1 + z2 + 1)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 0)

strictly orients the following dependency pairs:

  minus#(s(x),s(y)) -> minus#(x,y)
  plus#(s(x),y) -> plus#(x,y)

We remove them from the set of dependency pairs.

The weakly monotone algebra (N^3, >_lex) with

  quot#_A((x1,y1,z1),(x2,y2,z2)) = (x1, y1, z1)
  s_A((x1,y1,z1)) = (x1, y1 + 1, 2)
  minus_A((x1,y1,z1),(x2,y2,z2)) = (x1, y1, z1 + 1)
  0_A = (1, 1, 1)
  quot_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2 + 1, y1 + y2, 2)
  plus_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2 + 1, y1 + y2 + 1, z2 + 2)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 0)

strictly orients the following dependency pairs:

  quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))

We remove them from the set of dependency pairs.

No dependency pair remains.