YES

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

  pred(s(x)) -> x
  minus(x,0) -> x
  minus(x,s(y)) -> pred(minus(x,y))
  quot(0,s(y)) -> 0
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))

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

and Q consists of the rules:

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

The weakly monotone algebra (N^2, >_lex) with

  minus#_A((x1,y1),(x2,y2)) = (x2 + 1, y2 + 2)
  s_A((x1,y1)) = (x1, y1 + 1)
  pred#_A((x1,y1)) = (0, 4)
  minus_A((x1,y1),(x2,y2)) = (x1, y1)
  quot#_A((x1,y1),(x2,y2)) = (x2 + 2, y2)
  pred_A((x1,y1)) = (x1, y1)
  0_A = (1, 1)
  quot_A((x1,y1),(x2,y2)) = (x1 + x2 + 1, y1 + 1)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

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

We remove them from the set of dependency pairs.

The weakly monotone algebra (N^2, >_lex) with

  quot#_A((x1,y1),(x2,y2)) = (x1, y1)
  s_A((x1,y1)) = (x1, y1 + 1)
  minus_A((x1,y1),(x2,y2)) = (x1, y1)
  pred_A((x1,y1)) = (x1, y1)
  0_A = (1, 1)
  quot_A((x1,y1),(x2,y2)) = (x1 + x2 + 1, y1 + 1)
  rand_A((x1,y1)) = (x1 + 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.