YES

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

  le(0,y) -> true
  le(s(x),0) -> false
  le(s(x),s(y)) -> le(x,y)
  minus(0,y) -> 0
  minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
  if_minus(true,s(x),y) -> 0
  if_minus(false,s(x),y) -> s(minus(x,y))
  quot(0,s(y)) -> 0
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
  log(s(0)) -> 0
  log(s(s(x))) -> s(log(s(quot(x,s(s(0))))))

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

  le#(s(x),s(y)) -> le#(x,y)
  minus#(s(x),y) -> if_minus#(le(s(x),y),s(x),y)
  minus#(s(x),y) -> le#(s(x),y)
  if_minus#(false,s(x),y) -> minus#(x,y)
  quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
  quot#(s(x),s(y)) -> minus#(x,y)
  log#(s(s(x))) -> log#(s(quot(x,s(s(0)))))
  log#(s(s(x))) -> quot#(x,s(s(0)))

and Q consists of the rules:

  le(0,y) -> true
  le(s(x),0) -> false
  le(s(x),s(y)) -> le(x,y)
  minus(0,y) -> 0
  minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
  if_minus(true,s(x),y) -> 0
  if_minus(false,s(x),y) -> s(minus(x,y))
  quot(0,s(y)) -> 0
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
  log(s(0)) -> 0
  log(s(s(x))) -> s(log(s(quot(x,s(s(0))))))
  rand(x) -> x
  rand(x) -> rand(s(x))

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

  le#_A((x1,y1),(x2,y2)) = (0, 0)
  s_A((x1,y1)) = (x1, 1)
  minus#_A((x1,y1),(x2,y2)) = (1, 1)
  if_minus#_A((x1,y1),(x2,y2),(x3,y3)) = (1, 1)
  le_A((x1,y1),(x2,y2)) = (x2 + 2, 0)
  false_A = (1, 1)
  quot#_A((x1,y1),(x2,y2)) = (2, 0)
  minus_A((x1,y1),(x2,y2)) = (2, 2)
  log#_A((x1,y1)) = (3, 1)
  quot_A((x1,y1),(x2,y2)) = (2, 2)
  0_A = (1, 0)
  true_A = (1, 1)
  if_minus_A((x1,y1),(x2,y2),(x3,y3)) = (2, 1)
  log_A((x1,y1)) = (2, 1)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  minus#(s(x),y) -> le#(s(x),y)
  quot#(s(x),s(y)) -> minus#(x,y)
  log#(s(s(x))) -> quot#(x,s(s(0)))

We remove them from the set of dependency pairs.

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

  le#_A((x1,y1),(x2,y2)) = (x1 + x2, y1 + y2)
  s_A((x1,y1)) = (x1, y1 + 2)
  minus#_A((x1,y1),(x2,y2)) = (x1, y1 + 1)
  if_minus#_A((x1,y1),(x2,y2),(x3,y3)) = (x2, y2)
  le_A((x1,y1),(x2,y2)) = (2, 1)
  false_A = (1, 2)
  quot#_A((x1,y1),(x2,y2)) = (x1, y1)
  minus_A((x1,y1),(x2,y2)) = (x1, y1)
  log#_A((x1,y1)) = (x1, y1)
  quot_A((x1,y1),(x2,y2)) = (x1, y1)
  0_A = (0, 1)
  true_A = (1, 2)
  if_minus_A((x1,y1),(x2,y2),(x3,y3)) = (x2, y2)
  log_A((x1,y1)) = (x1 + 1, y1)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  le#(s(x),s(y)) -> le#(x,y)
  minus#(s(x),y) -> if_minus#(le(s(x),y),s(x),y)
  if_minus#(false,s(x),y) -> minus#(x,y)
  quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
  log#(s(s(x))) -> log#(s(quot(x,s(s(0)))))

We remove them from the set of dependency pairs.

No dependency pair remains.