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)))
  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

  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)
  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:

  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)))
  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^3, >_lex) with

  minus#_A((x1,y1,z1),(x2,y2,z2)) = (1, 1, 1)
  s_A((x1,y1,z1)) = (x1, y1, z1)
  pred#_A((x1,y1,z1)) = (0, 0, 0)
  minus_A((x1,y1,z1),(x2,y2,z2)) = (x1 + 1, y1 + 1, z1 + 1)
  quot#_A((x1,y1,z1),(x2,y2,z2)) = (2, 2, 2)
  log#_A((x1,y1,z1)) = (3, 3, 3)
  quot_A((x1,y1,z1),(x2,y2,z2)) = (2, 1, 1)
  0_A = (1, 2, 2)
  pred_A((x1,y1,z1)) = (x1, y1, z1)
  log_A((x1,y1,z1)) = (2, 1, 1)
  rand_A((x1,y1,z1)) = (x1 + 1, y1 + 1, 0)

strictly orients the following dependency pairs:

  minus#(x,s(y)) -> pred#(minus(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^3, >_lex) with

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

strictly orients the following dependency pairs:

  minus#(x,s(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.