YES

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

  half(0) -> 0
  half(s(s(x))) -> s(half(x))
  log(s(0)) -> 0
  log(s(s(x))) -> s(log(s(half(x))))

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

  half#(s(s(x))) -> half#(x)
  log#(s(s(x))) -> log#(s(half(x)))
  log#(s(s(x))) -> half#(x)

and Q consists of the rules:

  half(0) -> 0
  half(s(s(x))) -> s(half(x))
  log(s(0)) -> 0
  log(s(s(x))) -> s(log(s(half(x))))
  rand(x) -> x
  rand(x) -> rand(s(x))

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

  half#_A((x1,y1)) = (0, 0)
  s_A((x1,y1)) = (0, 2)
  log#_A((x1,y1)) = (1, 1)
  half_A((x1,y1)) = (2, 1)
  0_A = (1, 2)
  log_A((x1,y1)) = (2, 0)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  log#(s(s(x))) -> half#(x)

We remove them from the set of dependency pairs.

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

  half#_A((x1,y1)) = (x1, y1)
  s_A((x1,y1)) = (x1, y1 + 1)
  log#_A((x1,y1)) = (x1, y1)
  half_A((x1,y1)) = (x1, y1)
  0_A = (1, 1)
  log_A((x1,y1)) = (x1 + 1, y1)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  half#(s(s(x))) -> half#(x)
  log#(s(s(x))) -> log#(s(half(x)))

We remove them from the set of dependency pairs.

No dependency pair remains.