YES

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

  tests(0) -> true
  tests(s(x)) -> and(test(rands(rand(0),nil)),x)
  test(done(y)) -> eq(f(y),g(y))
  eq(x,x) -> true
  rands(0,y) -> done(y)
  rands(s(x),y) -> rands(x,::(rand(0),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

  tests#(s(x)) -> test#(rands(rand(0),nil))
  tests#(s(x)) -> rands#(rand(0),nil)
  test#(done(y)) -> eq#(f(y),g(y))
  rands#(s(x),y) -> rands#(x,::(rand(0),y))

and Q consists of the rules:

  tests(0) -> true
  tests(s(x)) -> and(test(rands(rand(0),nil)),x)
  test(done(y)) -> eq(f(y),g(y))
  eq(x,x) -> true
  rands(0,y) -> done(y)
  rands(s(x),y) -> rands(x,::(rand(0),y))
  rand(x) -> x
  rand(x) -> rand(s(x))

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

  tests#_A((x1,y1,z1)) = (4, 2, 2)
  s_A((x1,y1,z1)) = (x1, y1 + 1, z1 + 1)
  test#_A((x1,y1,z1)) = (x1, y1, z1)
  rands_A((x1,y1,z1),(x2,y2,z2)) = (x1 + 1, y1, z1)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 1)
  0_A = (1, 1, 1)
  nil_A = (1, 1, 1)
  rands#_A((x1,y1,z1),(x2,y2,z2)) = (x1, y1 + 2, z1 + 2)
  done_A((x1,y1,z1)) = (1, 2, 2)
  eq#_A((x1,y1,z1),(x2,y2,z2)) = (0, 0, 0)
  f_A((x1,y1,z1)) = (1, 1, 1)
  g_A((x1,y1,z1)) = (1, 1, 1)
  ::_A((x1,y1,z1),(x2,y2,z2)) = (x2 + 1, y2 + 1, z2 + 1)
  tests_A((x1,y1,z1)) = (x1 + 1, 1, 0)
  true_A = (0, 3, 3)
  and_A((x1,y1,z1),(x2,y2,z2)) = (0, 0, 1)
  test_A((x1,y1,z1)) = (2, 1, 1)
  eq_A((x1,y1,z1),(x2,y2,z2)) = (1, 2, 2)

strictly orients the following dependency pairs:

  tests#(s(x)) -> test#(rands(rand(0),nil))
  tests#(s(x)) -> rands#(rand(0),nil)
  test#(done(y)) -> eq#(f(y),g(y))
  rands#(s(x),y) -> rands#(x,::(rand(0),y))

We remove them from the set of dependency pairs.

No dependency pair remains.