YES

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

  minus(x,0) -> x
  minus(s(x),s(y)) -> minus(x,y)
  f(0) -> s(0)
  f(s(x)) -> minus(s(x),g(f(x)))
  g(0) -> 0
  g(s(x)) -> minus(s(x),f(g(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

  minus#(s(x),s(y)) -> minus#(x,y)
  f#(s(x)) -> minus#(s(x),g(f(x)))
  f#(s(x)) -> g#(f(x))
  f#(s(x)) -> f#(x)
  g#(s(x)) -> minus#(s(x),f(g(x)))
  g#(s(x)) -> f#(g(x))
  g#(s(x)) -> g#(x)

and Q consists of the rules:

  minus(x,0) -> x
  minus(s(x),s(y)) -> minus(x,y)
  f(0) -> s(0)
  f(s(x)) -> minus(s(x),g(f(x)))
  g(0) -> 0
  g(s(x)) -> minus(s(x),f(g(x)))
  rand(x) -> x
  rand(x) -> rand(s(x))

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

  minus#_A((x1,y1,z1),(x2,y2,z2)) = (0, 0, 0)
  s_A((x1,y1,z1)) = (x1, y1 + 1, 1)
  f#_A((x1,y1,z1)) = (x1 + 1, 1, 1)
  g_A((x1,y1,z1)) = (x1, y1 + 2, 1)
  f_A((x1,y1,z1)) = (x1, y1 + 2, z1 + 1)
  g#_A((x1,y1,z1)) = (x1 + 1, 1, 1)
  minus_A((x1,y1,z1),(x2,y2,z2)) = (x1, y1 + 1, 2)
  0_A = (1, 1, 0)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 0)

strictly orients the following dependency pairs:

  f#(s(x)) -> minus#(s(x),g(f(x)))
  g#(s(x)) -> minus#(s(x),f(g(x)))

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, 0)
  s_A((x1,y1,z1)) = (x1, y1 + 1, 2)
  f#_A((x1,y1,z1)) = (0, 0, 0)
  g#_A((x1,y1,z1)) = (0, 0, 0)
  f_A((x1,y1,z1)) = (x1 + 2, 4, 1)
  g_A((x1,y1,z1)) = (x1 + 2, y1 + 1, z1 + 1)
  minus_A((x1,y1,z1),(x2,y2,z2)) = (x1 + 1, y1 + 2, 4)
  0_A = (1, 4, 0)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 0)

strictly orients the following dependency pairs:

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

We remove them from the set of dependency pairs.

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

  f#_A((x1,y1,z1)) = (x1, y1 + 2, 2)
  s_A((x1,y1,z1)) = (x1, y1 + 5, z1 + 3)
  g#_A((x1,y1,z1)) = (x1, y1, z1)
  f_A((x1,y1,z1)) = (x1, y1 + 6, 1)
  g_A((x1,y1,z1)) = (x1, y1 + 2, 1)
  minus_A((x1,y1,z1),(x2,y2,z2)) = (x1, y1 + 1, 2)
  0_A = (1, 0, 0)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 0)

strictly orients the following dependency pairs:

  f#(s(x)) -> g#(f(x))
  f#(s(x)) -> f#(x)
  g#(s(x)) -> f#(g(x))
  g#(s(x)) -> g#(x)

We remove them from the set of dependency pairs.

No dependency pair remains.