YES

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

  rev(nil) -> nil
  rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
  rev1(0,nil) -> 0
  rev1(s(x),nil) -> s(x)
  rev1(x,cons(y,l)) -> rev1(y,l)
  rev2(x,nil) -> nil
  rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))

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

  rev#(cons(x,l)) -> rev1#(x,l)
  rev#(cons(x,l)) -> rev2#(x,l)
  rev1#(x,cons(y,l)) -> rev1#(y,l)
  rev2#(x,cons(y,l)) -> rev#(cons(x,rev2(y,l)))
  rev2#(x,cons(y,l)) -> rev2#(y,l)

and Q consists of the rules:

  rev(nil) -> nil
  rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
  rev1(0,nil) -> 0
  rev1(s(x),nil) -> s(x)
  rev1(x,cons(y,l)) -> rev1(y,l)
  rev2(x,nil) -> nil
  rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))
  rand(x) -> x
  rand(x) -> rand(s(x))

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

  rev#_A((x1,y1,z1)) = (x1, y1, 1)
  cons_A((x1,y1,z1),(x2,y2,z2)) = (x2 + 2, 1, 3)
  rev1#_A((x1,y1,z1),(x2,y2,z2)) = (0, 1, 0)
  rev2#_A((x1,y1,z1),(x2,y2,z2)) = (x2 + 1, y2 + 2, 0)
  rev2_A((x1,y1,z1),(x2,y2,z2)) = (x2, 3, 1)
  rev_A((x1,y1,z1)) = (x1, 2, 2)
  nil_A = (1, 1, 1)
  rev1_A((x1,y1,z1),(x2,y2,z2)) = (2, 1, 1)
  0_A = (1, 2, 2)
  s_A((x1,y1,z1)) = (0, 0, 0)
  rand_A((x1,y1,z1)) = (x1 + 1, y1 + 1, z1)

strictly orients the following dependency pairs:

  rev#(cons(x,l)) -> rev1#(x,l)
  rev#(cons(x,l)) -> rev2#(x,l)
  rev2#(x,cons(y,l)) -> rev#(cons(x,rev2(y,l)))
  rev2#(x,cons(y,l)) -> rev2#(y,l)

We remove them from the set of dependency pairs.

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

  rev1#_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2, y1 + y2, z1 + z2)
  cons_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2 + 1, y1 + 1, 1)
  rev_A((x1,y1,z1)) = (x1, y1 + 3, z1)
  nil_A = (1, 1, 2)
  rev1_A((x1,y1,z1),(x2,y2,z2)) = (0, 2, 1)
  rev2_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2, y1 + 5, 1)
  0_A = (0, 1, 2)
  s_A((x1,y1,z1)) = (0, 0, 0)
  rand_A((x1,y1,z1)) = (x1 + 1, y1 + 1, z1)

strictly orients the following dependency pairs:

  rev1#(x,cons(y,l)) -> rev1#(y,l)

We remove them from the set of dependency pairs.

No dependency pair remains.