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, >) with

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

strictly orients the following 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)

We remove them from the set of dependency pairs.

No dependency pair remains.