YES

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

  not(true) -> false
  not(false) -> true
  evenodd(x,0) -> not(evenodd(x,s(0)))
  evenodd(0,s(0)) -> false
  evenodd(s(x),s(0)) -> evenodd(x,0)

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

  evenodd#(x,0) -> not#(evenodd(x,s(0)))
  evenodd#(x,0) -> evenodd#(x,s(0))
  evenodd#(s(x),s(0)) -> evenodd#(x,0)

and Q consists of the rules:

  not(true) -> false
  not(false) -> true
  evenodd(x,0) -> not(evenodd(x,s(0)))
  evenodd(0,s(0)) -> false
  evenodd(s(x),s(0)) -> evenodd(x,0)
  rand(x) -> x
  rand(x) -> rand(s(x))

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

  evenodd#_A((x1,y1),(x2,y2)) = (1, 1)
  0_A = (1, 1)
  not#_A((x1,y1)) = (0, 0)
  evenodd_A((x1,y1),(x2,y2)) = (2, 1)
  s_A((x1,y1)) = (x1, y1 + 1)
  not_A((x1,y1)) = (x1, y1)
  true_A = (1, 2)
  false_A = (1, 2)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  evenodd#(x,0) -> not#(evenodd(x,s(0)))

We remove them from the set of dependency pairs.

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

  evenodd#_A((x1,y1),(x2,y2)) = (x1, y1)
  0_A = (1, 1)
  s_A((x1,y1)) = (x1, y1 + 1)
  not_A((x1,y1)) = (2, 2)
  true_A = (1, 3)
  false_A = (1, 3)
  evenodd_A((x1,y1),(x2,y2)) = (x1 + 3, y1 + 1)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  evenodd#(s(x),s(0)) -> evenodd#(x,0)

We remove them from the set of dependency pairs.

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

  evenodd#_A((x1,y1),(x2,y2)) = (x2, y2)
  0_A = (1, 2)
  s_A((x1,y1)) = (x1, 1)
  not_A((x1,y1)) = (x1, 1)
  true_A = (1, 1)
  false_A = (1, 1)
  evenodd_A((x1,y1),(x2,y2)) = (x1 + x2, 1)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

  evenodd#(x,0) -> evenodd#(x,s(0))

We remove them from the set of dependency pairs.

No dependency pair remains.