YES

We show the termination of the relative TRS R/S:

  R:
  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))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

Consider the non-minimal dependency pair problem (P, R), where P consists of

p1: tests#(s(x)) -> test#(rands(rand(|0|()),nil()))
p2: tests#(s(x)) -> rands#(rand(|0|()),nil())
p3: test#(done(y)) -> eq#(f(y),g(y))
p4: rands#(s(x),y) -> rands#(x,|::|(rand(|0|()),y))

and R consists of:

r1: tests(|0|()) -> true()
r2: tests(s(x)) -> and(test(rands(rand(|0|()),nil())),x)
r3: test(done(y)) -> eq(f(y),g(y))
r4: eq(x,x) -> true()
r5: rands(|0|(),y) -> done(y)
r6: rands(s(x),y) -> rands(x,|::|(rand(|0|()),y))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p4}


-- Reduction pair.

Consider the non-minimal dependency pair problem (P, R), where P consists of

p1: rands#(s(x),y) -> rands#(x,|::|(rand(|0|()),y))

and R consists of:

r1: tests(|0|()) -> true()
r2: tests(s(x)) -> and(test(rands(rand(|0|()),nil())),x)
r3: test(done(y)) -> eq(f(y),g(y))
r4: eq(x,x) -> true()
r5: rands(|0|(),y) -> done(y)
r6: rands(s(x),y) -> rands(x,|::|(rand(|0|()),y))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        rands#_A(x1,x2) = max{x1 + 7, x2 + 10}
        s_A(x1) = max{9, x1}
        |::|_A(x1,x2) = max{6, x1 - 27, x2}
        rand_A(x1) = max{33, x1 + 23}
        |0|_A = 6
        tests_A(x1) = max{77, x1 + 32}
        true_A = 23
        and_A(x1,x2) = max{x1 - 11, x2 + 32}
        test_A(x1) = x1 + 41
        rands_A(x1,x2) = max{x1 + 14, x2 + 7}
        nil_A = 1
        done_A(x1) = x1 + 7
        eq_A(x1,x2) = max{x1 + 24, x2 + 24}
        f_A(x1) = max{22, x1 + 1}
        g_A(x1) = x1 + 6
  
    precedence:
    
      rands# = s = |::| = |0| = tests = true = and = test = rands = nil = done = eq = f = g > rand
    
    partial status:
  
      pi(rands#) = [1]
      pi(s) = [1]
      pi(|::|) = []
      pi(rand) = []
      pi(|0|) = []
      pi(tests) = [1]
      pi(true) = []
      pi(and) = []
      pi(test) = [1]
      pi(rands) = [1, 2]
      pi(nil) = []
      pi(done) = []
      pi(eq) = [2]
      pi(f) = [1]
      pi(g) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.