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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        rands#_A(x1,x2) = x1
        s_A(x1) = ((1,0,0),(1,1,0),(0,0,1)) x1 + (0,2,1)
        |::|_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x2 + (0,1,1)
        rand_A(x1) = x1 + (1,1,1)
        |0|_A() = (1,1,1)
        tests_A(x1) = (1,1,6)
        true_A() = (0,0,1)
        and_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1
        test_A(x1) = x1 + (2,0,0)
        rands_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1 + ((0,0,0),(1,0,0),(1,1,0)) x2 + (1,0,1)
        nil_A() = (1,1,1)
        done_A(x1) = (1,1,1)
        eq_A(x1,x2) = x1 + (1,1,0)
        f_A(x1) = ((0,0,0),(1,0,0),(1,1,0)) x1 + (1,1,2)
        g_A(x1) = ((0,0,0),(1,0,0),(1,1,0)) x1 + (1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        done > eq > s > |::| > rands > test > true > f > |0| > nil > rands# > and > g > tests > rand
      
      argument filter:
    
        pi(rands#) = []
        pi(s) = []
        pi(|::|) = []
        pi(rand) = []
        pi(|0|) = []
        pi(tests) = []
        pi(true) = []
        pi(and) = []
        pi(test) = [1]
        pi(rands) = []
        pi(nil) = []
        pi(done) = []
        pi(eq) = 1
        pi(f) = []
        pi(g) = []
    

The next rules are strictly ordered:

  p1

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