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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          rands#_A(x1,x2) = max{x1 + 21, x2 + 19}
          s_A(x1) = max{6, x1}
          |::|_A(x1,x2) = x2
          rand_A(x1) = max{14, x1 + 7}
          |0|_A = 10
          tests_A(x1) = x1 + 48
          true_A = 3
          and_A(x1,x2) = max{x1, x2 + 48}
          test_A(x1) = 5
          rands_A(x1,x2) = max{x1 + 21, x2 + 18}
          nil_A = 0
          done_A(x1) = max{0, x1 - 1}
          eq_A(x1,x2) = max{4, x2}
          f_A(x1) = x1 + 5
          g_A(x1) = 5
    
      precedence:
      
        rands# = tests > |0| = true = and = test = g > f > s > nil > eq > rands > |::| = rand = done
      
      partial status:
    
        pi(rands#) = [1]
        pi(s) = [1]
        pi(|::|) = []
        pi(rand) = []
        pi(|0|) = []
        pi(tests) = [1]
        pi(true) = []
        pi(and) = [1, 2]
        pi(test) = []
        pi(rands) = [2]
        pi(nil) = []
        pi(done) = []
        pi(eq) = []
        pi(f) = []
        pi(g) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          rands#_A(x1,x2) = x1 + 18
          s_A(x1) = max{28, x1}
          |::|_A(x1,x2) = 2
          rand_A(x1) = 27
          |0|_A = 19
          tests_A(x1) = x1 + 6
          true_A = 24
          and_A(x1,x2) = max{x1 + 45, x2 + 7}
          test_A(x1) = 24
          rands_A(x1,x2) = max{22, x2 + 20}
          nil_A = 7
          done_A(x1) = 22
          eq_A(x1,x2) = 23
          f_A(x1) = 21
          g_A(x1) = 19
    
      precedence:
      
        rands# = s = |::| = rand = |0| = tests = true = and = test = rands = nil = eq = f > done = g
      
      partial status:
    
        pi(rands#) = [1]
        pi(s) = [1]
        pi(|::|) = []
        pi(rand) = []
        pi(|0|) = []
        pi(tests) = [1]
        pi(true) = []
        pi(and) = [1]
        pi(test) = []
        pi(rands) = []
        pi(nil) = []
        pi(done) = []
        pi(eq) = []
        pi(f) = []
        pi(g) = []
    

The next rules are strictly ordered:

  p1

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