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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rands#_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (5,2)
            s_A(x1) = x1 + (0,6)
            |::|_A(x1,x2) = x2 + (0,1)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,7)
            |0|_A() = (3,9)
            tests_A(x1) = x1 + (7,10)
            true_A() = (1,1)
            and_A(x1,x2) = (0,0)
            test_A(x1) = (6,7)
            rands_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,4)
            nil_A() = (0,0)
            done_A(x1) = (4,3)
            eq_A(x1,x2) = (2,2)
            f_A(x1) = (5,4)
            g_A(x1) = x1 + (1,4)
    
      precedence:
      
        rands# = s = |::| = rand = |0| = tests = true = and = test = rands = nil = done = eq = f = g
      
      partial status:
    
        pi(rands#) = []
        pi(s) = []
        pi(|::|) = []
        pi(rand) = []
        pi(|0|) = []
        pi(tests) = []
        pi(true) = []
        pi(and) = []
        pi(test) = []
        pi(rands) = []
        pi(nil) = []
        pi(done) = []
        pi(eq) = []
        pi(f) = []
        pi(g) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            rands#_A(x1,x2) = x1
            s_A(x1) = ((1,0),(1,1)) x1 + (5,4)
            |::|_A(x1,x2) = ((1,0),(1,1)) x2 + (4,6)
            rand_A(x1) = (1,1)
            |0|_A() = (3,5)
            tests_A(x1) = x1
            true_A() = (1,4)
            and_A(x1,x2) = (0,0)
            test_A(x1) = (4,3)
            rands_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (4,1)
            nil_A() = (1,1)
            done_A(x1) = (2,2)
            eq_A(x1,x2) = (3,5)
            f_A(x1) = (1,1)
            g_A(x1) = (1,1)
    
      precedence:
      
        test > rands > g > |::| > and > done > tests > eq > f > |0| > true > rands# = s = rand = nil
      
      partial status:
    
        pi(rands#) = []
        pi(s) = []
        pi(|::|) = []
        pi(rand) = []
        pi(|0|) = []
        pi(tests) = []
        pi(true) = []
        pi(and) = []
        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.