YES

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

  R:
  half(|0|()) -> |0|()
  half(s(s(x))) -> s(half(x))
  log(s(|0|())) -> |0|()
  log(s(s(x))) -> s(log(s(half(x))))

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

-- SCC decomposition.

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

p1: half#(s(s(x))) -> half#(x)
p2: log#(s(s(x))) -> log#(s(half(x)))
p3: log#(s(s(x))) -> half#(x)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(s(x))) -> s(half(x))
r3: log(s(|0|())) -> |0|()
r4: log(s(s(x))) -> s(log(s(half(x))))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}


-- Reduction pair.

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

p1: log#(s(s(x))) -> log#(s(half(x)))

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(s(x))) -> s(half(x))
r3: log(s(|0|())) -> |0|()
r4: log(s(s(x))) -> s(log(s(half(x))))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      log#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,0,0),(0,1,0,0)) x1
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,1,0,0)) x1 + (0,1,1,4)
      half_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,0)) x1 + (0,0,0,1)
      |0|_A() = (1,1,1,1)
      log_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,0,0)) x1 + (0,0,0,4)
      rand_A(x1) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (1,1,1,0)

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: half#(s(s(x))) -> half#(x)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(s(x))) -> s(half(x))
r3: log(s(|0|())) -> |0|()
r4: log(s(s(x))) -> s(log(s(half(x))))
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      half#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,1,1,0)) x1
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,0)) x1 + (0,1,1,1)
      half_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,0,1)) x1
      |0|_A() = (1,1,0,3)
      log_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,0)) x1 + (1,0,0,2)
      rand_A(x1) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (1,0,1,0)

The next rules are strictly ordered:

  p1

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