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-minimal dependency 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-minimal dependency 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          log#_A(x1) = max{2, x1}
          s_A(x1) = max{3, x1}
          half_A(x1) = max{3, x1}
          |0|_A = 6
          log_A(x1) = max{5, x1 + 4}
          rand_A(x1) = max{6, x1 + 3}
    
      precedence:
      
        log > s > log# = half = |0| = rand
      
      partial status:
    
        pi(log#) = [1]
        pi(s) = [1]
        pi(half) = [1]
        pi(|0|) = []
        pi(log) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          log#_A(x1) = 1
          s_A(x1) = x1 + 3
          half_A(x1) = max{0, x1 - 3}
          |0|_A = 8
          log_A(x1) = x1 + 8
          rand_A(x1) = 2
    
      precedence:
      
        log > half = |0| > log# = s > rand
      
      partial status:
    
        pi(log#) = []
        pi(s) = [1]
        pi(half) = []
        pi(|0|) = []
        pi(log) = [1]
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal dependency 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          half#_A(x1) = x1
          s_A(x1) = x1
          half_A(x1) = x1
          |0|_A = 1
          log_A(x1) = x1 + 2
          rand_A(x1) = x1 + 1
    
      precedence:
      
        log > s = |0| > half# = half = rand
      
      partial status:
    
        pi(half#) = [1]
        pi(s) = [1]
        pi(half) = [1]
        pi(|0|) = []
        pi(log) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          half#_A(x1) = x1 + 7
          s_A(x1) = x1 + 6
          half_A(x1) = x1
          |0|_A = 4
          log_A(x1) = max{3, x1 - 3}
          rand_A(x1) = 0
    
      precedence:
      
        half = log > |0| > half# = s > rand
      
      partial status:
    
        pi(half#) = [1]
        pi(s) = [1]
        pi(half) = [1]
        pi(|0|) = []
        pi(log) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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