YES

We show the termination of the TRS R:

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

-- SCC decomposition.

Consider the 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))))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}


-- Reduction pair.

Consider the 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))))

The set of usable rules consists of

  r1, r2

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > s > half > log#
      
      argument filter:
    
        pi(log#) = 1
        pi(s) = 1
        pi(half) = 1
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > s > half > log#
      
      argument filter:
    
        pi(log#) = 1
        pi(s) = [1]
        pi(half) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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

-- Reduction pair.

Consider the 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))))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > half#
      
      argument filter:
    
        pi(half#) = [1]
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > half#
      
      argument filter:
    
        pi(half#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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