YES

We show the termination of the TRS R:

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

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: half#(s(s(x))) -> s#(half(x))
p2: half#(s(s(x))) -> half#(x)
p3: s#(log(|0|())) -> s#(|0|())
p4: log#(s(x)) -> s#(log(half(s(x))))
p5: log#(s(x)) -> log#(half(s(x)))
p6: log#(s(x)) -> half#(s(x))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p5}
  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        log#_A(x1) = ((1,0,0),(0,1,0),(0,1,0)) x1
        s_A(x1) = x1 + (1,3,1)
        half_A(x1) = x1 + (0,2,0)
        |0|_A() = (1,1,1)
        log_A(x1) = (2,1,0)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > log# > s > half > log
      
      argument filter:
    
        pi(log#) = []
        pi(s) = []
        pi(half) = []
        pi(|0|) = []
        pi(log) = []
    

The next rules are strictly ordered:

  p1

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(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: s(log(|0|())) -> s(|0|())
r5: log(s(x)) -> s(log(half(s(x))))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        half#_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1
        s_A(x1) = ((1,0,0),(0,1,0),(0,1,1)) x1 + (1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        half# > s
      
      argument filter:
    
        pi(half#) = 1
        pi(s) = 1
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5

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