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 reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              log#_A(x1) = x1
              s_A(x1) = x1 + 5
              half_A(x1) = x1 + 1
              |0|_A() = 1
              log#!_A(x1) = 4
              s!_A(x1) = x1
              half!_A(x1) = 3
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            log#_A(x1) = 0
            s_A(x1) = max{1, x1 - 6}
            half_A(x1) = 4
            |0|_A = 4
            log#!_A(x1) = max{27, x1 - 6}
            s!_A(x1) = max{25, x1 - 10}
            half!_A(x1) = 26
            |0|!_A = 1
        
  
    partial status:
  
      pi(log#) = [1]
      pi(s) = [1]
      pi(half) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

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 reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              half#_A(x1) = x1
              s_A(x1) = x1 + 2
              half#!_A(x1) = x1 + 4
              s!_A(x1) = x1 + 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            half#_A(x1) = 0
            s_A(x1) = max{14, x1 - 9}
            half#!_A(x1) = max{3, x1 - 21}
            s!_A(x1) = max{1, x1 - 12}
        
  
    partial status:
  
      pi(half#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.