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, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        log#_A(x1) = max{12, x1 + 1}
        s_A(x1) = max{15, x1 + 11}
        half_A(x1) = max{6, x1 - 8}
        |0|_A = 5
        log_A(x1) = max{10, x1 - 29}
  
    base order 2:
  
      precedence:
      
        log# = s = log > half > |0|
      
  
    partial status:
  
      pi(log#) = [1]
      pi(s) = [1]
      pi(half) = []
      pi(|0|) = []
      pi(log) = []

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

  r3

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        half#_A(x1) = max{1, x1 - 1}
        s_A(x1) = x1 + 5
        log_A(x1) = max{4, x1 - 22}
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        half# = s > log > |0|
      
  
    partial status:
  
      pi(half#) = []
      pi(s) = [1]
      pi(log) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.