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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      log#_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,1,0,0)) x1
      s_A(x1) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (2,4,4,1)
      half_A(x1) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (0,1,1,2)
      |0|_A() = (1,1,1,1)
      log_A(x1) = (1,1,0,0)

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      half#_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,1,0,1)) x1
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,1,0,1)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5

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