YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: half#(s(s(x))) -> half#(x)
p2: bits#(s(x)) -> bits#(half(s(x)))
p3: bits#(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: bits(|0|()) -> |0|()
r5: bits(s(x)) -> s(bits(half(s(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: bits#(s(x)) -> bits#(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: bits(|0|()) -> |0|()
r5: bits(s(x)) -> s(bits(half(s(x))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        bits#_A(x1) = max{5, x1 + 3}
        s_A(x1) = x1 + 14
        half_A(x1) = max{8, x1 - 5}
        |0|_A = 7
  
    precedence:
    
      bits# = s > half = |0|
    
    partial status:
  
      pi(bits#) = [1]
      pi(s) = []
      pi(half) = []
      pi(|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: bits(|0|()) -> |0|()
r5: bits(s(x)) -> s(bits(half(s(x))))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        half#_A(x1) = x1 + 5
        s_A(x1) = x1 + 2
  
    precedence:
    
      half# = s
    
    partial status:
  
      pi(half#) = [1]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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