YES
We show the termination of the TRS R:

  half(|0|()) -> |0|()
  half(s(|0|())) -> |0|()
  half(s(s(x))) -> s(half(x))
  lastbit(|0|()) -> |0|()
  lastbit(s(|0|())) -> s(|0|())
  lastbit(s(s(x))) -> lastbit(x)
  conv(|0|()) -> cons(nil(),|0|())
  conv(s(x)) -> cons(conv(half(s(x))),lastbit(s(x)))


-- SCC decomposition.

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

p1: half#(s(s(x))) -> half#(x)
p2: lastbit#(s(s(x))) -> lastbit#(x)
p3: conv#(s(x)) -> conv#(half(s(x)))
p4: conv#(s(x)) -> half#(s(x))
p5: conv#(s(x)) -> lastbit#(s(x))

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: lastbit(|0|()) -> |0|()
r5: lastbit(s(|0|())) -> s(|0|())
r6: lastbit(s(s(x))) -> lastbit(x)
r7: conv(|0|()) -> cons(nil(),|0|())
r8: conv(s(x)) -> cons(conv(half(s(x))),lastbit(s(x)))

The estimated dependency graph contains the following SCCs:

  {p3}
  {p1}
  {p2}


-- Reduction pair.

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

p1: conv#(s(x)) -> conv#(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: lastbit(|0|()) -> |0|()
r5: lastbit(s(|0|())) -> s(|0|())
r6: lastbit(s(s(x))) -> lastbit(x)
r7: conv(|0|()) -> cons(nil(),|0|())
r8: conv(s(x)) -> cons(conv(half(s(x))),lastbit(s(x)))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        conv#_A(x1) = max{1, x1 - 5}
        s_A(x1) = x1 + 22
        half_A(x1) = max{3, x1 - 18}
        |0|_A = 1
        lastbit_A(x1) = max{30, x1 - 45}
  
    base order 2:
  
      precedence:
      
        lastbit > s > conv# > half > |0|
      
  
    partial status:
  
      pi(conv#) = []
      pi(s) = []
      pi(half) = []
      pi(|0|) = []
      pi(lastbit) = []

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(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: lastbit(|0|()) -> |0|()
r5: lastbit(s(|0|())) -> s(|0|())
r6: lastbit(s(s(x))) -> lastbit(x)
r7: conv(|0|()) -> cons(nil(),|0|())
r8: conv(s(x)) -> cons(conv(half(s(x))),lastbit(s(x)))

The set of usable rules consists of

  r1, r2, r4, r5, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        half#_A(x1) = max{8, x1 + 1}
        s_A(x1) = max{8, x1 + 6}
        half_A(x1) = max{5, x1 - 15}
        |0|_A = 3
        lastbit_A(x1) = max{13, x1 - 13}
        conv_A(x1) = max{4, x1 - 26}
        cons_A(x1,x2) = max{2, x1 - 5, x2 - 14}
        nil_A = 2
  
    base order 2:
  
      precedence:
      
        s = lastbit > half# = half > conv > |0| > cons = nil
      
  
    partial status:
  
      pi(half#) = [1]
      pi(s) = [1]
      pi(half) = []
      pi(|0|) = []
      pi(lastbit) = []
      pi(conv) = []
      pi(cons) = []
      pi(nil) = []

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: lastbit#(s(s(x))) -> lastbit#(x)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: lastbit(|0|()) -> |0|()
r5: lastbit(s(|0|())) -> s(|0|())
r6: lastbit(s(s(x))) -> lastbit(x)
r7: conv(|0|()) -> cons(nil(),|0|())
r8: conv(s(x)) -> cons(conv(half(s(x))),lastbit(s(x)))

The set of usable rules consists of

  r1, r2, r4, r5, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        lastbit#_A(x1) = max{15, x1 + 2}
        s_A(x1) = max{16, x1 + 7}
        half_A(x1) = max{15, x1 - 29}
        |0|_A = 10
        lastbit_A(x1) = max{22, x1 - 18}
        conv_A(x1) = max{10, x1 - 30}
        cons_A(x1,x2) = max{10, x1 - 11, x2 - 11}
        nil_A = 0
  
    base order 2:
  
      precedence:
      
        s > half > lastbit# = lastbit = conv > |0| > cons = nil
      
  
    partial status:
  
      pi(lastbit#) = [1]
      pi(s) = [1]
      pi(half) = []
      pi(|0|) = []
      pi(lastbit) = []
      pi(conv) = []
      pi(cons) = []
      pi(nil) = []

The following rules are removed:

  p1

Then no dependency pair remains.