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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          conv#_A(x1) = x1
          s_A(x1) = x1 + 4
          half_A(x1) = x1
          |0|_A() = 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          conv#_A(x1) = max{0, x1 - 9}
          s_A(x1) = max{2, x1}
          half_A(x1) = max{8, x1 - 14}
          |0|_A = 2
          conv#!_A(x1) = x1 + 2
          s!_A(x1) = max{19, x1 - 9}
          half!_A(x1) = max{16, x1 - 10}
          |0|!_A = 15
  
    partial status:
  
      pi(conv#) = [1]
      pi(s) = [1]
      pi(half) = []
      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(|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

  r4

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          half#_A(x1) = 0
          s_A(x1) = 1
          lastbit_A(x1) = 3
          |0|_A() = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          half#_A(x1) = 0
          s_A(x1) = max{4, x1 + 2}
          lastbit_A(x1) = max{9, x1 - 10}
          |0|_A = 7
          half#!_A(x1) = max{5, x1 + 2}
          s!_A(x1) = 4
          lastbit!_A(x1) = 4
          |0|!_A = 4
  
    partial status:
  
      pi(half#) = []
      pi(s) = []
      pi(lastbit) = []
      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: 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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          lastbit#_A(x1) = 0
          s_A(x1) = 3
          half_A(x1) = 4
          |0|_A() = 3
          lastbit_A(x1) = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          lastbit#_A(x1) = 0
          s_A(x1) = max{6, x1 + 2}
          half_A(x1) = max{8, x1 - 9}
          |0|_A = 3
          lastbit_A(x1) = max{5, x1 - 10}
          lastbit#!_A(x1) = max{1, x1 - 1}
          s!_A(x1) = max{0, x1 - 20}
          half!_A(x1) = 3
          |0|!_A = 2
          lastbit!_A(x1) = 3
  
    partial status:
  
      pi(lastbit#) = []
      pi(s) = []
      pi(half) = []
      pi(|0|) = []
      pi(lastbit) = []

The following rules are removed:

  p1

Then no dependency pair remains.