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)
  zero(|0|()) -> true()
  zero(s(x)) -> false()
  conv(x) -> conviter(x,cons(|0|(),nil()))
  conviter(x,l) -> if(zero(x),x,l)
  if(true(),x,l) -> l
  if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l))


-- 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#(x) -> conviter#(x,cons(|0|(),nil()))
p4: conviter#(x,l) -> if#(zero(x),x,l)
p5: conviter#(x,l) -> zero#(x)
p6: if#(false(),x,l) -> conviter#(half(x),cons(lastbit(x),l))
p7: if#(false(),x,l) -> half#(x)
p8: if#(false(),x,l) -> 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: zero(|0|()) -> true()
r8: zero(s(x)) -> false()
r9: conv(x) -> conviter(x,cons(|0|(),nil()))
r10: conviter(x,l) -> if(zero(x),x,l)
r11: if(true(),x,l) -> l
r12: if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l))

The estimated dependency graph contains the following SCCs:

  {p4, p6}
  {p1}
  {p2}


-- Reduction pair.

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

p1: conviter#(x,l) -> if#(zero(x),x,l)
p2: if#(false(),x,l) -> conviter#(half(x),cons(lastbit(x),l))

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: zero(|0|()) -> true()
r8: zero(s(x)) -> false()
r9: conv(x) -> conviter(x,cons(|0|(),nil()))
r10: conviter(x,l) -> if(zero(x),x,l)
r11: if(true(),x,l) -> l
r12: if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

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:
          conviter#_A(x1,x2) = 5
          if#_A(x1,x2,x3) = 5
          zero_A(x1) = 4
          false_A() = 4
          half_A(x1) = 8
          cons_A(x1,x2) = x2 + 8
          lastbit_A(x1) = 6
          |0|_A() = 2
          s_A(x1) = 3
          true_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          conviter#_A(x1,x2) = max{1, x1 - 2, x2 - 10}
          if#_A(x1,x2,x3) = max{1, x1 - 43, x2 - 2}
          zero_A(x1) = max{0, x1 - 10}
          false_A = 6
          half_A(x1) = max{2, x1 - 8}
          cons_A(x1,x2) = max{8, x1 - 70}
          lastbit_A(x1) = 44
          |0|_A = 1
          s_A(x1) = max{17, x1 + 8}
          true_A = 0
          conviter#!_A(x1,x2) = max{x1 + 12, x2 + 2}
          if#!_A(x1,x2,x3) = max{x1 + 9, x2 + 11, x3 + 1}
          zero!_A(x1) = 39
          false!_A = 15
          half!_A(x1) = x1 + 12
          cons!_A(x1,x2) = max{42, x1 - 2, x2 + 39}
          lastbit!_A(x1) = 0
          |0|!_A = 39
          s!_A(x1) = max{38, x1 + 1}
          true!_A = 1
  
    partial status:
  
      pi(conviter#) = []
      pi(if#) = []
      pi(zero) = []
      pi(false) = []
      pi(half) = []
      pi(cons) = [2]
      pi(lastbit) = []
      pi(|0|) = []
      pi(s) = []
      pi(true) = []

The following rules are removed:

  p1, p2

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: zero(|0|()) -> true()
r8: zero(s(x)) -> false()
r9: conv(x) -> conviter(x,cons(|0|(),nil()))
r10: conviter(x,l) -> if(zero(x),x,l)
r11: if(true(),x,l) -> l
r12: if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l))

The set of usable rules consists of

  r1, r2, r8

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) = x1
          s_A(x1) = x1 + 5
          half_A(x1) = x1 + 4
          |0|_A() = 2
          zero_A(x1) = x1 + 1
          false_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          half#_A(x1) = x1
          s_A(x1) = max{38, x1 + 14}
          half_A(x1) = max{42, x1 - 53}
          |0|_A = 14
          zero_A(x1) = max{50, x1 - 39}
          false_A = 1
          half#!_A(x1) = max{7, x1 - 44}
          s!_A(x1) = max{7, x1 - 44}
          half!_A(x1) = max{8, x1 - 29}
          |0|!_A = 5
          zero!_A(x1) = max{3, x1 - 15}
          false!_A = 1
  
    partial status:
  
      pi(half#) = []
      pi(s) = [1]
      pi(half) = [1]
      pi(|0|) = []
      pi(zero) = [1]
      pi(false) = []

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: zero(|0|()) -> true()
r8: zero(s(x)) -> false()
r9: conv(x) -> conviter(x,cons(|0|(),nil()))
r10: conviter(x,l) -> if(zero(x),x,l)
r11: if(true(),x,l) -> l
r12: if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l))

The set of usable rules consists of

  r1, r2, r4, r5

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) = x1 + 3
          s_A(x1) = x1 + 4
          half_A(x1) = x1 + 1
          |0|_A() = 2
          lastbit_A(x1) = 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{5, x1 - 14}
          half_A(x1) = max{13, x1 - 14}
          |0|_A = 5
          lastbit_A(x1) = max{7, x1 - 15}
          lastbit#!_A(x1) = max{1, x1 - 28}
          s!_A(x1) = max{5, x1 - 14}
          half!_A(x1) = 0
          |0|!_A = 4
          lastbit!_A(x1) = max{5, x1 - 28}
  
    partial status:
  
      pi(lastbit#) = []
      pi(s) = [1]
      pi(half) = [1]
      pi(|0|) = []
      pi(lastbit) = []

The following rules are removed:

  p1

Then no dependency pair remains.