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:
  
      max/plus interpretation on natural numbers:
      
        conviter#_A(x1,x2) = max{x1 + 5, x2 + 19}
        if#_A(x1,x2,x3) = max{x1 - 19, x2 + 4, x3 + 14}
        zero_A(x1) = max{18, x1 - 5}
        false_A = 54
        half_A(x1) = max{8, x1 - 6}
        cons_A(x1,x2) = max{15, x2 - 39}
        lastbit_A(x1) = x1 + 59
        |0|_A = 4
        s_A(x1) = x1 + 63
        true_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          conviter#_A(x1,x2) = max{0, x1 - 17, x2 - 30}
          if#_A(x1,x2,x3) = max{0, x1 - 30, x2 - 17, x3 - 30}
          zero_A(x1) = max{14, x1 - 14}
          false_A = 0
          half_A(x1) = 16
          cons_A(x1,x2) = max{1, x1 - 3, x2}
          lastbit_A(x1) = 13
          |0|_A = 1
          s_A(x1) = max{13, x1}
          true_A = 0
          conviter#!_A(x1,x2) = max{9, x1 - 17, x2 - 2}
          if#!_A(x1,x2,x3) = max{0, x2 - 18}
          zero!_A(x1) = max{8, x1 - 18}
          false!_A = 7
          half!_A(x1) = max{13, x1 - 12}
          cons!_A(x1,x2) = max{1, x1 - 16, x2 - 13}
          lastbit!_A(x1) = max{8, x1 - 11}
          |0|!_A = 9
          s!_A(x1) = max{14, x1 - 4}
          true!_A = 0
  
    partial status:
  
      pi(conviter#) = []
      pi(if#) = []
      pi(zero) = []
      pi(false) = []
      pi(half) = []
      pi(cons) = []
      pi(lastbit) = []
      pi(|0|) = []
      pi(s) = [1]
      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

  r4, r5, r6, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        half#_A(x1) = max{1, x1 - 33}
        s_A(x1) = max{19, x1 + 16}
        lastbit_A(x1) = max{20, x1 - 11}
        |0|_A = 2
        zero_A(x1) = max{7, x1 - 20}
        false_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          half#_A(x1) = 0
          s_A(x1) = max{2, x1 - 5}
          lastbit_A(x1) = 2
          |0|_A = 2
          zero_A(x1) = max{24, x1 - 3}
          false_A = 1
          half#!_A(x1) = max{1, x1 - 9}
          s!_A(x1) = max{3, x1 - 3}
          lastbit!_A(x1) = max{2, x1 - 9}
          |0|!_A = 1
          zero!_A(x1) = max{3, x1 - 3}
          false!_A = 1
  
    partial status:
  
      pi(half#) = []
      pi(s) = [1]
      pi(lastbit) = []
      pi(|0|) = []
      pi(zero) = []
      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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        lastbit#_A(x1) = max{7, x1 - 21}
        s_A(x1) = max{19, x1 + 10}
        half_A(x1) = max{3, x1 + 1}
        |0|_A = 10
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          lastbit#_A(x1) = 0
          s_A(x1) = max{1, x1 - 9}
          half_A(x1) = max{8, x1 - 20}
          |0|_A = 1
          lastbit#!_A(x1) = max{5, x1 - 18}
          s!_A(x1) = max{5, x1 - 9}
          half!_A(x1) = max{4, x1 - 10}
          |0|!_A = 4
  
    partial status:
  
      pi(lastbit#) = []
      pi(s) = [1]
      pi(half) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.