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, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            conviter#_A(x1,x2) = 1
            if#_A(x1,x2,x3) = max{1, x1 - 36}
            zero_A(x1) = 37
            false_A = 1
            half_A(x1) = 47
            cons_A(x1,x2) = max{7, x2 + 5}
            lastbit_A(x1) = max{28, x1}
            |0|_A = 4
            s_A(x1) = 47
            true_A = 1
            conviter#!_A(x1,x2) = max{22, x1 - 48}
            if#!_A(x1,x2,x3) = max{22, x2 - 48}
            zero!_A(x1) = max{21, x1 - 26}
            false!_A = 20
            half!_A(x1) = max{64, x1 + 13}
            cons!_A(x1,x2) = max{23, x1 - 5, x2 - 4}
            lastbit!_A(x1) = max{19, x1 + 15}
            |0|!_A = 18
            s!_A(x1) = 63
            true!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            conviter#_A(x1,x2) = max{70, x1 - 2}
            if#_A(x1,x2,x3) = max{70, x1 - 1, x2 - 2}
            zero_A(x1) = max{0, x1 - 5}
            false_A = 70
            half_A(x1) = max{63, x1 - 20}
            cons_A(x1,x2) = 100
            lastbit_A(x1) = 0
            |0|_A = 4
            s_A(x1) = max{102, x1 + 20}
            true_A = 0
            conviter#!_A(x1,x2) = x1 + 5
            if#!_A(x1,x2,x3) = max{x1 - 1, x2 + 4}
            zero!_A(x1) = 2
            false!_A = 110
            half!_A(x1) = 109
            cons!_A(x1,x2) = max{78, x2 - 1}
            lastbit!_A(x1) = 80
            |0|!_A = 108
            s!_A(x1) = 1
            true!_A = 1
        
  
    partial status:
  
      pi(conviter#) = []
      pi(if#) = []
      pi(zero) = [1]
      pi(false) = []
      pi(half) = [1]
      pi(cons) = [1]
      pi(lastbit) = [1]
      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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            half#_A(x1) = 0
            s_A(x1) = max{1, x1 - 6}
            half_A(x1) = max{5, x1 - 18}
            |0|_A = 1
            half#!_A(x1) = max{6, x1 - 12}
            s!_A(x1) = max{6, x1 - 6}
            half!_A(x1) = max{6, x1 - 12}
            |0|!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            half#_A(x1) = 0
            s_A(x1) = max{2, x1 - 7}
            half_A(x1) = max{6, x1 - 21}
            |0|_A = 2
            half#!_A(x1) = max{3, x1 - 38}
            s!_A(x1) = max{3, x1 - 31}
            half!_A(x1) = max{3, x1 - 38}
            |0|!_A = 2
        
  
    partial status:
  
      pi(half#) = [1]
      pi(s) = [1]
      pi(half) = [1]
      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: 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. 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 - 5}
            |0|_A = 4
            lastbit#!_A(x1) = max{6, x1 - 9}
            s!_A(x1) = max{6, x1 - 9}
            half!_A(x1) = max{6, x1 - 18}
            |0|!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            lastbit#_A(x1) = 0
            s_A(x1) = max{1, x1 - 6}
            half_A(x1) = max{5, x1 - 19}
            |0|_A = 1
            lastbit#!_A(x1) = max{3, x1 - 26}
            s!_A(x1) = max{3, x1 - 21}
            half!_A(x1) = max{3, x1 - 27}
            |0|!_A = 1
        
  
    partial status:
  
      pi(lastbit#) = [1]
      pi(s) = [1]
      pi(half) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.