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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          conviter#_A(x1,x2) = max{40, x1 + 37, x2 + 21}
          if#_A(x1,x2,x3) = max{x1 + 34, x2 + 33, x3 + 20}
          zero_A(x1) = max{2, x1 - 2}
          false_A = 6
          half_A(x1) = max{3, x1 - 5}
          cons_A(x1,x2) = max{11, x1, x2 - 2}
          lastbit_A(x1) = max{19, x1 + 9}
          |0|_A = 2
          s_A(x1) = max{19, x1 + 9}
          true_A = 1
    
      precedence:
      
        s > lastbit > cons = |0| > if# = true > conviter# = zero = false = half
      
      partial status:
    
        pi(conviter#) = [2]
        pi(if#) = [1, 2, 3]
        pi(zero) = []
        pi(false) = []
        pi(half) = []
        pi(cons) = [1]
        pi(lastbit) = [1]
        pi(|0|) = []
        pi(s) = [1]
        pi(true) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          conviter#_A(x1,x2) = max{11, x2 + 9}
          if#_A(x1,x2,x3) = max{12, x1 + 7}
          zero_A(x1) = 1
          false_A = 3
          half_A(x1) = 0
          cons_A(x1,x2) = 0
          lastbit_A(x1) = x1 + 22
          |0|_A = 10
          s_A(x1) = x1 + 4
          true_A = 11
    
      precedence:
      
        conviter# = if# = zero = false = half = cons = lastbit = |0| = s = true
      
      partial status:
    
        pi(conviter#) = [2]
        pi(if#) = []
        pi(zero) = []
        pi(false) = []
        pi(half) = []
        pi(cons) = []
        pi(lastbit) = [1]
        pi(|0|) = []
        pi(s) = [1]
        pi(true) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          half#_A(x1) = max{2, x1 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        half# = s
      
      partial status:
    
        pi(half#) = [1]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          half#_A(x1) = x1 + 1
          s_A(x1) = x1
    
      precedence:
      
        half# = s
      
      partial status:
    
        pi(half#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  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

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          lastbit#_A(x1) = max{2, x1 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        lastbit# = s
      
      partial status:
    
        pi(lastbit#) = [1]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          lastbit#_A(x1) = x1 + 1
          s_A(x1) = x1
    
      precedence:
      
        lastbit# = s
      
      partial status:
    
        pi(lastbit#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.