YES

We show the termination of the relative TRS R/S:

  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)))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

Consider the non-minimal 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)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p3}
  {p1}
  {p2}


-- Reduction pair.

Consider the non-minimal 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)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          conv#_A(x1) = x1 + 2
          s_A(x1) = x1
          half_A(x1) = x1
          |0|_A = 7
          lastbit_A(x1) = max{5, x1 + 3}
          conv_A(x1) = x1 + 4
          cons_A(x1,x2) = max{3, x1 - 1, x2 - 4}
          nil_A = 8
          rand_A(x1) = x1 + 2
    
      precedence:
      
        conv# = s = half = |0| = lastbit = conv = cons = nil = rand
      
      partial status:
    
        pi(conv#) = []
        pi(s) = []
        pi(half) = []
        pi(|0|) = []
        pi(lastbit) = []
        pi(conv) = []
        pi(cons) = []
        pi(nil) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          conv#_A(x1) = max{0, x1 - 11}
          s_A(x1) = max{16, x1 + 12}
          half_A(x1) = max{6, x1 - 5}
          |0|_A = 5
          lastbit_A(x1) = 43
          conv_A(x1) = x1 + 26
          cons_A(x1,x2) = 43
          nil_A = 8
          rand_A(x1) = 0
    
      precedence:
      
        cons > lastbit = conv = nil > s = half = |0| > conv# = rand
      
      partial status:
    
        pi(conv#) = []
        pi(s) = [1]
        pi(half) = []
        pi(|0|) = []
        pi(lastbit) = []
        pi(conv) = [1]
        pi(cons) = []
        pi(nil) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          half#_A(x1) = max{10, x1}
          s_A(x1) = max{11, x1}
          half_A(x1) = max{11, x1}
          |0|_A = 7
          lastbit_A(x1) = max{12, x1 + 3}
          conv_A(x1) = max{15, x1 + 13}
          cons_A(x1,x2) = max{6, x1 - 25}
          nil_A = 6
          rand_A(x1) = max{22, x1 + 11}
    
      precedence:
      
        half# = s > conv > half = |0| = lastbit = cons = nil = rand
      
      partial status:
    
        pi(half#) = [1]
        pi(s) = [1]
        pi(half) = [1]
        pi(|0|) = []
        pi(lastbit) = [1]
        pi(conv) = [1]
        pi(cons) = []
        pi(nil) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          half#_A(x1) = 25
          s_A(x1) = 10
          half_A(x1) = max{18, x1 + 13}
          |0|_A = 24
          lastbit_A(x1) = 9
          conv_A(x1) = max{18, x1 + 2}
          cons_A(x1,x2) = 11
          nil_A = 23
          rand_A(x1) = 1
    
      precedence:
      
        half# = s = half = |0| = lastbit = conv = cons = nil = rand
      
      partial status:
    
        pi(half#) = []
        pi(s) = []
        pi(half) = [1]
        pi(|0|) = []
        pi(lastbit) = []
        pi(conv) = [1]
        pi(cons) = []
        pi(nil) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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)))
r9: rand(x) -> x
r10: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          lastbit#_A(x1) = max{10, x1}
          s_A(x1) = max{11, x1}
          half_A(x1) = max{11, x1}
          |0|_A = 7
          lastbit_A(x1) = max{12, x1 + 3}
          conv_A(x1) = max{15, x1 + 13}
          cons_A(x1,x2) = max{6, x1 - 25}
          nil_A = 6
          rand_A(x1) = max{22, x1 + 11}
    
      precedence:
      
        lastbit# = s > conv > half = |0| = lastbit = cons = nil = rand
      
      partial status:
    
        pi(lastbit#) = [1]
        pi(s) = [1]
        pi(half) = [1]
        pi(|0|) = []
        pi(lastbit) = [1]
        pi(conv) = [1]
        pi(cons) = []
        pi(nil) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          lastbit#_A(x1) = 25
          s_A(x1) = 10
          half_A(x1) = max{18, x1 + 13}
          |0|_A = 24
          lastbit_A(x1) = 9
          conv_A(x1) = max{18, x1 + 2}
          cons_A(x1,x2) = 11
          nil_A = 23
          rand_A(x1) = 1
    
      precedence:
      
        lastbit# = s = half = |0| = lastbit = conv = cons = nil = rand
      
      partial status:
    
        pi(lastbit#) = []
        pi(s) = []
        pi(half) = [1]
        pi(|0|) = []
        pi(lastbit) = []
        pi(conv) = [1]
        pi(cons) = []
        pi(nil) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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