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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        conv#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,1,1,0)) x1
        s_A(x1) = x1 + (0,1,2,0)
        half_A(x1) = x1 + (0,0,1,0)
        |0|_A() = (1,1,0,0)
        lastbit_A(x1) = x1 + (1,1,1,0)
        conv_A(x1) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (1,2,2,1)
        cons_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,1,0)) x1 + ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + (0,1,0,0)
        nil_A() = (1,1,1,1)
        rand_A(x1) = x1 + (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        rand > half > s > conv# > lastbit > conv > |0| > nil > cons
      
      argument filter:
    
        pi(conv#) = []
        pi(s) = []
        pi(half) = []
        pi(|0|) = []
        pi(lastbit) = []
        pi(conv) = []
        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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        half#_A(x1) = x1
        s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,0,1)) x1 + (0,6,4,1)
        half_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,0,1)) x1 + (1,1,0,0)
        |0|_A() = (1,1,3,4)
        lastbit_A(x1) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,1,0)) x1 + (3,8,0,0)
        conv_A(x1) = x1 + (4,1,1,2)
        cons_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + (0,2,2,0)
        nil_A() = (1,1,1,1)
        rand_A(x1) = x1 + (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        half > lastbit > rand > nil > s > conv > cons > half# > |0|
      
      argument filter:
    
        pi(half#) = []
        pi(s) = []
        pi(half) = []
        pi(|0|) = []
        pi(lastbit) = []
        pi(conv) = []
        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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        lastbit#_A(x1) = x1
        s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,0,1)) x1 + (0,1,1,1)
        half_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,1)) x1 + (1,1,1,1)
        |0|_A() = (1,1,0,1)
        lastbit_A(x1) = x1 + (1,1,1,1)
        conv_A(x1) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + (2,1,0,4)
        cons_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x2
        nil_A() = (1,1,1,1)
        rand_A(x1) = x1 + (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        rand > nil > lastbit# > s > conv > lastbit > half > |0| > cons
      
      argument filter:
    
        pi(lastbit#) = 1
        pi(s) = 1
        pi(half) = []
        pi(|0|) = []
        pi(lastbit) = [1]
        pi(conv) = []
        pi(cons) = []
        pi(nil) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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