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-minimaldependency 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-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      conv#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,1)) x1
      s_A(x1) = x1 + (0,1,3,2)
      half_A(x1) = x1 + (0,0,2,1)
      |0|_A() = (3,1,1,2)
      lastbit_A(x1) = x1 + (1,3,4,3)
      conv_A(x1) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (1,2,1,1)
      cons_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2
      nil_A() = (1,0,1,1)
      rand_A(x1) = x1 + (1,0,1,0)

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimaldependency 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:

  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),(1,1,1,1)) x1 + (0,0,4,0)
      half_A(x1) = x1 + (1,1,1,1)
      |0|_A() = (1,2,1,1)
      lastbit_A(x1) = (2,1,2,6)
      conv_A(x1) = x1 + (3,2,2,1)
      cons_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x1 + x2 + (0,4,0,2)
      nil_A() = (1,1,0,1)
      rand_A(x1) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (1,0,1,0)

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimaldependency 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      lastbit#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,1,1)) x1
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(0,0,0,0)) x1 + (0,1,1,1)
      half_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,1,1)) x1 + (1,1,1,1)
      |0|_A() = (0,2,8,9)
      lastbit_A(x1) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,1,4,2)
      conv_A(x1) = x1 + (2,0,1,1)
      cons_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (0,0,2,2)
      nil_A() = (1,1,1,1)
      rand_A(x1) = x1 + (1,0,1,0)

The next rules are strictly ordered:

  p1

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