YES

We show the termination of the TRS R:

  half(|0|()) -> |0|()
  half(s(|0|())) -> |0|()
  half(s(s(x))) -> s(half(x))
  inc(|0|()) -> |0|()
  inc(s(x)) -> s(inc(x))
  zero(|0|()) -> true()
  zero(s(x)) -> false()
  p(|0|()) -> |0|()
  p(s(x)) -> x
  bits(x) -> bitIter(x,|0|())
  bitIter(x,y) -> if(zero(x),x,inc(y))
  if(true(),x,y) -> p(y)
  if(false(),x,y) -> bitIter(half(x),y)

-- SCC decomposition.

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

p1: half#(s(s(x))) -> half#(x)
p2: inc#(s(x)) -> inc#(x)
p3: bits#(x) -> bitIter#(x,|0|())
p4: bitIter#(x,y) -> if#(zero(x),x,inc(y))
p5: bitIter#(x,y) -> zero#(x)
p6: bitIter#(x,y) -> inc#(y)
p7: if#(true(),x,y) -> p#(y)
p8: if#(false(),x,y) -> bitIter#(half(x),y)
p9: if#(false(),x,y) -> half#(x)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: inc(|0|()) -> |0|()
r5: inc(s(x)) -> s(inc(x))
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(x)) -> x
r10: bits(x) -> bitIter(x,|0|())
r11: bitIter(x,y) -> if(zero(x),x,inc(y))
r12: if(true(),x,y) -> p(y)
r13: if(false(),x,y) -> bitIter(half(x),y)

The estimated dependency graph contains the following SCCs:

  {p4, p8}
  {p1}
  {p2}


-- Reduction pair.

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

p1: bitIter#(x,y) -> if#(zero(x),x,inc(y))
p2: if#(false(),x,y) -> bitIter#(half(x),y)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: inc(|0|()) -> |0|()
r5: inc(s(x)) -> s(inc(x))
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(x)) -> x
r10: bits(x) -> bitIter(x,|0|())
r11: bitIter(x,y) -> if(zero(x),x,inc(y))
r12: if(true(),x,y) -> p(y)
r13: if(false(),x,y) -> bitIter(half(x),y)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    bitIter#_A(x1,x2) = x1 + 3
    if#_A(x1,x2,x3) = max{x1 - 4, x2 + 2}
    zero_A(x1) = max{6, x1}
    inc_A(x1) = x1 + 7
    false_A = 13
    half_A(x1) = max{5, x1 - 2}
    |0|_A = 5
    s_A(x1) = max{20, x1 + 13}
    true_A = 6

The next rules are strictly ordered:

  p1, p2

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: 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: inc(|0|()) -> |0|()
r5: inc(s(x)) -> s(inc(x))
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(x)) -> x
r10: bits(x) -> bitIter(x,|0|())
r11: bitIter(x,y) -> if(zero(x),x,inc(y))
r12: if(true(),x,y) -> p(y)
r13: if(false(),x,y) -> bitIter(half(x),y)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    half#_A(x1) = x1
    s_A(x1) = x1 + 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: inc#(s(x)) -> inc#(x)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: inc(|0|()) -> |0|()
r5: inc(s(x)) -> s(inc(x))
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(x)) -> x
r10: bits(x) -> bitIter(x,|0|())
r11: bitIter(x,y) -> if(zero(x),x,inc(y))
r12: if(true(),x,y) -> p(y)
r13: if(false(),x,y) -> bitIter(half(x),y)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    inc#_A(x1) = x1
    s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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