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, r6, r7, r8, r9, r12

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        bitIter#_A(x1,x2) = x1 + 15
        if#_A(x1,x2,x3) = max{14, x1 - 13, x2 + 11}
        zero_A(x1) = max{12, x1 - 16}
        inc_A(x1) = 14
        false_A = 38
        half_A(x1) = max{10, x1 - 7}
        |0|_A = 9
        s_A(x1) = x1 + 55
        true_A = 11
        p_A(x1) = max{10, x1 - 16}
        if_A(x1,x2,x3) = max{40, x1 - 39, x2 + 9, x3 - 15}
  
    base order 2:
  
      precedence:
      
        bitIter# = if# = s = if > half > zero = false > inc = p > |0| > true
      
  
    partial status:
  
      pi(bitIter#) = []
      pi(if#) = []
      pi(zero) = []
      pi(inc) = []
      pi(false) = []
      pi(half) = []
      pi(|0|) = []
      pi(s) = []
      pi(true) = []
      pi(p) = []
      pi(if) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 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: 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, r6, r7, r8, r9, r12

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        half#_A(x1) = max{6, x1 - 1}
        s_A(x1) = x1 + 11
        half_A(x1) = max{10, x1 - 23}
        |0|_A = 6
        zero_A(x1) = max{10, x1 - 18}
        true_A = 1
        false_A = 9
        p_A(x1) = max{12, x1 - 8}
        if_A(x1,x2,x3) = max{12, x2 - 16, x3 - 8}
  
    base order 2:
  
      precedence:
      
        s = if > half = p > |0| > half# > zero > true = false
      
  
    partial status:
  
      pi(half#) = []
      pi(s) = [1]
      pi(half) = []
      pi(|0|) = []
      pi(zero) = []
      pi(true) = []
      pi(false) = []
      pi(p) = []
      pi(if) = []

The following rules are removed:

  p1

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

  r1, r2, r7, r8, r9, r12

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        inc#_A(x1) = max{4, x1 - 1}
        s_A(x1) = max{10, x1 + 6}
        half_A(x1) = max{5, x1 - 17}
        |0|_A = 2
        zero_A(x1) = max{2, x1 - 11}
        false_A = 1
        p_A(x1) = max{6, x1 - 4}
        if_A(x1,x2,x3) = max{6, x1 - 3, x2 - 2, x3 - 4}
        true_A = 1
  
    base order 2:
  
      precedence:
      
        if > half > inc# = s > zero = p > |0| = false > true
      
  
    partial status:
  
      pi(inc#) = []
      pi(s) = [1]
      pi(half) = []
      pi(|0|) = []
      pi(zero) = []
      pi(false) = []
      pi(p) = []
      pi(if) = []
      pi(true) = []

The following rules are removed:

  p1

Then no dependency pair remains.