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, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        bitIter#_A(x1,x2) = max{12, x1 - 8, x2 - 14}
        if#_A(x1,x2,x3) = max{12, x2 - 9, x3 - 13}
        zero_A(x1) = max{4, x1 - 4}
        inc_A(x1) = max{4, x1 - 17}
        false_A = 2
        half_A(x1) = max{11, x1 - 14}
        |0|_A = 3
        s_A(x1) = max{1, x1}
        true_A = 2
        p_A(x1) = 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          bitIter#_A(x1,x2) = 1
          if#_A(x1,x2,x3) = 1
          zero_A(x1) = max{19, x1 - 3}
          inc_A(x1) = x1 + 3
          false_A = 21
          half_A(x1) = max{4, x1 - 5}
          |0|_A = 4
          s_A(x1) = x1 + 24
          true_A = 19
          p_A(x1) = 4
          bitIter#!_A(x1,x2) = max{17, x1 - 5}
          if#!_A(x1,x2,x3) = max{14, x1 - 3, x2 - 10}
          zero!_A(x1) = max{16, x1 - 25}
          inc!_A(x1) = max{13, x1 - 23}
          false!_A = 1
          half!_A(x1) = max{13, x1 + 8}
          |0|!_A = 11
          s!_A(x1) = max{12, x1 - 25}
          true!_A = 1
          p!_A(x1) = max{12, x1 - 5}
  
    partial status:
  
      pi(bitIter#) = []
      pi(if#) = []
      pi(zero) = []
      pi(inc) = []
      pi(false) = []
      pi(half) = []
      pi(|0|) = []
      pi(s) = []
      pi(true) = []
      pi(p) = []

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        half#_A(x1) = x1 + 5
        s_A(x1) = x1 + 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          half#_A(x1) = 0
          s_A(x1) = max{3, x1 - 9}
          half#!_A(x1) = 0
          s!_A(x1) = max{4, x1 - 9}
  
    partial status:
  
      pi(half#) = [1]
      pi(s) = []

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        inc#_A(x1) = 1
        s_A(x1) = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          inc#_A(x1) = x1
          s_A(x1) = max{8, x1 + 6}
          inc#!_A(x1) = max{4, x1 - 1}
          s!_A(x1) = max{3, x1 - 22}
  
    partial status:
  
      pi(inc#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.