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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            bitIter#_A(x1,x2) = max{26, x1 - 1}
            if#_A(x1,x2,x3) = max{18, x1 - 3, x2 - 1}
            zero_A(x1) = max{4, x1 + 2}
            inc_A(x1) = 17
            false_A = 29
            half_A(x1) = max{20, x1 - 8}
            |0|_A = 18
            s_A(x1) = x1 + 33
            true_A = 0
            bitIter#!_A(x1,x2) = x1 + 3
            if#!_A(x1,x2,x3) = max{x1 - 5, x2 + 2}
            zero!_A(x1) = 31
            inc!_A(x1) = 10
            false!_A = 30
            half!_A(x1) = max{31, x1 - 27}
            |0|!_A = 11
            s!_A(x1) = 39
            true!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              bitIter#_A(x1,x2) = 0
              if#_A(x1,x2,x3) = 0
              zero_A(x1) = x1 + 1
              inc_A(x1) = 1
              false_A() = 3
              half_A(x1) = x1
              |0|_A() = 2
              s_A(x1) = 2
              true_A() = 3
              bitIter#!_A(x1,x2) = 2
              if#!_A(x1,x2,x3) = x1 + x2
              zero!_A(x1) = x1 + 3
              inc!_A(x1) = 11
              false!_A() = 9
              half!_A(x1) = x1 + 7
              |0|!_A() = 7
              s!_A(x1) = 8
              true!_A() = 6
        
  
    partial status:
  
      pi(bitIter#) = []
      pi(if#) = []
      pi(zero) = [1]
      pi(inc) = [1]
      pi(false) = []
      pi(half) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(true) = []

The following rules are removed:

  p1, p2

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 reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            half#_A(x1) = x1
            s_A(x1) = x1 + 34
            half#!_A(x1) = max{9, x1 - 1}
            s!_A(x1) = 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              half#_A(x1) = 0
              s_A(x1) = 2
              half#!_A(x1) = 3
              s!_A(x1) = 4
        
  
    partial status:
  
      pi(half#) = []
      pi(s) = [1]

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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            inc#_A(x1) = 0
            s_A(x1) = max{43, x1 + 9}
            inc#!_A(x1) = max{7, x1 - 1}
            s!_A(x1) = max{6, x1 - 37}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              inc#_A(x1) = x1
              s_A(x1) = x1 + 1
              inc#!_A(x1) = 6
              s!_A(x1) = x1 + 5
        
  
    partial status:
  
      pi(inc#) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.