YES
We show the termination of the TRS R:

  lt(|0|(),s(x)) -> true()
  lt(x,|0|()) -> false()
  lt(s(x),s(y)) -> lt(x,y)
  logarithm(x) -> ifa(lt(|0|(),x),x)
  ifa(true(),x) -> help(x,|1|())
  ifa(false(),x) -> logZeroError()
  help(x,y) -> ifb(lt(y,x),x,y)
  ifb(true(),x,y) -> help(half(x),s(y))
  ifb(false(),x,y) -> y
  half(|0|()) -> |0|()
  half(s(|0|())) -> |0|()
  half(s(s(x))) -> s(half(x))


-- SCC decomposition.

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

p1: lt#(s(x),s(y)) -> lt#(x,y)
p2: logarithm#(x) -> ifa#(lt(|0|(),x),x)
p3: logarithm#(x) -> lt#(|0|(),x)
p4: ifa#(true(),x) -> help#(x,|1|())
p5: help#(x,y) -> ifb#(lt(y,x),x,y)
p6: help#(x,y) -> lt#(y,x)
p7: ifb#(true(),x,y) -> help#(half(x),s(y))
p8: ifb#(true(),x,y) -> half#(x)
p9: half#(s(s(x))) -> half#(x)

and R consists of:

r1: lt(|0|(),s(x)) -> true()
r2: lt(x,|0|()) -> false()
r3: lt(s(x),s(y)) -> lt(x,y)
r4: logarithm(x) -> ifa(lt(|0|(),x),x)
r5: ifa(true(),x) -> help(x,|1|())
r6: ifa(false(),x) -> logZeroError()
r7: help(x,y) -> ifb(lt(y,x),x,y)
r8: ifb(true(),x,y) -> help(half(x),s(y))
r9: ifb(false(),x,y) -> y
r10: half(|0|()) -> |0|()
r11: half(s(|0|())) -> |0|()
r12: half(s(s(x))) -> s(half(x))

The estimated dependency graph contains the following SCCs:

  {p5, p7}
  {p1}
  {p9}


-- Reduction pair.

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

p1: ifb#(true(),x,y) -> help#(half(x),s(y))
p2: help#(x,y) -> ifb#(lt(y,x),x,y)

and R consists of:

r1: lt(|0|(),s(x)) -> true()
r2: lt(x,|0|()) -> false()
r3: lt(s(x),s(y)) -> lt(x,y)
r4: logarithm(x) -> ifa(lt(|0|(),x),x)
r5: ifa(true(),x) -> help(x,|1|())
r6: ifa(false(),x) -> logZeroError()
r7: help(x,y) -> ifb(lt(y,x),x,y)
r8: ifb(true(),x,y) -> help(half(x),s(y))
r9: ifb(false(),x,y) -> y
r10: half(|0|()) -> |0|()
r11: half(s(|0|())) -> |0|()
r12: half(s(s(x))) -> s(half(x))

The set of usable rules consists of

  r1, r2, r3, r10, r11, r12

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifb#_A(x1,x2,x3) = max{30, x1 + 15, x2 + 1}
        true_A = 17
        help#_A(x1,x2) = max{31, x1 + 10}
        half_A(x1) = max{19, x1 - 19}
        s_A(x1) = max{39, x1 + 27}
        lt_A(x1,x2) = max{9, x2 - 16}
        |0|_A = 18
        false_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          ifb#_A(x1,x2,x3) = max{8, x1 - 19, x3 - 19}
          true_A = 13
          help#_A(x1,x2) = max{8, x1 - 1, x2 - 1}
          half_A(x1) = 9
          s_A(x1) = max{3, x1 - 18}
          lt_A(x1,x2) = 17
          |0|_A = 0
          false_A = 17
          ifb#!_A(x1,x2,x3) = max{0, x1 - 18}
          true!_A = 1
          help#!_A(x1,x2) = max{2, x1 - 15}
          half!_A(x1) = max{2, x1 - 23}
          s!_A(x1) = max{4, x1 - 4}
          lt!_A(x1,x2) = 0
          |0|!_A = 3
          false!_A = 1
  
    partial status:
  
      pi(ifb#) = []
      pi(true) = []
      pi(help#) = []
      pi(half) = []
      pi(s) = [1]
      pi(lt) = []
      pi(|0|) = []
      pi(false) = []

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: lt#(s(x),s(y)) -> lt#(x,y)

and R consists of:

r1: lt(|0|(),s(x)) -> true()
r2: lt(x,|0|()) -> false()
r3: lt(s(x),s(y)) -> lt(x,y)
r4: logarithm(x) -> ifa(lt(|0|(),x),x)
r5: ifa(true(),x) -> help(x,|1|())
r6: ifa(false(),x) -> logZeroError()
r7: help(x,y) -> ifb(lt(y,x),x,y)
r8: ifb(true(),x,y) -> help(half(x),s(y))
r9: ifb(false(),x,y) -> y
r10: half(|0|()) -> |0|()
r11: half(s(|0|())) -> |0|()
r12: half(s(s(x))) -> s(half(x))

The set of usable rules consists of

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        lt#_A(x1,x2) = max{x1 + 12, x2 - 1}
        s_A(x1) = x1 + 11
        lt_A(x1,x2) = max{13, x1 - 13, x2 - 12}
        |0|_A = 4
        true_A = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          lt#_A(x1,x2) = 0
          s_A(x1) = max{22, x1 - 2}
          lt_A(x1,x2) = max{5, x1 - 19, x2 - 23}
          |0|_A = 18
          true_A = 0
          lt#!_A(x1,x2) = x2 + 8
          s!_A(x1) = max{4, x1 - 19}
          lt!_A(x1,x2) = max{5, x1 - 24, x2 - 23}
          |0|!_A = 5
          true!_A = 1
  
    partial status:
  
      pi(lt#) = []
      pi(s) = [1]
      pi(lt) = []
      pi(|0|) = []
      pi(true) = []

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: half#(s(s(x))) -> half#(x)

and R consists of:

r1: lt(|0|(),s(x)) -> true()
r2: lt(x,|0|()) -> false()
r3: lt(s(x),s(y)) -> lt(x,y)
r4: logarithm(x) -> ifa(lt(|0|(),x),x)
r5: ifa(true(),x) -> help(x,|1|())
r6: ifa(false(),x) -> logZeroError()
r7: help(x,y) -> ifb(lt(y,x),x,y)
r8: ifb(true(),x,y) -> help(half(x),s(y))
r9: ifb(false(),x,y) -> y
r10: half(|0|()) -> |0|()
r11: half(s(|0|())) -> |0|()
r12: half(s(s(x))) -> s(half(x))

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) = max{38, x1}
        s_A(x1) = x1 + 19
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          half#_A(x1) = 0
          s_A(x1) = max{9, x1 - 4}
          half#!_A(x1) = 1
          s!_A(x1) = max{4, x1 - 11}
  
    partial status:
  
      pi(half#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.