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:
  
      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:
          
            ifb#_A(x1,x2,x3) = max{30, x2 - 1}
            true_A = 32
            help#_A(x1,x2) = max{30, x1 + 15}
            half_A(x1) = max{15, x1 - 17}
            s_A(x1) = max{34, x1 + 17}
            lt_A(x1,x2) = x2 + 14
            |0|_A = 15
            false_A = 1
            ifb#!_A(x1,x2,x3) = max{14, x1 - 1, x2 - 1}
            true!_A = 11
            help#!_A(x1,x2) = x1 + 15
            half!_A(x1) = max{3, x1 - 2}
            s!_A(x1) = 13
            lt!_A(x1,x2) = max{x1 + 2, x2 + 16}
            |0|!_A = 12
            false!_A = 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ifb#_A(x1,x2,x3) = 0
              true_A() = 1
              help#_A(x1,x2) = 0
              half_A(x1) = x1 + 1
              s_A(x1) = x1 + 7
              lt_A(x1,x2) = 1
              |0|_A() = 13
              false_A() = 1
              ifb#!_A(x1,x2,x3) = 2
              true!_A() = 12
              help#!_A(x1,x2) = x1 + x2 + 3
              half!_A(x1) = x1 + 1
              s!_A(x1) = 14
              lt!_A(x1,x2) = 4
              |0|!_A() = 13
              false!_A() = 1
        
  
    partial status:
  
      pi(ifb#) = [3]
      pi(true) = []
      pi(help#) = [1, 2]
      pi(half) = []
      pi(s) = [1]
      pi(lt) = [1, 2]
      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

  (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:
          
            lt#_A(x1,x2) = 0
            s_A(x1) = max{4, x1 - 14}
            lt#!_A(x1,x2) = 11
            s!_A(x1) = x1 + 12
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              lt#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 5
              lt#!_A(x1,x2) = x2
              s!_A(x1) = x1 + 4
        
  
    partial status:
  
      pi(lt#) = [2]
      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: 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:
  
      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) = 0
            s_A(x1) = max{21, x1 + 15}
            half#!_A(x1) = max{4, x1 - 31}
            s!_A(x1) = 3
        
        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) = x1
              s_A(x1) = x1 + 2
              half#!_A(x1) = 6
              s!_A(x1) = x1 + 3
        
  
    partial status:
  
      pi(half#) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.