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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          ifb#_A(x1,x2,x3) = x2 + 4
          true_A() = 3
          help#_A(x1,x2) = x1 + 4
          half_A(x1) = x1
          s_A(x1) = x1 + 5
          lt_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 4
          false_A() = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          ifb#_A(x1,x2,x3) = max{8, x1 - 27, x2 + 1}
          true_A = 26
          help#_A(x1,x2) = max{8, x1 + 1}
          half_A(x1) = max{7, x1 - 37}
          s_A(x1) = x1 + 70
          lt_A(x1,x2) = max{11, x2 - 44}
          |0|_A = 7
          false_A = 1
          ifb#!_A(x1,x2,x3) = max{x1 - 12, x2 + 5}
          true!_A = 40
          help#!_A(x1,x2) = x1 + 6
          half!_A(x1) = max{41, x1 - 1}
          s!_A(x1) = 113
          lt!_A(x1,x2) = x1 + 32
          |0|!_A = 5
          false!_A = 4
  
    partial status:
  
      pi(ifb#) = []
      pi(true) = []
      pi(help#) = []
      pi(half) = [1]
      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

  r10

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          lt#_A(x1,x2) = x1 + x2
          s_A(x1) = x1 + 8
          half_A(x1) = 7
          |0|_A() = 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          lt#_A(x1,x2) = 0
          s_A(x1) = 13
          half_A(x1) = x1 + 12
          |0|_A = 26
          lt#!_A(x1,x2) = max{27, x2 - 14}
          s!_A(x1) = max{26, x1 + 15}
          half!_A(x1) = max{25, x1 + 12}
          |0|!_A = 24
  
    partial status:
  
      pi(lt#) = []
      pi(s) = [1]
      pi(half) = []
      pi(|0|) = []

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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          half#_A(x1) = x1
          s_A(x1) = x1 + 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          half#_A(x1) = 0
          s_A(x1) = 7
          half#!_A(x1) = max{1, x1 - 8}
          s!_A(x1) = 12
  
    partial status:
  
      pi(half#) = [1]
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.