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, r6, r10, r11, r12

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ifb#_A(x1,x2,x3) = max{2, x1 - 2, x2}
        true_A = 13
        help#_A(x1,x2) = x1 + 4
        half_A(x1) = max{7, x1 - 5}
        s_A(x1) = x1 + 14
        lt_A(x1,x2) = max{4, x2 + 1}
        |0|_A = 6
        false_A = 5
        ifa_A(x1,x2) = max{1, x1 - 22, x2 - 2}
        logZeroError_A = 0
  
    base order 2:
  
      precedence:
      
        ifb# > half > true > help# > lt > s > |0| = ifa > false > logZeroError
      
  
    partial status:
  
      pi(ifb#) = [2]
      pi(true) = []
      pi(help#) = [1]
      pi(half) = []
      pi(s) = [1]
      pi(lt) = []
      pi(|0|) = []
      pi(false) = []
      pi(ifa) = []
      pi(logZeroError) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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

  r2, r6, r10, r11

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        lt#_A(x1,x2) = max{x1 + 7, x2 + 9}
        s_A(x1) = x1 + 8
        lt_A(x1,x2) = max{4, x1 - 10, x2 - 9}
        |0|_A = 6
        false_A = 1
        ifa_A(x1,x2) = max{8, x1 - 11, x2 - 1}
        logZeroError_A = 0
        half_A(x1) = max{7, x1 - 17}
  
    base order 2:
  
      precedence:
      
        lt# = ifa > s = |0| > lt > false = half > logZeroError
      
  
    partial status:
  
      pi(lt#) = [2]
      pi(s) = []
      pi(lt) = []
      pi(|0|) = []
      pi(false) = []
      pi(ifa) = []
      pi(logZeroError) = []
      pi(half) = []

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

  r2, r6, r10, r11

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        half#_A(x1) = max{1, x1 - 1}
        s_A(x1) = x1 + 13
        lt_A(x1,x2) = max{14, x1 - 17, x2 - 14}
        |0|_A = 5
        false_A = 1
        ifa_A(x1,x2) = max{16, x1 - 15, x2 - 6}
        logZeroError_A = 0
        half_A(x1) = max{6, x1 - 19}
  
    base order 2:
  
      precedence:
      
        s = ifa > half# > half > lt = |0| > false > logZeroError
      
  
    partial status:
  
      pi(half#) = []
      pi(s) = [1]
      pi(lt) = []
      pi(|0|) = []
      pi(false) = []
      pi(ifa) = []
      pi(logZeroError) = []
      pi(half) = []

The following rules are removed:

  p1

Then no dependency pair remains.