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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ifb#_A(x1,x2,x3) = max{x1 + 5, x2 + 8}
          true_A = 7
          help#_A(x1,x2) = x1 + 8
          half_A(x1) = max{0, x1 - 8}
          s_A(x1) = x1 + 8
          lt_A(x1,x2) = x2 + 2
          |0|_A = 0
          false_A = 0
    
      precedence:
      
        half = |0| > help# > true = s = lt > ifb# = false
      
      partial status:
    
        pi(ifb#) = []
        pi(true) = []
        pi(help#) = [1]
        pi(half) = []
        pi(s) = []
        pi(lt) = [2]
        pi(|0|) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ifb#_A(x1,x2,x3) = 1
          true_A = 3
          help#_A(x1,x2) = x1 + 1
          half_A(x1) = 5
          s_A(x1) = 2
          lt_A(x1,x2) = 4
          |0|_A = 0
          false_A = 5
    
      precedence:
      
        |0| = false > ifb# = true = help# > lt > half > s
      
      partial status:
    
        pi(ifb#) = []
        pi(true) = []
        pi(help#) = []
        pi(half) = []
        pi(s) = []
        pi(lt) = []
        pi(|0|) = []
        pi(false) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- 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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          lt#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        lt# = s
      
      partial status:
    
        pi(lt#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          lt#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        lt# = s
      
      partial status:
    
        pi(lt#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          half#_A(x1) = max{2, x1 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        half# = s
      
      partial status:
    
        pi(half#) = [1]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          half#_A(x1) = x1 + 1
          s_A(x1) = x1
    
      precedence:
      
        half# = s
      
      partial status:
    
        pi(half#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.