YES
We show the termination of the TRS R:

  half(|0|()) -> |0|()
  half(s(|0|())) -> |0|()
  half(s(s(x))) -> s(half(x))
  le(|0|(),y) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  inc(|0|()) -> |0|()
  inc(s(x)) -> s(inc(x))
  log(x) -> log2(x,|0|())
  log2(x,y) -> if(le(x,s(|0|())),x,inc(y))
  if(true(),x,s(y)) -> y
  if(false(),x,y) -> log2(half(x),y)


-- SCC decomposition.

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

p1: half#(s(s(x))) -> half#(x)
p2: le#(s(x),s(y)) -> le#(x,y)
p3: inc#(s(x)) -> inc#(x)
p4: log#(x) -> log2#(x,|0|())
p5: log2#(x,y) -> if#(le(x,s(|0|())),x,inc(y))
p6: log2#(x,y) -> le#(x,s(|0|()))
p7: log2#(x,y) -> inc#(y)
p8: if#(false(),x,y) -> log2#(half(x),y)
p9: if#(false(),x,y) -> half#(x)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: inc(|0|()) -> |0|()
r8: inc(s(x)) -> s(inc(x))
r9: log(x) -> log2(x,|0|())
r10: log2(x,y) -> if(le(x,s(|0|())),x,inc(y))
r11: if(true(),x,s(y)) -> y
r12: if(false(),x,y) -> log2(half(x),y)

The estimated dependency graph contains the following SCCs:

  {p5, p8}
  {p1}
  {p2}
  {p3}


-- Reduction pair.

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

p1: log2#(x,y) -> if#(le(x,s(|0|())),x,inc(y))
p2: if#(false(),x,y) -> log2#(half(x),y)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: inc(|0|()) -> |0|()
r8: inc(s(x)) -> s(inc(x))
r9: log(x) -> log2(x,|0|())
r10: log2(x,y) -> if(le(x,s(|0|())),x,inc(y))
r11: if(true(),x,s(y)) -> y
r12: if(false(),x,y) -> log2(half(x),y)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

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:
          
            log2#_A(x1,x2) = max{1, x1 - 6, x2 - 10}
            if#_A(x1,x2,x3) = max{1, x1 - 1, x3 - 5}
            le_A(x1,x2) = max{0, x1 - 10, x2 - 5}
            s_A(x1) = max{1, x1}
            |0|_A = 4
            inc_A(x1) = max{4, x1 - 5}
            false_A = 0
            half_A(x1) = 7
            true_A = 0
            log2#!_A(x1,x2) = max{2, x1 - 8, x2 - 26}
            if#!_A(x1,x2,x3) = max{x1 + 2, x3 - 21}
            le!_A(x1,x2) = max{7, x1 - 25, x2 - 10}
            s!_A(x1) = max{3, x1 - 25}
            |0|!_A = 1
            inc!_A(x1) = max{3, x1 - 25}
            false!_A = 4
            half!_A(x1) = max{5, x1 - 2}
            true!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            log2#_A(x1,x2) = max{1, x1 - 1}
            if#_A(x1,x2,x3) = max{1, x1 - 24, x2 - 2}
            le_A(x1,x2) = x1 + 16
            s_A(x1) = x1 + 14
            |0|_A = 0
            inc_A(x1) = x1 + 11
            false_A = 22
            half_A(x1) = max{0, x1 - 4}
            true_A = 16
            log2#!_A(x1,x2) = x1 + 3
            if#!_A(x1,x2,x3) = max{3, x1 - 13, x2 + 1}
            le!_A(x1,x2) = max{16, x1 - 15, x2 - 12}
            s!_A(x1) = max{0, x1 - 13}
            |0|!_A = 14
            inc!_A(x1) = max{14, x1 - 8}
            false!_A = 11
            half!_A(x1) = 12
            true!_A = 15
        
  
    partial status:
  
      pi(log2#) = []
      pi(if#) = []
      pi(le) = []
      pi(s) = []
      pi(|0|) = []
      pi(inc) = []
      pi(false) = []
      pi(half) = []
      pi(true) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: log2#(x,y) -> if#(le(x,s(|0|())),x,inc(y))

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: inc(|0|()) -> |0|()
r8: inc(s(x)) -> s(inc(x))
r9: log(x) -> log2(x,|0|())
r10: log2(x,y) -> if(le(x,s(|0|())),x,inc(y))
r11: if(true(),x,s(y)) -> y
r12: if(false(),x,y) -> log2(half(x),y)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: inc(|0|()) -> |0|()
r8: inc(s(x)) -> s(inc(x))
r9: log(x) -> log2(x,|0|())
r10: log2(x,y) -> if(le(x,s(|0|())),x,inc(y))
r11: if(true(),x,s(y)) -> y
r12: if(false(),x,y) -> log2(half(x),y)

The set of usable rules consists of

  r1, r2

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{9, x1 - 14}
            half_A(x1) = max{13, x1 - 11}
            |0|_A = 9
            half#!_A(x1) = max{6, x1 - 14}
            s!_A(x1) = max{6, x1 - 14}
            half!_A(x1) = max{6, x1 - 28}
            |0|!_A = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            half#_A(x1) = 0
            s_A(x1) = max{9, x1 - 14}
            half_A(x1) = max{13, x1 - 11}
            |0|_A = 8
            half#!_A(x1) = max{6, x1 - 24}
            s!_A(x1) = max{6, x1 - 24}
            half!_A(x1) = max{6, x1 - 38}
            |0|!_A = 2
        
  
    partial status:
  
      pi(half#) = [1]
      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: le#(s(x),s(y)) -> le#(x,y)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: inc(|0|()) -> |0|()
r8: inc(s(x)) -> s(inc(x))
r9: log(x) -> log2(x,|0|())
r10: log2(x,y) -> if(le(x,s(|0|())),x,inc(y))
r11: if(true(),x,s(y)) -> y
r12: if(false(),x,y) -> log2(half(x),y)

The set of usable rules consists of

  r1, r2

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:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{8, x1 - 13}
            half_A(x1) = max{12, x1 - 10}
            |0|_A = 8
            le#!_A(x1,x2) = max{0, x2 - 22}
            s!_A(x1) = max{0, x1 - 22}
            half!_A(x1) = max{2, x1 - 27}
            |0|!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{8, x1 - 13}
            half_A(x1) = max{12, x1 - 13}
            |0|_A = 8
            le#!_A(x1,x2) = max{0, x2 - 46}
            s!_A(x1) = max{0, x1 - 46}
            half!_A(x1) = max{4, x1 - 32}
            |0|!_A = 3
        
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = [1]
      pi(half) = [1]
      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: inc#(s(x)) -> inc#(x)

and R consists of:

r1: half(|0|()) -> |0|()
r2: half(s(|0|())) -> |0|()
r3: half(s(s(x))) -> s(half(x))
r4: le(|0|(),y) -> true()
r5: le(s(x),|0|()) -> false()
r6: le(s(x),s(y)) -> le(x,y)
r7: inc(|0|()) -> |0|()
r8: inc(s(x)) -> s(inc(x))
r9: log(x) -> log2(x,|0|())
r10: log2(x,y) -> if(le(x,s(|0|())),x,inc(y))
r11: if(true(),x,s(y)) -> y
r12: if(false(),x,y) -> log2(half(x),y)

The set of usable rules consists of

  r1, r2

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:
          
            inc#_A(x1) = 0
            s_A(x1) = max{8, x1 - 13}
            half_A(x1) = max{12, x1 - 40}
            |0|_A = 8
            inc#!_A(x1) = 5
            s!_A(x1) = max{5, x1 - 13}
            half!_A(x1) = max{5, x1 - 26}
            |0|!_A = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            inc#_A(x1) = 0
            s_A(x1) = max{8, x1 - 13}
            half_A(x1) = max{12, x1 - 14}
            |0|_A = 8
            inc#!_A(x1) = max{6, x1 - 13}
            s!_A(x1) = max{6, x1 - 13}
            half!_A(x1) = 6
            |0|!_A = 2
        
  
    partial status:
  
      pi(inc#) = [1]
      pi(s) = [1]
      pi(half) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.