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, r11

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        log2#_A(x1,x2) = x1 + 7
        if#_A(x1,x2,x3) = max{x1 - 37, x2 + 4}
        le_A(x1,x2) = max{x1 + 38, x2 + 24}
        s_A(x1) = max{19, x1 + 10}
        |0|_A = 2
        inc_A(x1) = 6
        false_A = 49
        half_A(x1) = max{3, x1 - 8}
        true_A = 1
        if_A(x1,x2,x3) = max{33, x1 - 12, x2 + 24, x3 - 7}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          log2#_A(x1,x2) = max{7, x1 - 15}
          if#_A(x1,x2,x3) = max{7, x1 - 3, x2 - 15}
          le_A(x1,x2) = max{0, x1 - 12, x2 - 1}
          s_A(x1) = max{3, x1}
          |0|_A = 11
          inc_A(x1) = max{2, x1 - 5}
          false_A = 0
          half_A(x1) = max{14, x1 - 4}
          true_A = 0
          if_A(x1,x2,x3) = max{2, x1 - 11, x2 - 20, x3}
          log2#!_A(x1,x2) = 21
          if#!_A(x1,x2,x3) = max{x1 + 20, x2 + 22}
          le!_A(x1,x2) = max{22, x1 - 12, x2 - 12}
          s!_A(x1) = 0
          |0|!_A = 20
          inc!_A(x1) = max{15, x1 - 12}
          false!_A = 19
          half!_A(x1) = 0
          true!_A = 0
          if!_A(x1,x2,x3) = max{24, x1 - 11, x2 + 3, x3 + 20}
  
    partial status:
  
      pi(log2#) = []
      pi(if#) = [2]
      pi(le) = [1, 2]
      pi(s) = [1]
      pi(|0|) = []
      pi(inc) = []
      pi(false) = []
      pi(half) = []
      pi(true) = []
      pi(if) = []

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: 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, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        half#_A(x1) = max{11, x1 + 4}
        s_A(x1) = max{12, x1 + 5}
        half_A(x1) = max{25, x1 - 30}
        |0|_A = 24
        inc_A(x1) = max{3, x1}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          half#_A(x1) = 0
          s_A(x1) = max{8, x1 - 13}
          half_A(x1) = max{12, x1 - 10}
          |0|_A = 5
          inc_A(x1) = max{12, x1 - 10}
          half#!_A(x1) = 0
          s!_A(x1) = max{4, x1 - 35}
          half!_A(x1) = max{5, x1 - 20}
          |0|!_A = 2
          inc!_A(x1) = max{5, x1 - 9}
  
    partial status:
  
      pi(half#) = [1]
      pi(s) = [1]
      pi(half) = []
      pi(|0|) = []
      pi(inc) = [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: 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, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = max{4, x1 - 1, x2 - 1}
        s_A(x1) = x1 + 13
        half_A(x1) = max{12, x1 - 27}
        |0|_A = 7
        le_A(x1,x2) = max{5, x1 - 27, x2 - 45}
        false_A = 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{4, x1 - 10}
          half_A(x1) = max{8, x1 + 2}
          |0|_A = 9
          le_A(x1,x2) = max{0, x1 - 5, x2 - 4}
          false_A = 1
          le#!_A(x1,x2) = max{15, x2 + 11}
          s!_A(x1) = max{4, x1 - 22}
          half!_A(x1) = max{3, x1 - 33}
          |0|!_A = 2
          le!_A(x1,x2) = max{5, x1 - 13, x2 - 10}
          false!_A = 1
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(half) = []
      pi(|0|) = []
      pi(le) = []
      pi(false) = []

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:
  
      max/plus interpretation on natural numbers:
      
        inc#_A(x1) = x1 + 12
        s_A(x1) = x1 + 13
        half_A(x1) = max{10, x1 - 27}
        |0|_A = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          inc#_A(x1) = 0
          s_A(x1) = max{13, x1 - 18}
          half_A(x1) = max{17, x1 - 18}
          |0|_A = 1
          inc#!_A(x1) = 0
          s!_A(x1) = 5
          half!_A(x1) = max{4, x1 - 17}
          |0|!_A = 3
  
    partial status:
  
      pi(inc#) = [1]
      pi(s) = [1]
      pi(half) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.