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

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:
          log2#_A(x1,x2) = 5
          if#_A(x1,x2,x3) = 5
          le_A(x1,x2) = 7
          s_A(x1) = 3
          |0|_A() = 4
          inc_A(x1) = x1 + 6
          false_A() = 6
          half_A(x1) = x1 + 2
          true_A() = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          log2#_A(x1,x2) = max{1, x1 - 11}
          if#_A(x1,x2,x3) = max{1, x2 - 11}
          le_A(x1,x2) = max{28, x1 + 27, x2 + 1}
          s_A(x1) = x1 + 16
          |0|_A = 11
          inc_A(x1) = 0
          false_A = 41
          half_A(x1) = max{12, x1 - 2}
          true_A = 38
          log2#!_A(x1,x2) = x1 + 14
          if#!_A(x1,x2,x3) = max{x1 - 14, x2 + 14}
          le!_A(x1,x2) = max{11, x1 - 17, x2 - 33}
          s!_A(x1) = max{3, x1 - 15}
          |0|!_A = 4
          inc!_A(x1) = max{15, x1 - 14}
          false!_A = 12
          half!_A(x1) = 0
          true!_A = 5
  
    partial status:
  
      pi(log2#) = []
      pi(if#) = []
      pi(le) = []
      pi(s) = []
      pi(|0|) = []
      pi(inc) = [1]
      pi(false) = []
      pi(half) = [1]
      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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          half#_A(x1) = x1
          s_A(x1) = x1 + 4
          half_A(x1) = x1 + 2
          |0|_A() = 2
  
    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 - 12}
          |0|_A = 8
          half#!_A(x1) = max{5, x1 - 37}
          s!_A(x1) = max{5, x1 - 24}
          half!_A(x1) = max{4, x1 - 38}
          |0|!_A = 2
  
    partial status:
  
      pi(half#) = []
      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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          le#_A(x1,x2) = x1 + x2 + 5
          s_A(x1) = x1 + 4
          half_A(x1) = x1 + 2
          |0|_A() = 2
  
    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{8, x1 - 13}
          half_A(x1) = max{12, x1 - 40}
          |0|_A = 8
          le#!_A(x1,x2) = max{4, x2 - 9}
          s!_A(x1) = max{3, x1 - 54}
          half!_A(x1) = max{4, x1 - 68}
          |0|!_A = 2
  
    partial status:
  
      pi(le#) = [1]
      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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          inc#_A(x1) = x1
          s_A(x1) = x1 + 4
          half_A(x1) = x1 + 2
          |0|_A() = 2
  
    base order 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 - 12}
          |0|_A = 8
          inc#!_A(x1) = 0
          s!_A(x1) = max{5, x1 - 24}
          half!_A(x1) = max{2, x1 - 9}
          |0|!_A = 2
  
    partial status:
  
      pi(inc#) = []
      pi(s) = []
      pi(half) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.