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) = 1
            if#_A(x1,x2,x3) = 1
            le_A(x1,x2) = max{1, x1 - 11}
            s_A(x1) = x1 + 31
            |0|_A = 10
            inc_A(x1) = x1
            false_A = 8
            half_A(x1) = max{12, x1 - 19}
            true_A = 1
            log2#!_A(x1,x2) = max{7, x1}
            if#!_A(x1,x2,x3) = max{x1 + 5, x2 - 12}
            le!_A(x1,x2) = max{9, x1 + 1, x2 - 1}
            s!_A(x1) = x1 + 20
            |0|!_A = 16
            inc!_A(x1) = max{19, x1 - 11}
            false!_A = 8
            half!_A(x1) = 14
            true!_A = 12
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              log2#_A(x1,x2) = 0
              if#_A(x1,x2,x3) = 0
              le_A(x1,x2) = 1
              s_A(x1) = 3
              |0|_A() = 3
              inc_A(x1) = 3
              false_A() = 1
              half_A(x1) = x1
              true_A() = 1
              log2#!_A(x1,x2) = 4
              if#!_A(x1,x2,x3) = 3
              le!_A(x1,x2) = x1 + x2
              s!_A(x1) = 7
              |0|!_A() = 6
              inc!_A(x1) = x1 + 8
              false!_A() = 5
              half!_A(x1) = x1 + 11
              true!_A() = 4
        
  
    partial status:
  
      pi(log2#) = []
      pi(if#) = []
      pi(le) = [1, 2]
      pi(s) = [1]
      pi(|0|) = []
      pi(inc) = [1]
      pi(false) = []
      pi(half) = [1]
      pi(true) = []

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

  (no rules)

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{29, x1 - 16}
            half#!_A(x1) = max{16, x1 - 30}
            s!_A(x1) = x1 + 17
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              half#_A(x1) = x1
              s_A(x1) = x1 + 1
              half#!_A(x1) = 1
              s!_A(x1) = 2
        
  
    partial status:
  
      pi(half#) = [1]
      pi(s) = [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

  (no rules)

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) = max{x1, x2}
            s_A(x1) = max{12, x1 + 10}
            le#!_A(x1,x2) = max{8, x1 - 13, x2 - 1}
            s!_A(x1) = 7
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              le#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 2
              le#!_A(x1,x2) = x1 + x2 + 7
              s!_A(x1) = 6
        
  
    partial status:
  
      pi(le#) = []
      pi(s) = []

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

  (no rules)

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{12, x1 - 12}
            inc#!_A(x1) = 0
            s!_A(x1) = max{35, x1 + 23}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              inc#_A(x1) = x1
              s_A(x1) = x1 + 3
              inc#!_A(x1) = x1 + 7
              s!_A(x1) = 6
        
  
    partial status:
  
      pi(inc#) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.