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:

  max/plus interpretations on natural numbers:
  
    log2#_A(x1,x2) = x1 + 4
    if#_A(x1,x2,x3) = max{4, x1 - 9, x2 + 3}
    le_A(x1,x2) = max{x1 + 13, x2}
    s_A(x1) = max{12, x1 + 7}
    |0|_A = 3
    inc_A(x1) = x1 + 5
    false_A = 17
    half_A(x1) = max{3, x1 - 2}
    true_A = 0

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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

  (no rules)

Take the monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    half#_A(x1) = x1
    s_A(x1) = x1 + 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: 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:

  max/plus interpretations on natural numbers:
  
    le#_A(x1,x2) = max{x1 + 1, x2 + 1}
    s_A(x1) = x1 + 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: 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 monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    inc#_A(x1) = x1
    s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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