YES

Problem:
 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)

Proof:
 DP Processor:
  DPs:
   half#(s(s(x))) -> half#(x)
   le#(s(x),s(y)) -> le#(x,y)
   inc#(s(x)) -> inc#(x)
   log#(x) -> log2#(x,0())
   log2#(x,y) -> inc#(y)
   log2#(x,y) -> le#(x,s(0()))
   log2#(x,y) -> if#(le(x,s(0())),x,inc(y))
   if#(false(),x,y) -> half#(x)
   if#(false(),x,y) -> log2#(half(x),y)
  TRS:
   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)
  Usable Rule Processor:
   DPs:
    half#(s(s(x))) -> half#(x)
    le#(s(x),s(y)) -> le#(x,y)
    inc#(s(x)) -> inc#(x)
    log#(x) -> log2#(x,0())
    log2#(x,y) -> inc#(y)
    log2#(x,y) -> le#(x,s(0()))
    log2#(x,y) -> if#(le(x,s(0())),x,inc(y))
    if#(false(),x,y) -> half#(x)
    if#(false(),x,y) -> log2#(half(x),y)
   TRS:
    inc(0()) -> 0()
    inc(s(x)) -> s(inc(x))
    le(0(),y) -> true()
    le(s(x),s(y)) -> le(x,y)
    le(s(x),0()) -> false()
    half(0()) -> 0()
    half(s(0())) -> 0()
    half(s(s(x))) -> s(half(x))
   Matrix Interpretation Processor: dim=1
    
    usable rules:
     inc(0()) -> 0()
     inc(s(x)) -> s(inc(x))
     le(0(),y) -> true()
     le(s(x),s(y)) -> le(x,y)
     le(s(x),0()) -> false()
     half(0()) -> 0()
     half(s(0())) -> 0()
     half(s(s(x))) -> s(half(x))
    interpretation:
     [if#](x0, x1, x2) = x0 + 2x1 + 1/2x2 + 1/2,
     
     [log2#](x0, x1) = 3x0 + 1/2x1 + 2,
     
     [log#](x0) = 3x0 + 5/2,
     
     [inc#](x0) = 1/2x0 + 1/2,
     
     [le#](x0, x1) = 3x0 + 1,
     
     [half#](x0) = 2x0 + 1/2,
     
     [inc](x0) = x0 + 2,
     
     [false] = 2,
     
     [true] = 0,
     
     [le](x0, x1) = x0,
     
     [s](x0) = x0 + 2,
     
     [half](x0) = 1/2x0,
     
     [0] = 0
    orientation:
     half#(s(s(x))) = 2x + 17/2 >= 2x + 1/2 = half#(x)
     
     le#(s(x),s(y)) = 3x + 7 >= 3x + 1 = le#(x,y)
     
     inc#(s(x)) = 1/2x + 3/2 >= 1/2x + 1/2 = inc#(x)
     
     log#(x) = 3x + 5/2 >= 3x + 2 = log2#(x,0())
     
     log2#(x,y) = 3x + 1/2y + 2 >= 1/2y + 1/2 = inc#(y)
     
     log2#(x,y) = 3x + 1/2y + 2 >= 3x + 1 = le#(x,s(0()))
     
     log2#(x,y) = 3x + 1/2y + 2 >= 3x + 1/2y + 3/2 = if#(le(x,s(0())),x,inc(y))
     
     if#(false(),x,y) = 2x + 1/2y + 5/2 >= 2x + 1/2 = half#(x)
     
     if#(false(),x,y) = 2x + 1/2y + 5/2 >= 3/2x + 1/2y + 2 = log2#(half(x),y)
     
     inc(0()) = 2 >= 0 = 0()
     
     inc(s(x)) = x + 4 >= x + 4 = s(inc(x))
     
     le(0(),y) = 0 >= 0 = true()
     
     le(s(x),s(y)) = x + 2 >= x = le(x,y)
     
     le(s(x),0()) = x + 2 >= 2 = false()
     
     half(0()) = 0 >= 0 = 0()
     
     half(s(0())) = 1 >= 0 = 0()
     
     half(s(s(x))) = 1/2x + 2 >= 1/2x + 2 = s(half(x))
    problem:
     DPs:
      
     TRS:
      inc(0()) -> 0()
      inc(s(x)) -> s(inc(x))
      le(0(),y) -> true()
      le(s(x),s(y)) -> le(x,y)
      le(s(x),0()) -> false()
      half(0()) -> 0()
      half(s(0())) -> 0()
      half(s(s(x))) -> s(half(x))
    Qed