YES

Problem:
 half(0()) -> 0()
 half(s(s(x))) -> s(half(x))
 log(s(0())) -> 0()
 log(s(s(x))) -> s(log(s(half(x))))

Proof:
 DP Processor:
  DPs:
   half#(s(s(x))) -> half#(x)
   log#(s(s(x))) -> half#(x)
   log#(s(s(x))) -> log#(s(half(x)))
  TRS:
   half(0()) -> 0()
   half(s(s(x))) -> s(half(x))
   log(s(0())) -> 0()
   log(s(s(x))) -> s(log(s(half(x))))
  Usable Rule Processor:
   DPs:
    half#(s(s(x))) -> half#(x)
    log#(s(s(x))) -> half#(x)
    log#(s(s(x))) -> log#(s(half(x)))
   TRS:
    half(0()) -> 0()
    half(s(s(x))) -> s(half(x))
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     half(0()) -> 0()
     half(s(s(x))) -> s(half(x))
    interpretation:
     [log#](x0) = x0 + 0,
     
     [half#](x0) = x0 + 0,
     
     [s](x0) = 1x0 + 4,
     
     [half](x0) = x0 + 2,
     
     [0] = 0
    orientation:
     half#(s(s(x))) = 2x + 5 >= x + 0 = half#(x)
     
     log#(s(s(x))) = 2x + 5 >= x + 0 = half#(x)
     
     log#(s(s(x))) = 2x + 5 >= 1x + 4 = log#(s(half(x)))
     
     half(0()) = 2 >= 0 = 0()
     
     half(s(s(x))) = 2x + 5 >= 1x + 4 = s(half(x))
    problem:
     DPs:
      
     TRS:
      half(0()) -> 0()
      half(s(s(x))) -> s(half(x))
    Qed