YES

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

Proof:
 DP Processor:
  DPs:
   half#(s(s(x))) -> half#(x)
   half#(s(s(x))) -> s#(half(x))
   s#(log(0())) -> s#(0())
   log#(s(x)) -> half#(s(x))
   log#(s(x)) -> log#(half(s(x)))
   log#(s(x)) -> s#(log(half(s(x))))
  TRS:
   half(0()) -> 0()
   half(s(0())) -> 0()
   half(s(s(x))) -> s(half(x))
   s(log(0())) -> s(0())
   log(s(x)) -> s(log(half(s(x))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [log#](x0) = 2x0,
    
    [s#](x0) = x0,
    
    [half#](x0) = 3/2x0,
    
    [log](x0) = 2x0 + 1/2,
    
    [s](x0) = x0 + 1,
    
    [half](x0) = 1/2x0,
    
    [0] = 0
   orientation:
    half#(s(s(x))) = 3/2x + 3 >= 3/2x = half#(x)
    
    half#(s(s(x))) = 3/2x + 3 >= 1/2x = s#(half(x))
    
    s#(log(0())) = 1/2 >= 0 = s#(0())
    
    log#(s(x)) = 2x + 2 >= 3/2x + 3/2 = half#(s(x))
    
    log#(s(x)) = 2x + 2 >= x + 1 = log#(half(s(x)))
    
    log#(s(x)) = 2x + 2 >= x + 3/2 = s#(log(half(s(x))))
    
    half(0()) = 0 >= 0 = 0()
    
    half(s(0())) = 1/2 >= 0 = 0()
    
    half(s(s(x))) = 1/2x + 1 >= 1/2x + 1 = s(half(x))
    
    s(log(0())) = 3/2 >= 1 = s(0())
    
    log(s(x)) = 2x + 5/2 >= x + 5/2 = s(log(half(s(x))))
   problem:
    DPs:
     
    TRS:
     half(0()) -> 0()
     half(s(0())) -> 0()
     half(s(s(x))) -> s(half(x))
     s(log(0())) -> s(0())
     log(s(x)) -> s(log(half(s(x))))
   Qed