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))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [log#](x0) = 6x0 + 4,
    
    [half#](x0) = x0,
    
    [log](x0) = x0,
    
    [s](x0) = 2x0 + 1,
    
    [half](x0) = x0,
    
    [0] = 0
   orientation:
    half#(s(s(x))) = 4x + 3 >= x = half#(x)
    
    log#(s(s(x))) = 24x + 22 >= x = half#(x)
    
    log#(s(s(x))) = 24x + 22 >= 12x + 10 = log#(s(half(x)))
    
    half(0()) = 0 >= 0 = 0()
    
    half(s(s(x))) = 4x + 3 >= 2x + 1 = s(half(x))
    
    log(s(0())) = 1 >= 0 = 0()
    
    log(s(s(x))) = 4x + 3 >= 4x + 3 = s(log(s(half(x))))
   problem:
    DPs:
     
    TRS:
     half(0()) -> 0()
     half(s(s(x))) -> s(half(x))
     log(s(0())) -> 0()
     log(s(s(x))) -> s(log(s(half(x))))
   Qed