YES

Problem:
 double(0()) -> 0()
 double(s(x)) -> s(s(double(x)))
 half(0()) -> 0()
 half(s(0())) -> 0()
 half(s(s(x))) -> s(half(x))
 -(x,0()) -> x
 -(s(x),s(y)) -> -(x,y)
 if(0(),y,z) -> y
 if(s(x),y,z) -> z
 half(double(x)) -> x

Proof:
 DP Processor:
  DPs:
   double#(s(x)) -> double#(x)
   half#(s(s(x))) -> half#(x)
   -#(s(x),s(y)) -> -#(x,y)
  TRS:
   double(0()) -> 0()
   double(s(x)) -> s(s(double(x)))
   half(0()) -> 0()
   half(s(0())) -> 0()
   half(s(s(x))) -> s(half(x))
   -(x,0()) -> x
   -(s(x),s(y)) -> -(x,y)
   if(0(),y,z) -> y
   if(s(x),y,z) -> z
   half(double(x)) -> x
  Usable Rule Processor:
   DPs:
    double#(s(x)) -> double#(x)
    half#(s(s(x))) -> half#(x)
    -#(s(x),s(y)) -> -#(x,y)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [-#](x0, x1) = x0 + 1x1,
     
     [half#](x0) = x0,
     
     [double#](x0) = x0,
     
     [s](x0) = 1x0 + -2
    orientation:
     double#(s(x)) = 1x + -2 >= x = double#(x)
     
     half#(s(s(x))) = 2x + -1 >= x = half#(x)
     
     -#(s(x),s(y)) = 1x + 2y + -1 >= x + 1y = -#(x,y)
    problem:
     DPs:
      
     TRS:
      
    Qed