YES

Problem:
 double(0()) -> 0()
 double(s(x)) -> s(s(double(x)))
 +(x,0()) -> x
 +(x,s(y)) -> s(+(x,y))
 +(s(x),y) -> s(+(x,y))
 double(x) -> +(x,x)

Proof:
 DP Processor:
  DPs:
   double#(s(x)) -> double#(x)
   +#(x,s(y)) -> +#(x,y)
   +#(s(x),y) -> +#(x,y)
   double#(x) -> +#(x,x)
  TRS:
   double(0()) -> 0()
   double(s(x)) -> s(s(double(x)))
   +(x,0()) -> x
   +(x,s(y)) -> s(+(x,y))
   +(s(x),y) -> s(+(x,y))
   double(x) -> +(x,x)
  Usable Rule Processor:
   DPs:
    double#(s(x)) -> double#(x)
    +#(x,s(y)) -> +#(x,y)
    +#(s(x),y) -> +#(x,y)
    double#(x) -> +#(x,x)
   TRS:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [+#](x0, x1) = [0 0 0 1]x0 + [0 0 1 0]x1,
     
     [double#](x0) = [1 1 1 1]x0 + [1],
     
               [0 0 1 1]     [0]
               [0 0 1 0]     [0]
     [s](x0) = [1 0 1 0]x0 + [1]
               [1 1 0 1]     [1]
    orientation:
     double#(s(x)) = [2 1 3 2]x + [3] >= [1 1 1 1]x + [1] = double#(x)
     
     +#(x,s(y)) = [0 0 0 1]x + [1 0 1 0]y + [1] >= [0 0 0 1]x + [0 0 1 0]y = +#(x,y)
     
     +#(s(x),y) = [1 1 0 1]x + [0 0 1 0]y + [1] >= [0 0 0 1]x + [0 0 1 0]y = +#(x,y)
     
     double#(x) = [1 1 1 1]x + [1] >= [0 0 1 1]x = +#(x,x)
    problem:
     DPs:
      
     TRS:
      
    Qed