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)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+#](x0, x1) = 4x0 + x1 + 1,
    
    [double#](x0) = 5x0 + 2,
    
    [+](x0, x1) = x0 + x1,
    
    [s](x0) = x0 + 1,
    
    [double](x0) = 2x0 + 1,
    
    [0] = 0
   orientation:
    double#(s(x)) = 5x + 7 >= 5x + 2 = double#(x)
    
    +#(x,s(y)) = 4x + y + 2 >= 4x + y + 1 = +#(x,y)
    
    +#(s(x),y) = 4x + y + 5 >= 4x + y + 1 = +#(x,y)
    
    double#(x) = 5x + 2 >= 5x + 1 = +#(x,x)
    
    double(0()) = 1 >= 0 = 0()
    
    double(s(x)) = 2x + 3 >= 2x + 3 = s(s(double(x)))
    
    +(x,0()) = x >= x = x
    
    +(x,s(y)) = x + y + 1 >= x + y + 1 = s(+(x,y))
    
    +(s(x),y) = x + y + 1 >= x + y + 1 = s(+(x,y))
    
    double(x) = 2x + 1 >= 2x = +(x,x)
   problem:
    DPs:
     
    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)
   Qed