YES

Problem:
 f(s(X),Y) -> h(s(f(h(Y),X)))

Proof:
 DP Processor:
  DPs:
   f#(s(X),Y) -> f#(h(Y),X)
  TRS:
   f(s(X),Y) -> h(s(f(h(Y),X)))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = 2x0 + 0,
    
    [h](x0) = 0,
    
    [f](x0, x1) = 2x1 + 6,
    
    [s](x0) = x0 + 2
   orientation:
    f#(s(X),Y) = 2X + 4 >= 2 = f#(h(Y),X)
    
    f(s(X),Y) = 2Y + 6 >= 0 = h(s(f(h(Y),X)))
   problem:
    DPs:
     
    TRS:
     f(s(X),Y) -> h(s(f(h(Y),X)))
   Qed