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)))
  Usable Rule Processor:
   DPs:
    f#(s(X),Y) -> f#(h(Y),X)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0, x1) = x0 + 2x1,
     
     [h](x0) = 1x0,
     
     [s](x0) = 7x0
    orientation:
     f#(s(X),Y) = 7X + 2Y >= 2X + 1Y = f#(h(Y),X)
    problem:
     DPs:
      
     TRS:
      
    Qed