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:
    
   Restore Modifier:
    DPs:
     f#(s(X),Y) -> f#(h(Y),X)
    TRS:
     f(s(X),Y) -> h(s(f(h(Y),X)))
    SCC Processor:
     #sccs: 1
     #rules: 1
     #arcs: 1/1
     DPs:
      f#(s(X),Y) -> f#(h(Y),X)
     TRS:
      f(s(X),Y) -> h(s(f(h(Y),X)))
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [f#](x0, x1) = x0,
       
       [h](x0) = 0,
       
       [f](x0, x1) = 0,
       
       [s](x0) = 1
      orientation:
       f#(s(X),Y) = 1 >= 0 = f#(h(Y),X)
       
       f(s(X),Y) = 0 >= 0 = h(s(f(h(Y),X)))
      problem:
       DPs:
        
       TRS:
        f(s(X),Y) -> h(s(f(h(Y),X)))
      Qed