YES

Problem:
 h(f(x),y) -> f(g(x,y))
 g(x,y) -> h(x,y)

Proof:
 DP Processor:
  DPs:
   h#(f(x),y) -> g#(x,y)
   g#(x,y) -> h#(x,y)
  TRS:
   h(f(x),y) -> f(g(x,y))
   g(x,y) -> h(x,y)
  TDG Processor:
   DPs:
    h#(f(x),y) -> g#(x,y)
    g#(x,y) -> h#(x,y)
   TRS:
    h(f(x),y) -> f(g(x,y))
    g(x,y) -> h(x,y)
   graph:
    g#(x,y) -> h#(x,y) -> h#(f(x),y) -> g#(x,y)
    h#(f(x),y) -> g#(x,y) -> g#(x,y) -> h#(x,y)
   Subterm Criterion Processor:
    simple projection:
     pi(h#) = 0
     pi(g#) = 0
    problem:
     DPs:
      g#(x,y) -> h#(x,y)
     TRS:
      h(f(x),y) -> f(g(x,y))
      g(x,y) -> h(x,y)
    SCC Processor:
     #sccs: 0
     #rules: 0
     #arcs: 2/1