YES

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

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