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)
  EDG 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)
    Matrix Interpretation Processor:
     dimension: 1
     interpretation:
      [g#](x0, x1) = 1,
      
      [h#](x0, x1) = 0,
      
      [g](x0, x1) = 0,
      
      [h](x0, x1) = 0,
      
      [f](x0) = 0
     orientation:
      g#(x,y) = 1 >= 0 = h#(x,y)
      
      h(f(x),y) = 0 >= 0 = f(g(x,y))
      
      g(x,y) = 0 >= 0 = h(x,y)
     problem:
      DPs:
       
      TRS:
       h(f(x),y) -> f(g(x,y))
       g(x,y) -> h(x,y)
     Qed