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))
  Usable Rule Processor:
   DPs:
    g#(f(x),y) -> h#(x,y)
    h#(x,y) -> g#(x,f(y))
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [h#](x0, x1) = 2x0,
     
     [g#](x0, x1) = 1x0,
     
     [f](x0) = 6x0
    orientation:
     g#(f(x),y) = 7x >= 2x = h#(x,y)
     
     h#(x,y) = 2x >= 1x = g#(x,f(y))
    problem:
     DPs:
      
     TRS:
      
    Qed