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)
  Usable Rule Processor:
   DPs:
    h#(f(x),y) -> g#(x,y)
    g#(x,y) -> h#(x,y)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [g#](x0, x1) = 2x0,
     
     [h#](x0, x1) = 1x0,
     
     [f](x0) = 6x0 + -3
    orientation:
     h#(f(x),y) = 7x + -2 >= 2x = g#(x,y)
     
     g#(x,y) = 2x >= 1x = h#(x,y)
    problem:
     DPs:
      
     TRS:
      
    Qed