YES

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

Proof:
 DP Processor:
  DPs:
   f#(x,y) -> g#(x,y)
   g#(h(x),y) -> f#(x,y)
   g#(h(x),y) -> g#(x,y)
  TRS:
   f(x,y) -> g(x,y)
   g(h(x),y) -> h(f(x,y))
   g(h(x),y) -> h(g(x,y))
  KBO Processor:
   argument filtering:
    pi(f) = 0
    pi(g) = 0
    pi(h) = [0]
    pi(f#) = [0]
    pi(g#) = 0
   weight function:
    w0 = 1
    w(g#) = w(h) = w(f) = 1
    w(f#) = w(g) = 0
   precedence:
    g# ~ f# ~ h ~ g ~ f
   problem:
    DPs:
     
    TRS:
     f(x,y) -> g(x,y)
     g(h(x),y) -> h(f(x,y))
     g(h(x),y) -> h(g(x,y))
   Qed