YES

Problem:
 f(h(x)) -> f(i(x))
 g(i(x)) -> g(h(x))
 h(a()) -> b()
 i(a()) -> b()

Proof:
 DP Processor:
  DPs:
   f#(h(x)) -> i#(x)
   f#(h(x)) -> f#(i(x))
   g#(i(x)) -> h#(x)
   g#(i(x)) -> g#(h(x))
  TRS:
   f(h(x)) -> f(i(x))
   g(i(x)) -> g(h(x))
   h(a()) -> b()
   i(a()) -> b()
  Usable Rule Processor:
   DPs:
    f#(h(x)) -> i#(x)
    f#(h(x)) -> f#(i(x))
    g#(i(x)) -> h#(x)
    g#(i(x)) -> g#(h(x))
   TRS:
    i(a()) -> b()
    h(a()) -> b()
   CDG Processor:
    DPs:
     f#(h(x)) -> i#(x)
     f#(h(x)) -> f#(i(x))
     g#(i(x)) -> h#(x)
     g#(i(x)) -> g#(h(x))
    TRS:
     i(a()) -> b()
     h(a()) -> b()
    graph:
     
    Qed