YES

Problem:
 f(a()) -> f(b())
 g(b()) -> g(a())

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