YES

Problem:
 g(x,a(),b()) -> g(b(),b(),a())

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