YES

Problem:
 f(X,X) -> f(a(),b())
 b() -> c()

Proof:
 DP Processor:
  DPs:
   f#(X,X) -> b#()
   f#(X,X) -> f#(a(),b())
  TRS:
   f(X,X) -> f(a(),b())
   b() -> c()
  TDG Processor:
   DPs:
    f#(X,X) -> b#()
    f#(X,X) -> f#(a(),b())
   TRS:
    f(X,X) -> f(a(),b())
    b() -> c()
   graph:
    f#(X,X) -> f#(a(),b()) -> f#(X,X) -> f#(a(),b())
    f#(X,X) -> f#(a(),b()) -> f#(X,X) -> b#()
   CDG Processor:
    DPs:
     f#(X,X) -> b#()
     f#(X,X) -> f#(a(),b())
    TRS:
     f(X,X) -> f(a(),b())
     b() -> c()
    graph:
     
    Qed