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()
  CDG Processor:
   DPs:
    f#(x,x) -> b#()
    f#(x,x) -> f#(a(),b())
   TRS:
    f(x,x) -> f(a(),b())
    b() -> c()
   graph:
    
   Qed