YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),b()) -> f#(a(),c())
   f#(c(),d()) -> f#(b(),d())
  TRS:
   f(a(),b()) -> f(a(),c())
   f(c(),d()) -> f(b(),d())
  EDG Processor:
   DPs:
    f#(a(),b()) -> f#(a(),c())
    f#(c(),d()) -> f#(b(),d())
   TRS:
    f(a(),b()) -> f(a(),c())
    f(c(),d()) -> f(b(),d())
   graph:
    
   Qed