NO

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

Proof:
 Containment Processor: loop length: 1
                        terms:
                         f(x,c())
                        context: []
                        substitution:
                         x -> c()
  Qed