NO

Problem:
 a(x1) -> x1
 a(b(x1)) -> a(a(c(b(b(a(x1))))))
 b(x1) -> x1
 c(c(x1)) -> x1

Proof:
 Unfolding Processor:
  loop length: 7
  terms:
   a(b(b(x11064)))
   a(a(c(b(b(a(b(x11064)))))))
   a(a(c(b(a(b(x11064))))))
   a(a(c(a(b(x11064)))))
   a(a(c(a(a(c(b(b(a(x11064)))))))))
   a(a(c(a(c(b(b(a(x11064))))))))
   a(a(c(c(b(b(a(x11064)))))))
  context: a([])
  substitution:
   x11064 -> a(x11064)
  Qed