NO

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

Proof:
 Unfolding Processor:
  loop length: 10
  terms:
   a(b(b(b(c(b(b(x70479)))))))
   b(b(c(a(b(b(c(b(b(x70479)))))))))
   b(b(c(b(b(c(a(b(c(b(b(x70479)))))))))))
   b(b(c(b(c(a(b(c(b(b(x70479))))))))))
   b(b(c(c(a(b(c(b(b(x70479)))))))))
   b(a(a(b(c(b(b(x70479)))))))
   b(a(b(b(c(a(c(b(b(x70479)))))))))
   b(a(b(b(c(c(b(b(x70479))))))))
   b(a(b(a(b(b(x70479))))))
   b(a(b(b(b(c(a(b(x70479))))))))
  context: b([])
  substitution:
   x70479 -> c(a(x70479))
  Qed