NO

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

Proof:
 Unfolding Processor:
  loop length: 8
  terms:
   a(b(c(b(b(x51367)))))
   b(b(a(a(c(a(c(b(b(x51367)))))))))
   b(b(a(a(c(c(b(b(x51367))))))))
   b(b(a(a(b(b(x51367))))))
   b(b(a(b(b(a(a(c(a(b(x51367))))))))))
   b(b(b(b(a(a(c(a(b(a(a(c(a(b(x51367))))))))))))))
   b(b(b(b(a(a(c(a(b(a(c(a(b(x51367)))))))))))))
   b(b(b(b(a(a(c(a(b(c(a(b(x51367))))))))))))
  context: b(b(b(b(a(a(c([])))))))
  substitution:
   x51367 -> a(a(c(a(x51367))))
  Qed