Equivalence of different versions of Dynkin's lemma Dynkin's Lemma.

(1) For any $\pi$-class $P$, $\lambda(P)=\sigma(P)$.


(2) Let $P$ be a $\pi$-class and $D$ be a $\lambda$-class. IF $P\subseteq D$, then $\sigma(P)\subseteq D$.

These are two versions of the same lemma that I know. I have the proofs of both and I undertsand them as separate statements. But, I can't figure how to prove their "equivalence". They seem to be two separate lemmas. In fact the 2nd lemma seems to be a corollary of the first. Any help?
 A: Dynkin's Lemma.

(1) For any $\pi$-class $P$, $\lambda(P)=\sigma(P)$.
(2) Let $P$ be a $\pi$-class and $D$ be a $\lambda$-class. IF $P\subseteq D$, then $\sigma(P)\subseteq D$.

Let us prove they are equivalent. Let us assume we are considering sets of subsets of $\Omega$ (where $\Omega$ is a set).
($1 \Rightarrow 2$) Note that if $ A \subseteq B$, then $\lambda(A) \subseteq \lambda(B)$. So, if $P$ be a $\pi$-class, $D$ be a $\lambda$-class and $P\subseteq D$, then, using ($1$), we have:
$ \sigma(P) = \lambda(P) \subseteq \lambda(D) = D$.
($2 \Rightarrow 1$) Let $P$ be a $\pi$-class. Since $\lambda(P)$ is a  $\lambda$-class and $P \subseteq \lambda(P)$, then using ($2$), we have : $\sigma(P) \subseteq \lambda(P)$.
Now, note that if $L$ is any $\lambda$-class containing $P$, then $\lambda(P) \subseteq L$. Since all $\sigma$-algebras are $\lambda$-classes, we have that $\sigma(P)$ is a $\lambda$-class containing $P$, so $\lambda(P) \subseteq \sigma(P)$. So we have $\lambda(P)=\sigma(P)$.
