$\textbf{Definitions:}$


Let $M$ be an $A$-module. The collection $\mathcal{S}(M)$ of all submodules of $M$ is partially ordered by the inclusion relation. The collection $\mathcal{S}(M)$ is said to be distributive if and only if the following equivalent identities hold for all choices of $N,P,Q\in\mathcal{S}(M)$:


(a) $N\cap(P+Q)=(N\cap P)+(N\cap Q)$


(b) $N+(P\cap Q)=(N+P)\cap (N+Q)$


(c) $(N\cap P)+(P\cap Q)+(Q\cap N)= (N+P)\cap (P+Q)\cap (Q+N)$


$\textbf{Question:}$


Prove that $\mathcal{S}(M)$ is distributive if and only if for $N,P,Q\in\mathcal{S}(M)$, if $N\cap P=N\cap Q$ and $N+P=N+Q$, then $P=Q$.


$\textit{Proof}:$($\implies$) Let $\mathcal{S}(M)$ be distributive and $N,P,Q\in\mathcal{S}(M)$ havethe given properties.


Then \begin{equation}
P = P\cap(N+P)
 =P\cap(N+Q)
 =P\cap N + P\cap Q\\
 = Q\cap N + P\cap Q
 =Q\cap(P+N)
 =Q\cap(Q+N)
 =Q.
\end{equation} 


I'm facing problem with proving the reverse implication. My idea of the proof is using contradiction.


If possible assume that $\mathcal{S}(M)$ is not distributive. So there exist $N,P,Q\in\mathcal{S}(M)$ such that $(N\cap P)+(N\cap Q)\subsetneq N\cap(P+Q)$. If I can come up with some submodule in between these two which will act as $'N'$ in the property, then I'll be done. 


I would like to get a hint about this problem. You can suggest me another ways for the same, as well. Thanks in advance. 



Equivalent condition for distributive property of $\mathcal{S}(M)$
Let $H$ be an $A$-submodule of $M$. Note that the condition cannot hold if there exist two distinct $A$-submodules $P$ and $Q$ of $M$ containing $H$ such that $P/H$ and $Q/H$ are simple and isomorphic. Therefore the desired conclusion follows from Theorem 1 of the following paper by V. Camillo: <https://www.sciencedirect.com/science/article/pii/0021869375901519>


