I'll write $\to$ instead of $\supset$, and $\bot$ instead of false, below.
Since Law of Excluded Middle is given, I'll argue using classical propositional logic.

Since $M\to\bigcirc M$ is already given as Axiom $\bigcirc$R, let's prove $$\bigcirc M\to M.$$
We are given
$$\neg\bigcirc\bot\tag{*}$$
$$\bigcirc\bigcirc M\to\bigcirc M\tag{$\dagger$}$$
$$\bigcirc M\wedge\bigcirc N\to\bigcirc (M\wedge N)$$
We need to show
$$\neg M\to\neg\bigcirc M.$$
In other words
$$A\to\neg\bigcirc\neg A.$$
So it suffices to show
$$A\wedge\bigcirc\neg A\to\bot$$
so by (*) it suffices to obtain
$$A\wedge\bigcirc\neg A\to \bigcirc\bot$$
But we have
$$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc(\bot)$$
so we're done. I guess I didn't use ($\dagger$).