Consider Propositional Lax Logic ($PLL$)


* <https://www.uni-bamberg.de/fileadmin/uni/fakultaeten/wiai_professuren/grundlagen_informatik/papersMM/pll.pdf>


The Hilbert system of $PLL$ takes as axiom schemata all theorems of (or a complete set of axioms for) the Intuitionistic propositional calculus plus the modal axiom schemata $\bigcirc R, \bigcirc M, \bigcirc S$ below. The inference rules are Modus Ponens and the rule "from $M \supset N$ infer $\bigcirc M \supset \bigcirc N$": 


$$\text{Axiom} \bigcirc R: \hspace{0.5cm}M \supset \bigcirc M$$
$$\text{Axiom} \bigcirc M: \hspace{0.8cm} (\thinspace \bigcirc \bigcirc M\thinspace) \supset \bigcirc M$$
$$\text{Axiom} \bigcirc S: \hspace{0.8cm}(\bigcirc M \land \bigcirc N) \supset \bigcirc(M \land N)$$


The authors of the above article write that the modal $\bigcirc$ becomes trivial if we add the law of the excluded middle and $\neg \bigcirc false$ to the logic:



> 
> "...if we add the axiom of the Excluded Middle (EM) and $\neg \bigcirc false$ which is valid for both $\Diamond$ and $\Box$ to the modal system $\bigcirc R, \bigcirc M, \bigcirc S$ then $\bigcirc$ becomes trivial. We can derive both $\bigcirc M \supset M$ and $M \supset \bigcirc M$
>  In other words there is no classical Kripke semantics for $\bigcirc$." (p.4, para 1 of above article)
> 
> 
> 


I do not understand how we are able to derive both $\bigcirc M \supset M$ and $M \supset \bigcirc M$ if we add the law of the excluded middle and $\neg \bigcirc false$ to the logic. 


Can anyone help me with this matter?



Modal collapse upon addition of the law of the excluded middle to an Intuitionistic modal logic
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
$$\neg\bigcirc\bot.\tag{\*}$$
First, by Axiom $\bigcirc R$,
$$\neg M\wedge\bigcirc M\to\bigcirc \neg M\wedge\bigcirc M.$$
Therefore by $\bigcirc S$, 
$$\neg M\wedge\bigcirc M\to\bigcirc (\neg M\wedge M)$$
Therefore by definition of $\bot$,
$$\neg M\wedge\bigcirc M\to\bigcirc\bot$$
By (\*),
$$\neg(\neg M\wedge\bigcirc M)$$
By de Morgan and law of excluded middle,
$$M\vee \neg\bigcirc M$$
So,
$$\bigcirc M\to M$$



> 
> Note that Axiom $\bigcirc M$ was not needed.
> 
> 
> 


