How to prove that " necessarily (A OR ~A) " does not imply " necessarily A OR necessarily ~A".

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In De Interpretatione , Aristotle criticizes logical fatalism ( a metaphysical doctrine professed in the Megarian School, in particular by Diodorus Cronus).

Aristotle reconstructs the reasoning of logical fatalists as follow :

(1) Necessarily ( there will be a sea battle tomorrow or there will not be a sea battle tomorrow).

(2) Therefore, necessarily a sea battle will happen tomorrow or necessarily it will not happen.

One can say informally that the logical mistake consists in the fact that " the necessity of the disjunction does not imply the disjunction of the necessities".

However, how to prove rigorously using the tools of modal logic that the reasoning is not valid?

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You can provide a semantics for 'necessarily $A$' by saying that in all worlds out of some collection of worlds, $A$ is true.

As such, we can show that we can make $(1)$ true, but $(2)$ false: create a collection of two worlds, $w_1$ and $w_2$. Then assume that $A$ is true in $w_1$, but $A$ is false in $w_2$.

Now, $(1)$ is true, since in $w_1$, we have that $A$ is true, and hence $A \lor \neg A$ is truew in $w_1$ as well. In $w_2$ we have that $\neg A$ is true, and hence $A \lor \neg A$ is true in $w_2$ is true as well. So, $A \lor A$ is true in all worlds, and hence it is necessarily true. So, $(1)$ is true.

$(2)$, however is false. We don't have that $A$ is necessarily true, since $A$ is not true in all worlds: $A$ is not true in $w_2$. Likewise, $\neg A$ is not true in $w_1$, and hence $\neg A$ is not necessarily true. Hence, both disjuncts from $(2)$ are false, so $(2)$ is false.

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For unknown $n\in\Bbb N$, let $A$ denote $n=1$. Neither $A$ nor $\neg A$ is necessarily true; each has a counterexample.