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So in LTL we have the following elementary modalities

$ \Diamond = $ "eventually"

$ \square = $ "always"

I am aware the following distributivity laws hold

i) $ \Diamond (a \lor b) = \Diamond a \lor \Diamond b $

ii) $ \square (a \land b) = \square a \land \square b $

iii) $ \Diamond (a \land b) \neq \Diamond a \land \Diamond b $

iv) $ \square (a \lor b) \neq \square a \lor \square b $

But what about when you combine the two elementary modalities? For example

$ \square \Diamond(a \lor b) $ and $ \Diamond \square(a \land b) $

I would assume you could distribute as follows

$ \square \Diamond(a \lor b) = \square (\Diamond a \lor \Diamond b) \neq \square \Diamond a \lor \square \Diamond b $

$ \Diamond \square(a \land b) = \Diamond (\square a \land \square b) \neq \Diamond \square a \land \Diamond \square b $

But I can't figure out any example series to prove the last inequalities. Is it the case when combining modalities you simply consider them both as a whole and distribute based on the modality furthest to the right?

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When you write $\Box(a\lor b) \ne \Box a \lor \Box b$, what you actually mean (or should mean) is that it is not guaranteed that $\Box (a \lor b)$ and $\Box a \lor \Box b$ will mean the same thing. There may still be cases where they do mean the same thing -- for example, when $a$ and $b$ are the same formula.

Another case where they have the same meaning is when $a$ and $b$ both happen to start with a $\Diamond$.

For this to be true it is important that we're talking about linear temporal logic; a logic that allowed for branching time would not have this property.

It is clear that either of $\Box\Diamond a$ or $\Box\Diamond b$ will imply $\Box(\Diamond a \lor \Diamond b)$.

On the other hand, if $\Box\Diamond a$ is false it means that there's a certain amount of time we can wait, after which $a$ will never be true again -- and similarly for $\Box\Diamond b$, so if $\Box\Diamond a \lor \Box\Diamond b$ is false, then there's a point in the future after which neither $a$ nor $b$ will ever be true. But this means exactly that $\Box\Diamond(a\lor b)$ is false!