Model for formula $\Box (\psi \vee \phi) \Rightarrow (\Box \psi) \vee (\Box \phi)$ I am new to LTL and I am trying to understand how it works. My question is: is there such $\sigma$ that: 

$ \sigma \models [\Box (\psi \vee \phi) \Rightarrow (\Box \psi) \vee (\Box \phi)]$

I know that:

$\sigma \models \Box \psi: \forall k \, [(\sigma, k) \models \psi] \\
\sigma \models [\psi \Rightarrow \phi]: (\sigma \models \neg \psi) \vee (\sigma \models \phi) \\
\sigma \models \Box \psi \iff \sigma \models \neg\Diamond \neg \psi$

So 

$\sigma \models [\Box (\psi \vee \phi) \Rightarrow (\Box \psi) \vee (\Box \phi)] \iff \\(\sigma \models \Diamond (\neg\psi \land \neg\phi)) \vee [(\sigma \models (\Box \psi)) \vee (\sigma \models (\Box \phi))]$

I would say there is such $\sigma$ since I don't see any contrarguments but I'm not sure if I can say anything about $\psi$ or $\phi$ except that both are formulas (so I don't know how to show any specific example).
If you're just asking for any model satisfying the formula, you simply try to make one of these the case


The left hand side false
The right hand side true


Let's just pick number 2.  You have to make a model (with a state $\sigma$) where $\Box \psi \lor \Box \phi$ is the case.  Even easier is $\Box \phi$.

Create a (trivial) model where every state $\sigma$ has $\sigma \models p$ for all propositions $p$.



If you want to see if the formula is valid then the answer is no.  You should now be able to create a model that falsifies the formula.