Hi guys,
In modal logic i.e. propositional logic with box and diamond, are then any laws to get a box or a diamond from outside a bracket to inside?
I.e. $\Box (x \rightarrow \Box x)$
I want the box inside the brackets :).
Hi guys, In modal logic i.e. propositional logic with box and diamond, are then any laws to get a box or a diamond from outside a bracket to inside? I.e. $\Box (x \rightarrow \Box x)$ I want the box inside the brackets :). 


Thanks for your clarification. If you think about Kripke frames, the logics under consideration are normal modal logics and hence you have the Distribution Axiom $\Box(p\rightarrow q)\rightarrow(\Box p\rightarrow\Box q)$. Since every normal modal logic is also closed under substitution and Modus Ponens, you can derive the rule that from $\Box(A\rightarrow B)$ you can conclude $(\Box A\rightarrow\Box B)$, so in your case from $\Box(x\rightarrow\Box x)$ you can derive $(\Box x\rightarrow\Box\Box x)$. 


Some further cases. Since tautologies are provable, we have $\vdash p\wedge q\rightarrow q$ and hence by the T axiom $\vdash\square (p\wedge q\rightarrow q)$. So in the context Stefan Geschke describes, $$\vdash\square (p\wedge q)\quad\Longrightarrow\quad \vdash\square p\wedge \square q\tag{1}$$ is a valid inference. On the other hand, $$\vdash\square (p\vee q)\quad\Longrightarrow\quad \vdash\square p\vee \square q\tag{2}$$ is not a valid inference; consider for example $$\vdash\square (p\vee \neg p),\quad\text{but}\quad \not\vdash\square p\vee \square \neg p$$ So $\square$ works essentially like a $\forall$ quantifier. 


[](p > []p) <> (<>p > []p). []([]p > p) <> ([]p > []p). <>(p > []p) <> ([]p > []p). <>([]p > p) <> ([]p > <>p). [](<>p > p) <> (<>p > []p). [](p > <>p) <> (<>p > <>p). <>(<>p > p) <> (<>p > <>p). <>(p > <>p) <> ([]p > <>p). [](<>p > []p) <> <>(<>p > []p) <> (<>p > []p). <>(p v q) <> (<>p v <>q). [](p & q) <> ([]p & []q). The above are theorems of S5. 

