I know this is an amateur question but is it true to say that for any three nonempty languages $L_{1},L_{2},L_{3}$ over an alphabet $\Sigma$ we have $L_{1}(L_{2} \cap L_{3}) = L_{1}L_{2} \cap L_{1}L_{3}$ if and only if $L_{2} \subseteq L_{3}$ or $L_{3} \subseteq L_{2}$? Some theorems in mathematics lead me ask this. If not, what is the necessary and sufficient condition for the equality to hold?

  • 1
    $\begingroup$ The condition has to depend on $L_1$, since if $L_1=\emptyset$ then the equality trivially holds for any two sets $L_2,L_3$. $\endgroup$
    – nir shahar
    Sep 25, 2021 at 15:35
  • $\begingroup$ @nirshahar What if we consider the languages to be nonempty? Is there actually any necessary and sufficient condition? $\endgroup$
    – Emad
    Sep 25, 2021 at 15:40
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    $\begingroup$ Another case: $L_1=(L_2\cup L_3)^*$. Or $L_1=\Sigma^*$. What mathematical theorems having to do with concatenation are you thinking of? $\endgroup$
    – rici
    Sep 25, 2021 at 18:37
  • $\begingroup$ @rici There are a few (to some extent) similar theorems about vectors spaces, groups, etc: Like the union of two subspaces is a subspace if and only if one is contained in the other. I saw one direction is correct, so I made the guess. $\endgroup$
    – Emad
    Sep 25, 2021 at 18:50
  • $\begingroup$ @rici Also one other thing is that we have a monoid here so it's natural for these ideas to strike the mind. $\endgroup$
    – Emad
    Sep 25, 2021 at 18:57


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