Closure set of closure set intersection - formal languages Can we say...
Given L1 and L2.
Is the following true?
$$
L_{1}^{*} \cap L_{2}^{*} = (L_{1}^{*} \cap L_{2}^{*})^{*}
$$
I think it is true but I can't be sure. What permutation of either wouldn't be contained in each other. Sorry I don't know how to format on Math Exchange to make things look better.
-Steve

Left to right is easy, because the set closure operator preserves all elements of the set. So now it's left to prove the right-to-left case. For this, assume an element $u$ in $(L_1* \cap L_2*)*$. It is of the form $u_1u_2\dots u_n$ with each $u_i \in L_1*$ and $u_i \in L_2*$. Now, consider the case of $u_i \in L_1*$. Since this is true for all $u_i$, and the sets are closed under concatenation, then $u = u_1u_2\dots u_n \in L_1*$. Same for $L_2*$. So it's in the intersection, and we're done.
