We have languages $A= \{a\}$ and $B = \{b\}$. If we consider $(A\cup B)^*$, where ${}^*$ means Kleene star, we have a set of words like $\{\lambda, a,b,aa,ab,aaa,\dots\}$, where $\lambda$ is the empty word. Another example we do is $A^*(BA^*)^*$. If we concatenate two languages, $\lambda$ will not be in the set, but the Kleene star adds the $\lambda$. Question is, is this set same as the set with union so $\{\lambda,a,b,aa,ab,aaa,\dots\}$ or is the set for the second example $\{a,b,aa,ab,aaa,\dots\}$, in which case the langauages would not be equal. My opinion is that concatenation is the last operation so sets are not equal but I'm not sure.
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1$\begingroup$ "If we concatenate two languages, $\lambda$ will not be in the set" - This is not quite right. The concatenation $L_1 L_2$ does include $\lambda$, if $\lambda \in L_1$ and $\lambda \in L_2$. $\endgroup$– D.W. ♦Commented Oct 6, 2015 at 23:53
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$\begingroup$ "is this set same as the set" - Which set? I'm confused by your pronouns. I suggest you give the sets names (e.g., $C$, $D$, etc.) so you can refer to them clearly. $\endgroup$– D.W. ♦Commented Oct 6, 2015 at 23:54
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1 Answer
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Unless parentheses say otherwise, the customary precedence is ${}^*$ (like exponentiation), then concatenation (like multiplication), then union (like addition).
Your second example is the concatenation of starred expressions, so it does contain the empty string.
