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Let L be CFL and M a regular language. Language L ⋂ M is always
(a) always regular (b) never regular
(c) always DCFL (d) always context free language
in Theory of Computation by Junior (951 points) | 98 views
none of these

1 Answer

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Let L be context free language lets take L = {$a^{n}b^{n} | n > 1$}
Let M be regular language, lets take L = { $a^{*}b^{*}$ }

L  ⋂ M = {$a^{n}b^{n} | n > 1$} which is CFL

Only case i can think of regular is when you get $\phi$ as intersection, anyhow all regular language is CFL by default.

So we can say answer should be always CFL which is option D.
by Loyal (7.8k points)
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