L3’ = complement of L3 which is a$^{n}$ >=2

L$_{3}$.L$_{2}$.(L$_{1}$)* here L1={phi}*=epsilon

so L3’.L2.L1=a$^{n}$ >=2 --------------------eq1

now 2^{nd} part L1.L3’= {phi}.L3’=phi

so from eq 1 i think option D match..

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abhinowKatore
asked
in Theory of Computation
Oct 18, 2022

241 views
1 vote

Best answer

This could be solved using some basic Regular Expression Identities.

1)$\phi$*=$\varepsilon$

2)R.$\phi$=$\phi$.R=$\phi$

3)R.$\varepsilon$=$\varepsilon$.R=R

Where R is some Regular Expression.

Let's find *L3'=Complement of L3=a*-L3*

Since LE is defined over Alphabet With only symbol a.

L3'=a*-{a,$\varepsilon$}

L3'={$\varepsilon$,$a$,${a^2}$,$a^3$,$a^4$,.......} - {$a$,$\varepsilon$}

L3'={$a^2$,$a^3$ ,$a^4$,$a^5$,.......}

L3'={$a^n$/ $n$>=2}

Now,Solving the Expression

L3'.L2.L1*+L1.L3'=L3'.$\varepsilon$.$\phi$*+$\phi$.L3'

=L3'.$\varepsilon$.$\varepsilon$ +$\phi$(using $1$ and $2$)

=L3'+$\phi$

=L3'

={$ a^2$/$n$>=2}

____

Thus ,option D holds.