# test series

1 vote
61 views
If L1 and L2 are non-regular, then L1⋃ L2 is also not-regular.

true or false?
0
False.

Take counter examples to disapprove in such questions
Let a^nb^m where n!=m be L1 and a^nb^m where n=m be L2.
Both are non regular infact dcfl to be precise but their union is
a*b* a regular language.

## Related questions

1
390 views
$P_{1}:$ {$<M>|M$ is a TM that accepts atleast $2$ strings of different length} $P_{2}:$ {$<M>|M$ is a TM and there exists an input whose length less than $100,$ on which $M$ halts } The number of problem which is $RE$ but not $REC$ _____________
How many number of $DFA$ states(minimal DFA) required which accepts the language $L=\left \{ a^{n}:n=\text{3 or n>= 2m for all m>= 1} \right \}$ ___________ Answer will be $3$ or $6?$
Consider $\left \langle M \right \rangle$ be the encoding of a turing machine as a string over alphabet $\Sigma =\left \{ 0,1 \right \}$. Consider $D=${$\left \langle M \right \rangle$ $M$ ... Recursive $(B)$ Non-Recursive $(C)$ Recursively enumerable $(D)$ Not Recursively enumerable My question is Is it not a Halting Problem they are asking for?