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Let L1 be the recursive language. Let L2 and L3 be languages that are recursively enumerable but not recursive. Which of the following statements is not necessarily true?

a.) L2−L1 is recursively enumerable.

b.)L1−L3is recursively enumerable.

c.)L2∩L3is recursively enumerable.

d.)L2∪L3is recursively enumerable.

My approach:- I am able to falsify a,c and d option. Now for b option i took one example where my L1 is universal language so when i do L1-L3,i get the $L3^c$(L3 complement).So complement of RE but Not REC is NOT RE always.So it is not required that L1-L3 will be RE. But on the other hand if i solve b option as follow:-

L1-L3

= L1 ∩ $(L3)^c$

= REC ∩ $(RE but Not REC)^c$

=REC ∩ NOT RE

So the final expression will give me empty language and it is regular.Is it possible for the final expression to have non empty set above?

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The intersection of a recursive language and a non-RE language isn't necessarily empty. Take the recursive langauge to be $\Sigma^*$, as you've already done once.

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  • $\begingroup$ Thank you.I didn't notice that.I was just thinking how can something be both REC and NOT RE. $\endgroup$ – rahul sharma Aug 8 '17 at 0:36
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    $\begingroup$ A language can't be both recursive and non-RE. But that's not what you're doing here: you're intersecting languages (sets of strings), not language classes. Any particular string could be in a recursive language and in a non-RE one, as you've now realised. $\endgroup$ – David Richerby Aug 8 '17 at 9:38

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