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L = {<M1, M2>M1 and M2 are two TMs, and ε ∈ L(M1) \ L(M2) }.

is it RECURSIVE OR RECURSIVE ENUMERABLE OR NOT EVEN RECURSIVE ENUM.
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$(M_1)$ \ $L(M_2)= L(M_1) \cap \bar{ L(M_2)}$

M2 is a TM so L(M2) is RE and $\bar{L(M_2)}$  is not RE.

Whether ∈ belongs to L(M1) which is RE is an undecidable problem. But whether ∈ belongs to $\bar{L(M_2)}$ which is non RE also should be undecidable.

But L whether RE or not RE is the question :/

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@MiNiPanda

I think it depends on if M1 and M2 RE or non RE on that

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@srestha

M1 and M2 are TMs (given) that means L(M1) and L(M2) are RE..am i going wrong?

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Is it halting TM always?
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No I guess..
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Then how will you telling it will be RE always

It could be Rec too, or non RE too
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night i got some reason behind it plzz see if it is correct....

from halting problem i reduced -----1st point to prove can we say yes if it has epsilon or not ............see we have one turing machine tm1 and epsilon is not there till now  , and language may be infinite  so we came across epsilon now can we say that yes it has epsilon ? no because we still have to see l2 that it have epsilon or not ......so we cant yes ............ hence not RE   is it correct ?

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@Deepanshu Didn't get you.. :(

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that is one question that how is L(r1) / L(r2) = L(r1) Intersection    ( complement  of )L(r2)

because if its so then L(r1)- L(r2) = L(r1) / L(r2)
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@MiNiPanda

chk below

I'm assuming the given operation is set difference.

The given problem is, given two Turing machines $M_1$ and $M_2$ if $\epsilon$ is accepted by $M_1$ and not accepted by $M_2.$

Lets assume this problem is semi-decidable or equivalently the corresponding language is recursively enumerable. i.e., we have a Turing Machine say $N$ to semi-decide it (given a string which is in $L$ TM $N$ will always say "yes")

Now, consider another known problem which is not even semi-decidable  say $P$

• given a TM $A$ if $\epsilon$ is not accepted by it.

Using our assumed TM $N$, we can make a TM for $P$ as follows:

• Take $M_1$ as a TM which accepts $\epsilon$ (its language can be anything but must contain $\epsilon)$
• Take $M_2$ as $A$ the given input TM

Now, if our TM $N$ outputs "yes" it means $M_2$ or equivalently $A$ must not accept $\epsilon$ as we know that $M_1$ accepts $\epsilon.$ i.e., using TM $N$ we semi-decide a known not even semi-decidable problem. So, such a TM $N$ cannot exist. i.e., the original given problem is not even semi-decidable.

by Veteran (431k points)
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@Arjun Sir, please clarify few things..

1) is this operator left quotient?

2) you have reduced this problem to complement of halting problem which is well known Non RE. Am i right?

3) Can we prove this language by Rice's second theorem?

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1. No, I suppose it is set difference

2. No. I reduced from complement of halting (accepting) problem to the given problem and not the other way

3. Not easy here as we don't know the strings in $L$
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It could be RE and non RE both?
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Sir,

If the machine is like $M_{1}-M_{2}$

And as u said $M_{2}$ does not accept $\epsilon$ . Say it is like $\Sigma ^{+}$

Then will it not even R.E.?

I am not getting this

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And why we only checking $M_{2}$ part and telling, it is not even RE? Why not at the mean time we looking after $M_{1}$ too?