(who is I? what is know?)

Read my previous answer. ]]>

but this is not what i'm asking - my question is - just to be clear,

if M_0 is fixed, does it imply that **I KNOW** what L(M_0) is?

Maybe this will convince you:

to show that L is decidable we have to prove that there exists a TM that decides it.

I provided two TMs and we all agree that one of them decides L2.

So L2 is decidable.

("If L(Mo) is regularâ€¦" is not a part of the algorithm!)

]]>describing the algorithm:

"If L(Mo) is regular…" - how can I continue executing - this is not decidable.

Unless U ment that a part fo the input is : "Given M0, a TM and L(Mo) is regular(/Not regular)". ]]>

The algorithm is built according to M_0.

If L(M0) is regular we suggest one decider (checking if regular languages are equal),

and otherwise we suggest a different decider (always reject). ]]>

give a decider D the input <A>, what D sould do?

if he guesses that M0 is regular - it has to compare L(M0), L(A).

if he guesses L(Mo) is not regular - it should reject all inputs.

the different algorithm depends on L(M0).

can someone describe a Decider ??? ]]>

Either way, we get the same result - L2 is decidable.

Notice that in REGtm, we can't say that we get the same result no matter what - if L(M) is regular, it belongs to REGtm, otherwise, it does not belong to REGtm.

Hope it helps.

L2={<A> | A is DFA , L(A)=L(M0)}

solution: L2 is decidable:

"If L(Mo) regular - checking if regular languages are equal,

if not, L(M0) is not regular and L2= empty "

we saw REGtm = {<M> | M is Tm and L(M) is regular} is not decidable in class,

how does it works in this solution?

TNX

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