however h0 (meaning halts on epsilon ).

L={<M>| m halts on epsilon} that group is in RE\R however m' that u returned belong to it but isnt the whole h0.

meaning L(m') isnt halt0 !

beacuse halt0 is all the machines and u just made one that group is final and therfore regular and the reduction doesnt work.

u need a reduction from halt0 needs to be a on input <N> blah blah blah we have <m'> which L(m') in R the machine u returned is in h0 and doesnt halt on every input. ]]>

First we'll show L2 is not in Co-RE; i.e L1 is not in RE.

Reduction from halt0 to L2 (the complement of L1):

On input <M> (input to halt0), f builds M', TM that works as follow:

On input <N>

1. Run M on epsilon

2. If it halts then Run N on epsilon

3. If N halts then accepts.

Correctness: If M halts on epsilon then we get to check N which halts on epsilon iff N is a TM that is in Halt0, L(M')=halt0, a language in RE\R, L(M') is in RE\R, its in L2. (if M doesnt halt then L(M')=empty because it doesnt accepts any N. L(M') is in R, its in L1).

Second we'll show L1 is not in CoRE; i.e L2 is not in RE

On input <M> (input to halt0), f builds M', TM that works as follow:

On input <N>

1. Run Both M on epsilon and N on epsilon

2. If one of these halts then accept.

Correctness: If M halts on epsilon then M' accepts all N, L(M')= sigma* which is in R, L(M') is in L1.

If M doesnt halt on epsilon then N halts iff N is a TM in halt0. as above then L(M')=halt0, halt0 is in RE\R, L(M') is in RE, its in L2.

תודה מראש! ]]>

Alex: When doing these reduction, try to get used to the idea: 1. on input x M' simulates M on w if M halts (accepts….whatever) then….It means that if M halts on w then it doesnt matter which x it was. i.e for any x = for any input… Hope it helps.

]]>for example m loopy a machine that always loops clearly doesnt belong to htm* because it never stops how ever belongs to L1 because L(m) is the empty language which belongs to R.

from rice theorem we can see that its not in R and one of the conclusions from rice is that if the empty language belongs to L then its in co-RE otherwise its in RE.

plus u can see that its semantic and non trivial quite easily.

i think the hard thing is finding a a machine that not in the language: the machine accpeting the language L= (atm complement intersection with htm) (from benny chor test and there are probarly more). ]]>

But how do you make the reduction from halt when you have to check for every possible input?

]]>Try and prove it using reduction from Htm to L1 and another reduction from Htm to L2 (L1complement). Its easier to think of it as the Halting problem (or to be precise halt* problem): Given <m>, does m halts on any input w?? Because if it does then it either accepts or rejects any input w, hence decides == in R.

]]>Don't take my word for it though :)

]]>1) L = {<M>| L(M) is in R}

2) L = {<M>| L(M) is NOT in R}

Thanks!

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