Another question:

is ALL-DFA's status the same? of course it is in coRE just as ALLCFG. But what reduction can we do to prove that it isn't R? We can't do a reduction as the ALLCFG

But we

We need to show a machine that accepts the complement of ALL-CFG and then we will know that ALL-CFG is in coRE\R. ]]>

reduction from Htm (so it isn't in coRE):

<M,w> -> <M'>

M' given x: run M on w for |x| steps. if stopped naturally, accept. else check if x is of the form <M1,w1> and run M1 on w1.

<M,w> e Htm ==> L(M') = sigma* \ A , s.t. A is finite , so L(M') is Regular ( L(M')^c is regular and regular lang. is closed under complement )

<M,w> !e Htm ==> L(M') = Htm

this reduction is good to both CFLtm , REGtm

]]>- $ALL_{CFG}\in co-\mathcal{R}\mathcal{E}/\mathcal{R}$, as you can do a controlled run on all inputs and know when it's not sigma-star.
- $EMPTY_{TM}\in co-\mathcal{R}\mathcal{E}/\mathcal{R}$, as you can do a controlled run and reject if any input is ever accepted by the TM
- $CFL_{TM}, REG_{TM} \not\in co-\mathcal{R}\mathcal{E}\cup\mathcal{R}\mathcal{E}$, one direction by Rice and the other by a reduction I can't find right now.

Will of course be glad to be corrected if I'm wrong.

]]>For each of the above languages, we have stated that they aren't in R. Are they in Re? coRe?

Thanks

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