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Proof or Disproof

if $ L $ is a Regular language then it has to be that $ L\leq HP $

$ HP=\{<M,x> | M \ halts \ on \ x \} $ Regular language is a language that can be expressed with a regular expression or a deterministic or non-deterministic finite automata

I know that HP is in RE\R and I know the regular language is in R and I know that due to the reduction principle, if $ L1 \leq L2 $ then if $ L2 \in RE $ then L1 must also be in RE and also if L1 is not in RE then L2 also not in RE but this is not the case...

$ f(x) = <M',x> $ lets create M' such that for all x $in \Sigma^* $

  1. run $ M_{regular} $ on x if accepts then accepts
  2. if rejects then go to infinite loop

L is a decidable language then exist a decidable turning machine such that for all x will halt

HP will only halt on x in the language,

if I try to build a reduction from L to HP lets look at the conditions

$x\in L \rightarrow M_L(x)=1 \rightarrow $ M halt on x $ <M',x> \in HP $

$x\notin L \rightarrow M_L(x)=0 \rightarrow $ I can send here to an infinite loop so that M doesn't halt on x $ <M',x> \notin HP $

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    $\begingroup$ What have you tried? Did you try to suggest a reduction? Where you got stuck? We can't just solve your exercises for you. $\endgroup$ Jan 27 at 20:03
  • $\begingroup$ I ask for help this is why I am stuck, from my knowledge I don't know what to do because all the Data in the question does not help me, and I do not understand how to continue because the reduction principle doesn't fit here $\endgroup$
    – maya cohen
    Jan 27 at 20:06
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    $\begingroup$ You need to show some effort. Use the fact that $L$ is decidable, and the fact that HP is non-trivial. Proceed from there. $\endgroup$ Jan 27 at 20:07
  • $\begingroup$ Hey @BaderAbuRadi I added what I tried but with what I think but then it looks like the reduction is valid what I did wrong? $\endgroup$
    – maya cohen
    Jan 27 at 20:13
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    $\begingroup$ "it looks like", in this case, try to prove that it is correct, and see how things go... $\endgroup$ Jan 27 at 20:37

1 Answer 1

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From your attempt, I see that you got the intuition, but you need to define the reduction formally, and then prove that it works. In your solution, it is not clear how $M'$ is defined, and what is $M_{regular}$ -- for me, I just saw your thoughts/attempts, not a proof.

Here, I showed a slightly more general claim, specifically, every non-trivial language is $R$-hard (harder than every language in $R$). Since $L\in R$, and $HP$ is non-trivial, then what you're asking for follows immediately. Try to prove that the reduction there is computable (there is a TM that computes it), and that it is correct.

The idea essentially is as follows. Since we can decide $L$, then we can define a reduction that checks whether its input is in $L$, and then outputs a word inside or outside $HP$, accordingly. So languages in $R$ are too easy, w.r.t mapping reductions, in the sense that we (the reduction) can solve/decide them, and then output whatever we want.

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  • $\begingroup$ Oh i re-read this question and answer and I see that I didn't mention I was thinking it is a disproof , but from your answer I read you think I was understanding its a proof ... sorry for that ... $\endgroup$
    – maya cohen
    Feb 2 at 22:09
  • $\begingroup$ so if i understood correctly then I need to use the ability that regular languages are in R, and define a reduction that will do the same as what I said in my reduction? $\endgroup$
    – maya cohen
    Feb 2 at 22:10
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    $\begingroup$ Your reduction is not clear to me. If you want we can discus it in chat sometime. $\endgroup$ Feb 3 at 0:41
  • $\begingroup$ do you have time tommorow at 19:00 ? how can we have a chat in this website? $\endgroup$
    – maya cohen
    Feb 4 at 20:49

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