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2 votes
2 votes
if we construct a halting turing machine then it is decidable(recursive). when we give an input of turing machine then if W$\varepsilon$ L THEN it says yes. if w$\varepsilon !$L THEN IT SAY NO OR IT MAT GOES TO INFINITE LOOP. SO WE can not construct halting turing machine so it is undecidable.
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0 votes

Let A be a HTM & L(A)={m/m accepts w}

As we do not know the behaviour of m, we have to consider all possibility.

if m accepts w, then A will accept m & will go to accept state.

If m rejects w,  then A will reject w & will go to reject state.

if m goes to looping, A will neither go to accept nor to reject state. 

But as A is a HTM it must end up either in accept or reject. It means our contradiction is wrong that A is HTM. It means m cannot have a HTM. So for m, membership is undecidable.

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