Let T = {<M> | M is a TM that accepts $w^R$ whenever it accepts w}. Show that T is undecidable

Question

Let T = {<M> | M is a TM that accepts w ^r whenever it accepts w}.
Show that T is undecidable.

I have two answers to this question - San Diego :

5.9
Let T = { <M> | M is a TM that accepts w ^r whenever it accepts w }.

Assume T is decidable and let decider R decide T. Reduce from A _TM by constructing a TM S as follows:

• S: on input <M,w>
  1. create a TM Q as follows:
     On input x:
     1. if x does not have the form 01 or 10 reject.
     2. if x has the form 01, then accept.
     3. else (x has the form 10), Run M on w and accept if M accepts w.
  2. Run R on
  3. Accept if R accepts, reject if R rejects.

Because S decides A _TM , which is known to be undecidable, we then know that T is not decidable

Undisclosed source:

• 5.12 We show that A _TM ≤ _m S by mapping ‹ M , w › to ‹ M' › where M' is the following TM:

  • M' = “On input x :
    1. If x = 01 then accept .
    2. If x ≠ 10 then reject .
    3. If x = 10 simulate M on w .
       If M accepts w then accept ; if M halts and rejects then reject .”

  If ‹ M , w › ∈ A _TM then M accepts w and L ( M' ) = {01,10}, so ‹ M' › ∈ S .
  Conversely, if ‹ M , w › ∉ A _TM then L ( M' ) = {01}, so ‹ M' › ∉ S . Therefore,
  ‹ M , w › ∈ A _TM ⇔ ‹ M' › ∈ S .

But I do not understand the following:

1- what is the relation between x and w?

2- why we consider the 2 cases ‹ M , w › ∈ A _TM and ‹ M , w › ∉ A _TM ?

3- why if A is mapping reducible to S this makes S undecidable?

could anyone clarify these points for me?

Answer 1

I think it is not suitable for asking in SO because it is not a educational website, but I answered it.

1- what is the relation between x and w?

Answer 1: x is a symbol that used for using a symbol for operate. This symbol should not be in alphabet of language, just it. It hasn't any relation to w.

2- why we consider the 2 cases ‹M, w› ∈ ATM and ‹M, w› ∉ ATM?

Answer 2: For proofing a language like L is decidable or not, we need to determine a string like w is member of language or not. So we have to consider two type of string w∉L and w∈L.

3- why if A is mapping reducible to S this makes S undecidable?

Answer 3: It means the process of checking a string is in language in A and S is similar and if we can't find a algorithm for checking this for A, we can't find any algorithm for S.