Let is a $TM$ and $M$ accepts $\{$:$w\}$

Let's consider the following hypothesis:

$(H_0)A_{TM}$ is decidable and the TM for "Powerful decider"$)$ is a decider for $A_{TM}$ This means that $P$ takes as an input any encoded pair where $M$ is a TM and $w$ an input.

$=>$ If $M$ accepts $w,$ then $T$ accepts

$=>$ If $M$ rejects $w,$ then $T$ rejects

$P$ is a decider so $P$ never loops for ever.

We now construct a new machine for "Contradictory machine"$)$ running $P$ as a subroutine:

$C$ takes an encoded $TM$ as an input, say $,$ and run $P$ on $wPMCCP>P=>>P>,C=>>P>,CCPPP{A}_{TM},w>$ such that $w=.In$ other words, $Pis$ executed $to$ simulate a machine $Mon$ its own encoding $asan$ input. That's strange but not inconceivable. Let's just imagine running a Python program that takes strings $as$ inputs and here $C$ will impose that the Python program takes its own code $asan$ input. Maybe $it$ won't make much sense $in$ the case $of$ this program. But $in$ the case $of$ a compiler, $it$ would make sense.

$SoC$ takes $asan$ input and runs $Pon>$ but $it$ will always return the opposite answer from $P.$

$=>>ifP$ accepts $>,C$ rejects.

$=>>ifP$ rejects $>,C$ accepts.

$So$ basically $CisaTM$ that will take another $TMasan$ input and, using the powerful decider $P,$ execute that $TMon$ it's own encoding $(like$ a program executed $on$ its own code$)$ and then reject $ifP$ accepts, accept $ifP$ rejects.

$So$ far $so$ good.

Finish the proof $to$ show that $A_{TM}is$ undecidable.