Is reduction from A_TM to EQ_TM possible to prove EQ_TM is undecidable?

\begin{align} EQ_{\mathrm{TM}} &= {\{ \langle M,N\rangle : L(M)=L(N) \}}\\ A_{\mathrm{TM}} &= {\{ \langle M,w\rangle : \textrm{TM M accepts w}\}} \end{align}

I can do it using $$E_{\mathrm{TM}}$$ but I want to know if it is possible using $$A_{\mathrm{TM}}$$ because I am not able to do it.

I thought the following:

Input to $$A_{\mathrm{TM}}$$ is $$\langle M,w\rangle$$

Run $$EQ_{\mathrm{TM}}$$ on $$\langle M_1,N\rangle$$ where $$M_1$$ is a modified version of $$M$$ that only runs if input is $$w$$. Hence $$L(M_1) = \{w\}$$.

$$N$$ is a simple TM that accepts only $$w$$.

Is this correct?

• Yes, your approach is correct except that the terms are not appropriate. Fore example, you meant to run the decider for $EQ_{\mathrm{TM}}$ on $\langle M_1,N\rangle$ instead of "Run $EQ_{\mathrm{TM}}$". $M_1$ should still run if the input is not $w$; it just rejects. – Apass.Jack Apr 16 at 21:59