Diagonalization and the Halting Problem: Examples and Counterexamples

Applying [L1] with $M=D_H$ itself gives $D_H\text{ accepts }\overline{<mi\; mathvariant="normal">⌜</mi>D_H<mi\; mathvariant="normal">⌝</mi>}⟺D_H\text{ does not accept }\overline{<mi\; mathvariant="normal">⌜</mi>D_H<mi\; mathvariant="normal">⌝</mi>}$. So the special input is the recoded description $\overline{<mi\; mathvariant="normal">⌜</mi>D_H<mi\; mathvariant="normal">⌝</mi>}$, not an informal self-reference slogan.

The contradiction in step 1.1 is exactly the obstruction recorded in [L2]. This concrete trace is the diagonal core of the undecidability proof.

If $M$ halts on $w$, then the simulation inside $E_{M,w}$ eventually terminates on every input, after which $E_{M,w}$ accepts that input. So [L1] gives $L(E_{M,w})={\mathrm{\Sigma }}^{\ast }$, the same language recognized by an always-accept machine.

If $M$ does not halt on $w$, then the simulation inside $E_{M,w}$ never terminates, so $E_{M,w}$ accepts no input at all. Again by [L1], $L(E_{M,w})=\mathrm{\varnothing }$, which is not the language of the always-accept machine. This sharp dichotomy is why deciding equivalence would decide halting, contradicting [L2].

The machine $M_{\mathrm{y}\mathrm{e}\mathrm{s}}$ accepts the one-symbol input word $1$ on its first step. Therefore [L1] gives $⟨<mi\; mathvariant="normal">⌜</mi>M_{\mathrm{y}\mathrm{e}\mathrm{s}}<mi\; mathvariant="normal">⌝</mi>,\widehat{1}⟩\in A_{TM}$.

This specific membership question has a completely transparent yes-answer, so it contradicts the slogan named in [L2]. Hence the slogan is false.