FALSE: universality decides halting

Let $M_{\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{p}}=(3,0,1,{\tau }_{\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{p}})$, where the unique transition is ${\tau }_{\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{p}}(0,0)=(0,0,1)$. Its input alphabet is empty, so its unique input word is $\epsilon$. On that input the machine keeps state $0$ forever, moves right at every step, and never reaches a halting state. Hence it has no halting computation history on $\epsilon$, so by [L2] it diverges on $\epsilon$.

By [L3], there exists at least one universal machine $U$. By [L1], the run of $U$ on the coded input $⟨<mi\; mathvariant="normal">⌜</mi>M_{\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{p}}<mi\; mathvariant="normal">⌝</mi>,\widehat{\epsilon }⟩$ must reproduce the divergence from step 1.1. Hence $U$ does not halt on that instance.

A machine that fails to halt on some instance does not decide a yes-or-no halting problem on all instances. Step 2.1 therefore contradicts the false claim [A1].