The Turing-machine acceptance problem is recognizable

Build a deterministic one-tape machine $R$ that, on input $z$, first parses $z$ as a candidate pair $⟨x,y⟩$. If the pair parse fails, it rejects. It next runs the decider from [L4] on $x$; if $x$ is not a well-formed machine code, it rejects. If $x$ is valid, it uses [L2] to recover the coded machine $M=(m,s,n,\tau )$. It then parses $y$ as a candidate word code $c(\mathrm{bin}(\mathrm{\ell }))c(\mathrm{bin}(a_1))\cdots c(\mathrm{bin}(a_{\mathrm{\ell }}))$, rejects if that parse fails, and also rejects if some recovered symbol $a_i$ lies outside the input alphabet ${\mathrm{\Sigma }}_s=\{1,\dots ,s\}$. From the recovered word $w=a_1\cdots a_{\mathrm{\ell }}$, the machine writes the canonical initial configuration code of $M$ on input $w$.

The machine $R$ now repeatedly invokes the one-step interpreter from [L3] on the pair consisting of the recovered code $x=<mi\; mathvariant="normal">⌜</mi>M<mi\; mathvariant="normal">⌝</mi>$ and the current configuration code. After each return, it reads the decoded state component of that configuration code. If the state is accepting, $R$ accepts. If the state is rejecting, $R$ rejects. Otherwise it feeds the returned successor configuration back into the next interpreter call.

By [L3], each iteration in step 2.1 maintains exactly the current configuration of $M$ on $w$. Therefore $R$ accepts precisely when $M$ eventually reaches an accepting configuration on $w$, rejects when $M$ reaches a rejecting configuration, and diverges when $M$ diverges. By [L1] and [L5], this means that $R$ recognizes $A_{TM}$.