A universal Turing machine exists for the chosen coding

Build a deterministic multitape machine $W$ that, on input $⟨<mi\; mathvariant="normal">⌜</mi>M<mi\; mathvariant="normal">⌝</mi>,\widehat{w}⟩$, first decodes the coded machine $M=(m,s,n,\tau )$ from the first component and the input word $w\in {\mathrm{\Sigma }}_s^{\ast }$ from the second component block-by-block. It then copies $<mi\; mathvariant="normal">⌜</mi>M<mi\; mathvariant="normal">⌝</mi>$ to one work tape, constructs the encoded initial configuration of $M$ on input $w$ on a second work tape, and reserves a third work tape for running the one-step interpreter from [L3].

The machine $W$ now iterates the interpreter from [L3]. Before each cycle it checks whether the current code represents an accepting or rejecting configuration and stops the iteration if so; otherwise it replaces that code by the one-step successor returned by the interpreter. This state check, rather than equality of successive codes, detects halting.

Induction on the number of interpreter cycles shows that the maintained configuration code is always the code of the current configuration of $M$ on input $w$. Therefore, if $M$ halts with output $u$, then $W$ eventually reaches the halting configuration code, decodes the represented output word $u$, writes its binary code $\widehat{u}$ in its own output convention, and halts. If $M$ diverges, the iteration never reaches a halting configuration code, so $W$ diverges as well. Thus $W$ has exactly the behavior required by [L1] and [L2].

Finally, [L4] converts the deterministic multitape interpreter $W$ into an equivalent deterministic one-tape machine. By step 3.1, that one-tape machine is universal for the chosen coding.