Proposition: Complete two-slot streaming decoder

Receive carriers $1,2,3$ in that order. Retain carrier $1$ until carrier $2$ arrives, apply $B_{12}$, destructively read carriers $1,2$ as $t,s$, then receive carrier $3$, apply $C_{ts}{R_3}^{\mathrm{rot}}$, and return it. This is a complete CPTP decoder and satisfies

Its peak is two live quantum carriers, excluding the independent validation reference and the incoming carriers before admission to the decoder apparatus.

Proof. Conjugation gives

and hence

The third-qubit rotation maps $Z\mapsto X$, $Y\mapsto Z$, $X\mapsto Y$.

Thus, before the classical correction, the branch $t,s$ has probability $1/4$ and normalized output

Conjugation by $C_{ts}$ removes all three signs. Each of the four unnormalized branch maps is therefore $\rho \mapsto \rho /4$ on the encoded input. Summing the branches returns $\rho$. Linearity proves the equality on every matrix and every referenced input.

On arbitrary physical triple inputs, the program consists of unitaries, a complete two-qubit measurement, conditional unitaries, and quantum return. It is therefore CPTP without postselection. At most carriers $1,2$ are simultaneously live; after their consumption only carrier $3$ is live.