PromiseP is contained in EQP, EQP in ZQP, and ZQP in BQP

Let $\mathrm{\Pi }=(Y,N)\in \mathrm{P}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{P}$ and choose its deterministic polynomial-time decider $M$ from [F1]. The Boolean circuit family of [F2] computes $M$; [F3] gives a polynomial-time full-list generator, and [F4] compiles each list into a QC14 family whose one-bit target is $1$ exactly when $M$ accepts. The two polynomial-time stages compose to one fixed deterministic multitape generator $G$ on input $1^n$: it emits the Boolean list and compiles that list to the complete QC14 description. Its intermediate list length and total running time are polynomial in $n$. Convert $G$ to one tape by storing the finitely many simulated tape contents in aligned tracks of a finite product alphabet, with one head marker per tape. During a computation of at most $P(n)$ steps, each track uses at most $n+P(n)+O(1)$ cells; a constant number of sweeps across this interval simulates each transition, so the one-tape time is $O(P(n)(n+P(n)))$. Write the generated circuit description on an output track starting at cell $0$; at halt, a cleanup sweep replaces each product symbol by its output-track component, leaving the exact binary description followed by blanks on the final tape. This is one fixed deterministic one-tape polynomial-time generator as required by [F5]. Use the target bit as the acceptance predicate. For every $x\in Y$ the acceptance probability is $1$, and for every $x\in N$ it is $0$, so $\mathrm{\Pi }\in \mathrm{E}\mathrm{Q}\mathrm{P}$.

If $\mathrm{\Pi }\in \mathrm{E}\mathrm{Q}\mathrm{P}$, take its witnessing family and define $D_n(x,z)=\mathrm{Y}\mathrm{E}\mathrm{S}$ when its acceptance predicate accepts and $D_n(x,z)=\mathrm{N}\mathrm{O}$ otherwise. This is a total polynomial-time decoder. The exact probabilities in [F5] make the wrong definite label have probability zero and the correct definite label have probability one on each promised side; therefore the same family witnesses $\mathrm{\Pi }\in \mathrm{Z}\mathrm{Q}\mathrm{P}$.

If $\mathrm{\Pi }\in \mathrm{Z}\mathrm{Q}\mathrm{P}$, keep its full-list family and define the BQP acceptance predicate to accept exactly when $D_n(x,z)=\mathrm{Y}\mathrm{E}\mathrm{S}$. On $Y$ this event has probability at least $2\mathrm{/}3$. On $N$ it has probability zero, since a wrong definite YES label is forbidden. By [F8] these event probabilities are normalized, and [F7] gives $\mathrm{\Pi }\in \mathrm{B}\mathrm{Q}\mathrm{P}$.