ZPP equals RP intersection coRP

Suppose $L\in \mathrm{Z}\mathrm{P}\mathrm{P}$ and let $M$ be a zero-error expected-time machine for $L$ with $\mathbb{E}[T_x]\le p(\mathrm{\mid }x\mathrm{\mid })$. Run $M$ for exactly $2p(\mathrm{\mid }x\mathrm{\mid })$ steps and accept iff it has accepted by then. Because $M$ never errs, this truncated machine never accepts a no-instance. By [L2], $\mathbb{P}(T_x>2p(\mathrm{\mid }x\mathrm{\mid }))\le 1\mathrm{/}2$, so on yes-instances it accepts with probability at least $1\mathrm{/}2$. Thus $L\in \mathrm{R}\mathrm{P}$. By the same truncation with accept and reject swapped, also $L\in \mathrm{c}\mathrm{o}\mathrm{R}\mathrm{P}$.

Conversely, suppose $L\in \mathrm{R}\mathrm{P}\cap \mathrm{c}\mathrm{o}\mathrm{R}\mathrm{P}$. Choose an RP machine $M_{\mathrm{y}\mathrm{e}\mathrm{s}}$ for $L$ and a coRP machine $M_{\mathrm{n}\mathrm{o}}$ for $L$. On input $x$, perform independent rounds; in each round run both machines with fresh randomness. If $M_{\mathrm{y}\mathrm{e}\mathrm{s}}$ accepts, accept. If $M_{\mathrm{n}\mathrm{o}}$ rejects, reject. Otherwise repeat. By [L1], on a yes-instance the first event occurs with probability at least $1\mathrm{/}2$ and the second never occurs, while on a no-instance the second occurs with probability at least $1\mathrm{/}2$ and the first never occurs. So the algorithm is zero-error and the expected number of rounds is at most $2$.

Each round of step 1.2 uses polynomial time, so the expected runtime is polynomial. Hence $L\in \mathrm{Z}\mathrm{P}\mathrm{P}$, proving the reverse inclusion. Therefore $\mathrm{Z}\mathrm{P}\mathrm{P}=\mathrm{R}\mathrm{P}\cap \mathrm{c}\mathrm{o}\mathrm{R}\mathrm{P}$.