“Boltzmann acceptance with energy $E$”. This involves accepting the move ($\Gamma'$ is the new state) with probability $M(x)$ and otherwise rejecting the move ($\Gamma$ is the new state). Here $M(x)$ is a function satisfying $0 < M(x) \le 1$ and $M(x)/M(-x) = \exp(-x)$ with $x
= \beta(E(\Gamma') - E(\Gamma))$. Usually we take $M(x) =
\min(1,\exp(-x))$; however other choices, e.g., the Fermi function, $M(x) = 1/(1+\exp(x))$, are possible [@bennett76].

Consider an $E_1$ move from $\Gamma$ to $\Gamma'$ followed by an Boltzmann acceptance using $E_2$. This compound move is an $(E_1 +
E_2)$ move. The proof follows as a special case of the “multiple time-step” (MTS) method [@hetenyi02 §II] or, alternatively, as a special case of early rejection [@frenkel02 §14.3.2]. If the $E_1$ move was already a “rejected” move, i.e., $\Gamma'=\Gamma$, then the Boltzmann test involving $E_2$ automatically “succeeds” ($M(0) =
1$). Thus $E_2$ does not need to be evaluated in this case.

These results allow us to generalize the MTS method by splitting the energy into $m$ terms (instead of just two), $$E(\Gamma) = \sum_{l=1}^m E_l(\Gamma).$$ The method is defined recursively as follows: a level-$0$ move is defined to be a zero move; a level-$l$ move, with $l>0$, is defined to 