 Solution

For the coordinatewise lattice operations $\omega \vee \eta$ and $\omega \wedge \eta$, Holley's condition isThis sufficient condition implies ${\mu }_1\ge {\mu }_2$ by the Holley inequality. It is often stated first for strictly positive probability weights, as will hold for the random-cluster application, but the displayed condition is also sufficient for nonnegative weights.

For clarity about the zero-weight case, apply the four functions theorem to an increasing event $A$, with the first two functions ${\mu }_1{1}_{A^c}$ and ${\mu }_2{1}_A$, and the last two ${\mu }_1{1}_A$ and ${\mu }_2{1}_{A^c}$. If $\omega \notin A$ and $\eta \in A$, then $\omega \vee \eta \in A$ and $\omega \wedge \eta \notin A$. The pointwise hypothesis is exactly the displayed cross-lattice inequality. Its summed conclusion giveswhich rearranges to ${\mu }_1(A)\ge {\mu }_2(A)$. This establishes the same statement without a positivity-of-support assumption.