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Order $\Omega =\{0,1{\}}^E$ coordinatewise. The statement ${\mu }_1{\ge }_{\mathrm{st}}{\mu }_2$ meansfor every real-valued order-preserving function $f$; equivalently, ${\mu }_1(A)\ge {\mu }_2(A)$ for every increasing event $A$. This is stochastic domination of probability measures. For strictly positive laws, the Holley condition is the sufficient conditionHere join and meet are coordinatewise maximum and minimum. The Holley inequality states that this condition implies the stated stochastic domination.

For a single positive law, the FKG lattice condition isWe prove that it gives positive association of random variables, namely ${\mathrm{Cov}}_{\mu }(f,g)\ge 0$ for all increasing $f,g$. Fix such a $g$, and for $t\ge 0$ form the strictly positive tilted lawThe lattice condition and $g(\omega \vee \eta )\ge g(\omega )$ implyThus the Holley condition applies to ${\mu }_t,\mu$, and ${\mathbb{E}}_{{\mu }_t}f\ge {\mathbb{E}}_{\mu }f$. Equality holds at $t=0$, so the right derivative there is nonnegative. Differentiating the finite sums givesHencewhich proves the requested positive association of random variables. Neither $f$ nor $g$ needs to be nonnegative; the finite space makes all differentiations justified. This is the exponential-tilt proof of positive association.

On the coordinatewise ordered configuration space, ${\mu }_1$ stochastically dominates ${\mu }_2$ when $\int fd{\mu }_1\ge \int fd{\mu }_2$ for every increasing real function $f$. Equivalently, ${\mu }_1(A)\ge {\mu }_2(A)$ for every increasing event. The Holley condition is the sufficient cross-lattice inequalityHere join and meet are coordinatewise maximum and minimum. This condition implies ${\mu }_1{\ge }_{\mathrm{st}}{\mu }_2$ by the Holley inequality; it is not asserted to be necessary. It can be checked on unnormalized weights, since their normalization constants cancel.

Use the weightsFor the upper bound, take ${\mu }_1={\varphi }_{p,1}$ and ${\mu }_2={\varphi }_{p,q}$. Since open-edge count is modular, the Bernoulli factors cancel in the Holley ratio, leavingThe last inequality uses $q\ge 1$ and the fact that removing edges cannot reduce the number of components. Thus ${\varphi }_{p,q}{\le }_{\mathrm{st}}{\varphi }_{p,1}$.

For the lower bound, let $p^{\prime }=p/[p+q(1-p)]$. Its odds satisfySet $a=o(\omega \vee \eta )-o(\omega )=o(\eta )-o(\omega \wedge \eta )$, the number of edges added to $\omega$ in the union. For ${\mu }_1={\varphi }_{p,q}$ and ${\mu }_2={\varphi }_{p^{\prime },1}$, the corresponding ratio isAdding one edge either merges two components or leaves the count unchanged. Therefore adding $a$ edges reduces $k$ by at most $a$, proving the exponent nonnegative. This also proves the hinted monotonicity of $k(\omega )+o(\omega )$ rather than assuming it. The Holley condition now proves both Bernoulli bounds for the random-cluster model:

The same argument holds with wired boundary conditions: identify the wired boundary vertices first and count components on that quotient graph. Taking finite-volume wired limits preserves every increasing cylinder-event inequality. Approximating $\{0\leftrightarrow \infty \}$ by the decreasing events $\{0\leftrightarrow \partial [-n,n{]}^d\}$ consequently givesLet $a=p_c(1)$. For $p<a$, the upper bound makes the wired percolation probability zero, so $p_c(q)\ge a$. If $a<1$ and $p^{\prime }>a$, the lower bound makes it positive. Solving this strict inequality,Taking infima provesFor $a=1$, the lower bound and the trivial upper bound $p_c(q)\le 1$ give the same conclusion. No assumption about percolation at the critical parameter is required.

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.

Identify a configuration with its set $S$ of open edges. Let $k(S)$ count the connected components of a graph on the full graph vertex set, including isolated graph vertices. The random-cluster measure isFor $p\in (0,1)$ and finite $q\ge 1$ all its weights are positive. We compare $1\le q_1\le q_2$, taking ${\mu }_1={\mu }_{p,q_1}$ and ${\mu }_2={\mu }_{p,q_2}$.

The key fact is supermodularity of graph component count:To prove it, orient the graph edges arbitrarily and let $U_S$ be the real span of their incidence vectors for edges in $S$. Its dimension of a vector space is $r(S)=|V|-k(S)$: on each connected component of a graph these vectors span the vectors whose coordinates sum to zero. Since $U_{S\cup T}=U_S+U_T$ and $U_{S\cap T}\subset U_S\cap U_T$, the dimension formula for a sum of subspaces giveswhich is exactly the component-count inequality.

Put $a=k(S)-k(S\cup T)\ge 0$ and $b=k(S\cap T)-k(T)\ge 0$. The preceding inequality gives $b\ge a$. In the ratio of the two sides of Holley's condition, the normalization constants and Bernoulli edge factors cancel, because $|S\cup T|+|S\cap T|=|S|+|T|$. The ratio is thereforeBy the Holley inequality, ${\mu }_{p,q_1}$ stochastically dominates ${\mu }_{p,q_2}$. Thus increasing cluster weight suppresses every increasing event:The random-cluster single-edge conditional probability confirms the mechanism: it is $p$ when the endpoints are already connected, and $p/(p+q(1-p))$ otherwise. Larger $q$ favours configurations with more separate components. The condition $q\ge 1$ enters explicitly in the last factor $q_2^{b-a}$.