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Choosing an intermediate radius uniformly and exploring the open cluster meeting that sphere gives a decision tree for a one-arm event. Translation and a union bound control each revealment by a constant times the average of the one-arm probabilities over the possible radii.

The one-arm event $A_R(a)$ depends on the field values in the finite ball $B_R$. Replace $V_x$ by $U_x+a$ for $x\in B_R$. In the new variables the event has threshold zero, while the product normal distribution density is ${\prod }_{x\in B_R}\phi (U_x+a)$. Differentiating this finite-dimensional integral under the integral sign gives the Gaussian shift identityThis is also an instance of Gaussian integration by parts. In particular, the asserted inequality holds, in fact with equality.

Apply the OSSS inequality to the independent coordinates $(V_x,W_x)$ and to the indicator of the one-arm event $A_R(a)$. Use the randomized OSSS exploration of a one-arm event: choose $k$ uniformly from $\{1,\dots ,R\}$ and reveal the variables needed to explore the superlevel cluster meeting $\partial B_k$. A coordinate can be revealed only if a nearby vertex has an open connection over the relevant distance. Translation invariance, finite-range dependence and the union bound therefore give the revealment estimatefor both kinds of coordinates, after enlarging the explored neighbourhood by a distance depending only on $d$.

For an increasing Gaussian threshold event, the resampling influence of $V_x$ is bounded by a universal constant times $\mathbb{E}[V_x{1}_{A_R(a)}]$. The influence of $W_x$ is bounded by the sum of the corresponding $V$ influences at the $2d$ neighbours of $x$: indeed Gaussian integration by parts givesfirst for smooth increasing approximations and then by a limit. Consequently the OSSS bound becomesPart (c) identifies the final sum with $-{\theta }_R^{\prime }(a)$. Dividing and putting $c=C_d^{-1}>0$ proves