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Work edge by edge. Let $\omega ,\eta$ have two-bit restrictions to an edge. If these restrictions are comparable, taking their meet in a lattice and join in a lattice only reorders the two pairs, so their total equality indicators do not change. If they are incomparable, they are $01$ and $10$, while their meet in a lattice and join in a lattice are $00$ and $11$. In that case the sum of equality indicators increases from zero to two. ConsequentlyThe partition functions cancel from the required ratio, leavingThis is the ferromagnetic Ising lattice inequality in binary-spin normalization, so the measure satisfies the FKG lattice condition and is positively associated. If $\sigma (x)=2\omega (x)-1$, then ${\delta }_{\omega (x),\omega (y)}=(1+\sigma (x)\sigma (y))/2$: the conventional $\{-1,1\}$ coupling is $\beta /2$, which explains the factor two in the edge calculation.

The law is a ferromagnetic Ising model with unit couplings and no field. Its masses are strictly positive. Put $H(\sigma )={\sum }_{\{x,y\}\in E}{\sigma }_x{\sigma }_y$, so $\pi (\sigma )=Z^{-1}{e}^{H(\sigma )}$. The order is coordinatewise with $-1<1$, which identifies the configuration space with a Boolean lattice.

Use the permitted reduction to pairs that disagree at no more than two graph vertices. If such a pair is comparable, its join and meet are the two original configurations, so the lattice inequality is equality. In the incomparable case, let the two differing graph vertices be $x,y$, with one configuration taking $(1,-1)$ there and the other $(-1,1)$. Their join takes $(1,1)$ and their meet $(-1,-1)$.

Every edge not joining $x$ to $y$ cancels ina fixed edge has the same contribution throughout, while an edge from a changed graph vertex to a fixed graph vertex has the same sum of its two endpoint-spin products before and after. An edge between $x$ and $y$ contributes $1+1-(-1)-(-1)=4$. Thus the difference is nonnegative, whether or not the two changed graph vertices are adjacent. Exponentiation, with the normalizing constants cancelling, proves the FKG lattice condition for these pairs, and hence for all pairs by the given reduction.

Equivalently one can see the full ferromagnetic Ising lattice inequality directly. Let $U=\{x:{\sigma }_x=1,{\tau }_x=-1\}$ and $W=\{x:{\sigma }_x=-1,{\tau }_x=1\}$. Exactly the edges between $U$ and $W$ contribute, each by four. ThereforeThe preceding part now givesso the measure is positively associated. The sign of the coupling is essential to this argument: positive couplings favor agreement and give the nonnegative lattice difference.