 Solution

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.