= 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. Consequently
$$
\Delta_e=\delta_{(\omega\vee\eta)(x),(\omega\vee\eta)(y)}
+\delta_{(\omega\wedge\eta)(x),(\omega\wedge\eta)(y)}
-\delta_{\omega(x),\omega(y)}-\delta_{\eta(x),\eta(y)}\ge0.
$$
The <partition functions> cancel from the required ratio, leaving
$$
\frac{\mu_\Lambda(\omega\vee\eta)\mu_\Lambda(\omega\wedge\eta)}
{\mu_\Lambda(\omega)\mu_\Lambda(\eta)}
=\exp\left(\beta\sum_e\Delta_e\right)\ge1.
$$
This is the <ferromagnetic Ising lattice inequality> in binary-spin normalization, so \b[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.
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