= Solution
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 in
$$
H(\sigma\vee\tau)+H(\sigma\wedge\tau)-H(\sigma)-H(\tau):
$$
a 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. Therefore
$$
\frac{\pi(\sigma\vee\tau)\pi(\sigma\wedge\tau)}{\pi(\sigma)\pi(\tau)}
=\exp\bigl(4\,\#\{\text{edges between }U\text{ and }W\}\bigr)\ge1.
$$
The preceding part now gives
$$
\boxed{\pi(fg)\ge\pi(f)\pi(g)\quad\text{for every pair of increasing functions }f,g,}
$$
so 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.
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