First suppose \(V=A\sqcup S\sqcup B\). Separation implies
\(Q_{AB}=0\). With \(\phi _S=s\) fixed, the quadratic energy in \(a,b\)
is the sum
\[\begin {aligned}
&\frac 12a^TQ_{AA}a+a^TQ_{AS}s\\
&\qquad +\frac 12b^TQ_{BB}b+b^TQ_{BS}s
+\frac 12s^TQ_{SS}s.
\end {aligned}\]
The last term is constant. The two remaining positive Gaussian
factors normalize separately, proving independence at every \(s\).
Equivalently, the conditional precision is
\(\operatorname {diag}(Q_{AA},Q_{BB})\); its inverse has zero cross block.
The square completion is the coordinate case of
Proposition [fdip-0060].
For smaller disjoint \(A,B,S\), no connected component of the graph
on \(V\setminus S\) meets both \(A\) and \(B\). Assign its components
to two unions \(A',B'\) with \(A\subseteq A'\), \(B\subseteq B'\) and
\(A'\sqcup B'=V\setminus S\); components meeting neither may be
assigned arbitrarily. The preceding argument gives conditional
independence of \(\phi _{A'}\) and \(\phi _{B'}\) given \(s\). Taking
coordinate marginals gives the asserted independence of \(\phi _A\)
and \(\phi _B\).
For the pairwise statement, fixing all other coordinates leaves
conditional precision and cross covariance
\[\begin {pmatrix}Q_{aa}&Q_{ab}\\Q_{ab}&Q_{bb}\end {pmatrix},
\qquad
\operatorname {Cov}(\phi _a,\phi _b\mid \phi _{V\setminus \{a,b\}})
=\frac {-Q_{ab}}{Q_{aa}Q_{bb}-Q_{ab}^2}.\]
The denominator is positive. Independence implies zero covariance,
and zero \(Q_{ab}\) makes the conditional density a product. This
proves both directions.