Take
A={N↔S} and let
s=1−p↓0 with
Then
sq→∞,
s2q→0, and
s4q3=s−1/2→∞. Part
d gives
PFKp,q(A)→0.
On
a four-cycle,
A□A occurs exactly when all four edges are open, because the two
length-two paths from
N to
S are the only disjoint witnesses. Hence
PFKp,q(A□A∣A)=p4q+4p3sq+2p2s2q2p4q⟶1.
Choosing
s sufficiently small gives the two numerical
inequalities in the question.