= Solution
For finite $A\subset\mathbb Z$, define the occupation-product function
$$
D(\eta,A)=\prod_{x\in A}\eta(x)=1_{\{\eta\equiv1\text{ on }A\}},
$$
with empty product one. Use the <stirring representation of symmetric exclusion> on the time interval $[0,t]$. Trace the labels at the sites of $A$ backwards to time zero, and call their set of ancestors $B_t$. Labels remain distinct under swaps, so $|B_t|=|A|$. Pathwise,
$$
D(\eta_t,A)=D(\eta,B_t).
$$
Reverse time in the Poisson clocks. Their joint law is unchanged, and each interchange is its own inverse. Thus the backward set $B_t$ has the same law as the finite-particle symmetric exclusion set $A_t$ started from $A$. Taking <expectations> gives the <product self-duality of symmetric exclusion>
$$
\boxed{\mathbb P^\eta(\eta_t\equiv1\text{ on }A)=\mathbb P^A(\eta\equiv1\text{ on }A_t).}
$$
The argument holds for every deterministic initial configuration $\eta$ and every <finite set> $A$, including the empty set, and makes no stationarity assumption.
The generator identity confirms the same mechanism. For any <edge> $xy$, let $A^{xy}$ interchange membership of $x,y$. Then
$$
D(\eta^{xy},A)=D(\eta,A^{xy}).
$$
Summing over <edges> gives $L_\eta D=L_A D$. In the finite-particle generator only <edges> joining an occupied to a vacant site contribute; in the occupation generator the same transpositions act on the other argument. Symmetry supplies the same rate in both operations.
To see why this qualification matters, suppose right jumps have rate $a$ and left jumps rate $b\ne a$. Take $A=\{0\}$ and an initial configuration whose only particle is at $1$. The derivative at zero of the left-hand <probability> is $b$, since that particle must jump left into zero. The derivative of the right-hand <probability> for the same asymmetric finite-particle process is $a$, since its particle must jump right from zero to the initially occupied site $1$. Thus the printed same-process identity would be false in that asymmetric interpretation. Under the symmetric convention established in the preceding part, the proof is complete.
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