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Let the common vertex degree be $r>0$. For each unordered edge place an independent Poisson process of rate $1/r$, and at a mark interchange the two endpoint occupations. If exactly one endpoint is occupied, this moves its particle across that edge at rate $1/r$; if both or neither are occupied, the occupied set stays unchanged. Thus each particle proposes jumps at total rate one, uniformly among its neighbours, with occupied-target proposals suppressed. This is the stated symmetric simple exclusion process. An edgeless graph has the identity dynamics and both conclusions below hold trivially.

In the stirring representation of symmetric exclusion, put a distinct label at every vertex and interchange labels at every edge mark, even when the occupations agree. There are finitely many marks on every finite time interval. Their chronological composition is a random permutation ${\pi }_t$ of the vertex set, and simultaneously for every initial set,Each edge swap is its own inverse. Reflecting the edge-mark times in $[0,t]$ preserves their joint Poisson process law and reverses the order of the swaps. Hence ${\pi }_t^{-1}$ and ${\pi }_t$ have the same distribution. Pathwise,Taking probabilities and then using the inverse-permutation distribution proves the product self-duality of symmetric exclusion in set form:No assumption about disjointness or equal cardinality of $A$ and $B$ is required; impossible inclusions have probability zero on both sides.

For invariance, independently choose the initial occupied set $A$ under the product measure ${\mu }_{\rho }$. The occupations are independent identically distributed Bernoulli distribution variables, so permuting their sites by any deterministic permutation preserves their joint law. Conditional on the independent stirring permutation ${\pi }_t$, the output therefore still has law ${\mu }_{\rho }$. Averaging proves the Bernoulli invariant laws of finite symmetric exclusion:There is also a direct check from the duality. For every $B\subseteq V$,because the exclusion process conserves particle number. These are exactly the joint occupation moments of ${\mu }_{\rho }$; on a finite set they determine the law by the inclusion-exclusion principle. The endpoint densities give the absorbing empty and fully occupied configurations, and the empty-set moment is one also at $\rho =0$.