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If the probability of occupation at every site in a finite set depends only on its cardinality, the law is exchangeable. The product self-duality of symmetric exclusion and conservation of the finite dual particle count imply that all these occupation-product moments are unchanged by evolution. Inclusion-exclusion principle reconstructs every finite cylinder probability from them, proving invariance. This direct argument does not require a representation theorem for exchangeable laws.

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 $\rho =0$.

For finite $A\subset \mathbb{Z}$, define the occupation-product functionwith 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,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 exclusionThe 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 $x,y$. ThenSumming 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.

For the given exchangeable law $\mu$, putwhich is well-defined by the stated assumption. Integrate the product self-duality of symmetric exclusion against $\mu$, and use boundedness to interchange expectation and integration:The finite-particle process conserves its cardinality. Hence the right side is ${\mathbb{E}}^A{m}_{|A_t|}=m_{|A|}$. All occupation-product moments are therefore unchanged by evolution.

These moments determine the law, not merely its one-site marginals. For disjoint finite sets $B,C$, the inclusion-exclusion principle givesEvery term is unchanged, so every finite cylinder probability is unchanged. Such cylinder events generate the product sigma-algebra on $\{0,1{\}}^{\mathbb{Z}}$, and probability laws agreeing on them agree everywhere. ConsequentlyThus every exchangeable probability law is invariant for the symmetric exclusion process. This proof of exchangeable invariant laws of symmetric exclusion needs only the given cardinality dependence and duality; no representation theorem for exchangeable measures is required.

Give every site a distinct label and interchange labels at the edge marks defining the symmetric simple exclusion process. Each forward or backward label path makes only finitely many jumps in a bounded time interval, since its total rate is $2c$. Occupations are transported along these paths. The backward ancestors of a finite set have the law of the finite-particle symmetric exclusion process, by time reversal of the Poisson marks. This proves the product self-duality of symmetric exclusion.

On $\mathbb{Z}$, put independent rate-$c$ Poisson processes on unoriented nearest-neighbour edges, and exchange their endpoint occupations at each mark. Its generator on cylinder functions is $Lf(\eta )=c{\sum }_x[f({\eta }^{x,x+1})-f(\eta )]$. Equal occupations do not change the state, and a particle next to a vacancy jumps across at rate $c$. The stirring representation of symmetric exclusion constructs it for arbitrary initial configurations. The simple occupation-product product self-duality of symmetric exclusion requires symmetry and is not the ordinary duality of an asymmetric process.