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Group the martini lattice into independent three-terminal cells. A cell has terminals $A,B,C$, an inner triangle with vertices $a,b,c$, and the three spokes $Aa,Bb,Cc$. All six bonds are independently open with parameter $p$. The cells arise from the vertices in the decorated class of the honeycomb lattice; they have disjoint bond sets and meet only at terminals. The terminal positions form a triangular lattice, with cells occupying one orientation of its triangular faces.

A cell induces one of five connection partitions of its terminals. Let $P_3(p)$ be the probability that all three terminals are connected, and $P_0(p)$ the probability that no pair is connected. Each of the three remaining pair partitions has equal probability by symmetry. The three-terminal cell partition duality gives a particularly useful exact criterion. Draw dual terminals on the three intervening boundary arcs. The dual cell has a triangle of these three terminals and a central vertex joined to all three: its outer bonds cross the primal spokes, and its inner spokes cross the primal triangle bonds. A primal connection separates the corresponding dual boundary arcs, and conversely a dual connection separates primal terminals. Consequently duality swaps the all-connected and all-separate partitions, and permutes the three pair partitions. For example, if $A,B$ connect and $C$ is isolated, the dual terminal between $A,B$ is isolated while the other two dual terminals connect. It follows that $P_3=P_0$ makes the entire symmetric cell partition law invariant under this duality.

The triangular arrangement of cells is itself self-dual: the dual terminals lie in the complementary triangular gaps, and reflection and translation identify their cell arrangement with the original one. The induced partitions of distinct cells are independent in both models. At a parameter satisfying $P_3=P_0$, the primal model at $p$ and the complementary dual bond percolation model at $1-p$ therefore have the same coarse connection law. Replacing a cell by its terminal partition preserves infinite connectivity, since the cells have uniformly bounded size. This gives a rigorous link between the finite calculation and the infinite model, rather than assuming that a symmetric-looking cell is critical.

For all three terminals to connect, all three spokes must be open and the inner triangle must be connected. The latter event requires two or three of its edges. ThereforeTo compute $P_0$, condition on the number of open spokes. Zero or one open spoke cannot connect a pair of terminals. If exactly two spokes are open, the two attached inner vertices must not connect: their direct triangle bond is closed, and the two-edge route through the third inner vertex is not fully open. These independent requirements have probability $(1-p)(1-p^2)$. If all three spokes are open, all three triangle bonds must be closed. HenceSubtracting and factoring givesThe second factor equals $1+p^2(p-1)(p-2)$ and is at least $1$ on $[0,1]$. Thus the unique equality parameter is $p_0=1/\sqrt{2}$.

Finally part (i) proves $c+c^{\ast }=1$. If $p_0>c$, the primal model percolates while $1-p_0<c^{\ast }$ makes the dual model subcritical, contradicting their identical coarse connection laws. If $p_0<c$, the same contradiction arises with the roles reversed. Therefore the critical probability isThe local partition transformation is the mechanism underlying this exact value; see Ziff's original cell–dual-cell paper.

Let $G=\Lambda$ be the martini lattice and $G^{\ast }$ its planar dual graph. Write $c=p_c^b(G)$ and $c^{\ast }=p_c^b(G^{\ast })$. Both are periodic, connected, bounded-degree plane graphs with bounded face sizes. Dual bond percolation declares a dual bond open exactly when its crossing primal bond is closed, giving independent dual parameter $1-p$.

First, $c+c^{\ast }\le 1$. Suppose instead that the sum exceeds $1$, and choose $1-c^{\ast }<p<c$. Both the primal model at $p$ and the dual model at $1-p$ are strictly subcritical. Sharpness of the percolation transition gives exponential connection-radius decay on each periodic graph, uniformly over its finitely many vertex types. A crossing of a rectangle of side lengths proportional to $R$ starts at one of $O(R)$ boundary vertices and travels distance of order $R$. A union bound therefore shows that each crossing probability tends to zero. But planar duality for rectangle crossings says that exactly one of an open primal left-to-right crossing and an open dual top-to-bottom crossing occurs, using the complementary boundary convention. Their probabilities sum to $1$. This contradiction proves the first inequality.

For the opposite inequality, we need noncoexistence of primal and dual infinite percolation clusters. The four-fold version in Q3(iii) cannot be assumed without change: this decorated honeycomb lattice has three-fold rotation symmetry. Here is the necessary three-fold symmetric primal-dual noncoexistence argument.

Assume both models percolate at complementary parameters $p,1-p$. For a large disc, the probability of having no exterior infinite arm is at most $\epsilon$ in each model. Divide its boundary into six arcs by splitting each $120$-degree sector at the same relative angle. In the primal model let $a,b$ be the failure probabilities for the two alternating arc types. Rotation symmetry and the FKG inequality giveSlide the splitting point across one sector. Initially the first arc is empty; finally the second is empty. A jump occurs only when an arc gains or loses an edge-crossing point. An arm using that point requires its crossing bond to be open. The FKG inequality therefore ensures that a failure probability changes by at most a factor $(1-p{)}^{-1}$ at a single jump: for a newly added arm event $L_x$, $\mathbb{P}(L_x^c)\ge 1-p$ and $\mathbb{P}(L^c\cap L_x^c)\ge (1-p)\mathbb{P}(L^c)$. At the first reversal of the order of $a,b$, their ratio is consequently at most $(1-p{)}^{-2}$. At that splitting point,Thus every one of the six primal arcs has an exterior arm with probability tending to $1$. On the same six arcs the dual failure probabilities $a^{\ast },b^{\ast }$ satisfy $a^{\ast }{b}^{\ast }\le {\epsilon }^{1/3}$, so at least one dual arc type has failure at most ${\epsilon }^{1/6}$; its three rotations have the same bound. Take two of those dual arcs separated by one arc, and the two intervening primal arcs. A union bound gives four alternating primal/dual infinite exterior arms with positive probability when $\epsilon$ is sufficiently small. The separation and finite-closure argument of Q3(iii), together with percolation uniqueness on an amenable quasi-transitive graph, gives a contradiction. This proves noncoexistence. The arc balancing is the extra step needed for three-fold symmetry; it is also developed in Bollobás and Riordan's original paper on dual lattices.

If $c+c^{\ast }<1$, choose $c<p<1-c^{\ast }$. Both complementary models would then percolate, contradicting noncoexistence. Hence $c+c^{\ast }\ge 1$. Combining the two inequalities proves the dual threshold relation:

For independent bond percolation on a periodic plane lattice with three-fold rotation symmetry, the primal model at $p$ and the dual bond percolation model at $1-p$ cannot both percolate. To adapt Zhang's argument, divide a large boundary circle into six arcs, repeating the division in each $120$-degree sector. Vary the division point until the failure probabilities $a,b$ of the two alternating primal arc types are comparable within a factor $(1-p{)}^{-2}$. Moving past a single crossing edge changes a failure probability by at most a factor $(1-p{)}^{-1}$, by the FKG inequality. If the probability of no primal exterior arm is at most $\epsilon$, then $a^3{b}^3\le \epsilon$ and both failures are at most $(1-p{)}^{-1}{\epsilon }^{1/6}$. For the dual model, at least one arc type has failure at most ${\epsilon }^{1/6}$; its three rotations have the same bound. Four alternating primal/dual arms now occur with positive probability, contradicting the alternating exterior-arm separation lemma and uniqueness after a finite modification. This extends the four-fold argument to the martini lattice.

A planar finite network with three boundary terminals induces one of five connection partitions: all three connected, all three separate, or one connected pair and an isolated third terminal. Put three dual terminals on the intervening boundary arcs. Passing to the planar dual graph interchanges the all-connected and all-separate partitions and permutes the three pair partitions. If the network is symmetric under all terminal permutations, equality $P_3=P_0$ of the first two probabilities makes its whole partition law dual-invariant. Independent cells on a self-dual triangular terminal arrangement then give identical primal and dual coarse connection laws. Infinite connectivity is preserved because cells are uniformly finite. For a martini lattice cell with six independent bonds of common parameter $p$, $P_3=3p^5-2p^6$ and $P_3-P_0=(2p^2-1)(p^4-3p^3+2p^2+1)$. The unique zero in $[0,1]$ is $1/\sqrt{2}$.