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Write $T_a=\Theta (a)$ for the total of $H^{\prime }$ assigned to a point $a\in H$. The duad-syntheme duality on six points is constructed entirely from incidence, as follows.

For a duad $d=\{a,b\}$, define $\Theta (d)$ to be the unique syntheme common to $T_a$ and $T_b$. This is a bijection between the fifteen duads and the fifteen synthemes of $H^{\prime }$, by the last incidence count in part (a).

For a duad $d^{\prime }$ of $H^{\prime }$, the three synthemes containing $d^{\prime }$ each belong to two totals. Each total contains exactly one of these synthemes, since its five matchings cover every duad exactly once. Their three pairs of totals therefore partition all six totals. Pulling these pairs back to $H$ gives a syntheme $s$. Different $d^{\prime }$ give different $s$, since two distinct synthemes of $H^{\prime }$ containing $d^{\prime }$ have intersection exactly that duad. There are fifteen of each, so this construction is bijective. Define $\Theta (s)=d^{\prime }$ by its inverse. Equivalently, the three synthemes $\Theta (d)$ for $d\in s$ have common duad $\Theta (s)$. In particular,

For a total $T$ of $H$, map its five synthemes to five duads of $H^{\prime }$. Any two of the original synthemes are disjoint. Their image duads must intersect: if two image duads were disjoint, the unique syntheme containing both in $H^{\prime }$ would give a common duad in the original two synthemes through the pairs-of-totals construction. Conversely intersecting duads cannot lie together in a syntheme and give disjoint original synthemes. Five distinct pairwise-intersecting edges must form the full star at one point. Indeed two edges meeting at a point either force every other edge through that point or leave only the three edges of a triangle, which cannot contain five edges. Define $\Theta (T)$ to be the star's center. Distinct totals give distinct stars; since there are six of each, this is a bijection to the points of $H^{\prime }$.

It remains to extend to unordered three-versus-three partitions. Start with a partition $B\mid B^c$ of $H^{\prime }$. Its six cross synthemes are the perfect matchings between the two triples, parametrized by permutations in $S_3$. Two of these are disjoint exactly when the quotient of their permutations is a three-cycle. Hence the six cross synthemes split into two classes of three: within a class any two are disjoint, and between classes any pair shares a duad. This unordered division into two classes is independent of the chosen orderings of the triples.

Every total has exactly two cross synthemes. To see this, any syntheme has either one or three cross duads. If a total has $h$ all-cross synthemes, it covers $3h+(5-h)$ cross duads; the whole complete graph has nine, so $h=2$. Its two cross synthemes belong to the same parity class. Conversely any pair in one class extends to a unique total. Thus the six totals split into two triples, the three totals arising from pairs in each parity class. Pulling them back through the original point-total bijection defines a partition $A\mid A^c$ of $H$.

Under the duad mapping, its six internal duads become exactly the six cross synthemes of $B\mid B^c$: a syntheme in a parity class belongs to the two totals formed by pairing it with the other two members. These incidences give the three edges of a triangle on each triple of totals. Therefore the partition is characterized byThis correspondence is injective: the six cross synthemes determine all nine cross duads of $K_{3,3}$, whose bipartition is unique up to interchange. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$ partitions on each side, so it is bijective. Define $\Theta (A\mid A^c)=B\mid B^c$ by the inverse of the construction above.

The inverse incidence rule is also useful. If a duad $d^{\prime }$ is internal to $B\mid B^c$, none of the cross synthemes contains it, so its inverse syntheme has no internal duad of $A\mid A^c$ and is entirely cross. If $d^{\prime }$ is cross, exactly two cross synthemes contain it, so the inverse syntheme has two internal duads and one cross duad. Thus internal duads and cross synthemes exchange roles in both directions. All the extensions are natural: they use intersections and incidence, with no auxiliary ordering left in the answer.