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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.

A Steiner system $S(l,m,n)$ is an $n$-point set together with $m$-element blocks such that every $l$-element subset is contained in exactly one block. On $X=H\cup H^{\prime }$, define the following six-element blocks using the incidence extensions of $\Theta$.

Take the two blocks $H$ and $H^{\prime }$. For each duad $d$ of $H$ and each duad $d^{\prime }$ in the syntheme $\Theta (d)$, takeThese give forty-five blocks of type $(2,4)$ and forty-five of type $(4,2)$. Finally, for every corresponding partition pair $A\mid A^c$ and $B\mid B^c$, take all four unionsThere are ten partition pairs and forty blocks of type $(3,3)$. Distinct indexing data give distinct blocks within each family, and different types have different intersection sizes with $H$. Thus the total is

To prove the defining property, fix a five-element subset $Y$ and set $r=|Y\cap H|$.

If $r=5$ or $0$, only $H$ or $H^{\prime }$ can contain it. If $r=1$, write $Y\cap H=\{a\}$ and $Y\cap H^{\prime }=C^{\prime }$ with $|C^{\prime }|=4$. A containing block must have type $(2,4)$ and its omitted duad is $d^{\prime }=H^{\prime }\setminus C^{\prime }$. The total $T_a$ has exactly one syntheme containing $d^{\prime }$; the other total containing that syntheme determines a unique second point $b$. Thus the unique block is $B(\{a,b\},d^{\prime })$.

If $r=4$, write $Y\cap H=C$ and $Y\cap H^{\prime }=\{a^{\prime }\}$. A containing block must have type $(4,2)$, with omitted duad $d=H\setminus C$. Exactly one duad $d^{\prime }\in \Theta (d)$ contains $a^{\prime }$, so $(H\setminus d)\cup d^{\prime }$ is the unique block.

If $r=2$, write $Y\cap H=d$ and $Y\cap H^{\prime }=C^{\prime }$, where $|C^{\prime }|=3$. There are only two possible types. A $(2,4)$ block exists exactly when the matching $\Theta (d)$ has a duad contained in $H^{\prime }\setminus C^{\prime }$. There is then exactly one such duad, since two disjoint duads cannot fit inside a triple. For a perfect matching on two triples, either all three pairs are cross, or there is one internal pair in each triple and one cross pair. Thus a $(2,4)$ block exists precisely when $\Theta (d)$ is not entirely cross for $C^{\prime }\mid (H^{\prime }\setminus C^{\prime })$.

On the other hand, a $(3,3)$ block containing $Y$ must use the unique partition of $H$ corresponding to $C^{\prime }\mid (H^{\prime }\setminus C^{\prime })$. It exists precisely when $d$ is contained in one of that partition's triples, and is then unique. By the partition incidence rule in part (b), this happens precisely when $\Theta (d)$ is entirely cross. Hence exactly one of the two possible block types exists, always uniquely.

For $r=3$, interchange the roles of $H$ and $H^{\prime }$ and use the inverse partition incidence rule proved in part (b). More explicitly, the duad $Y\cap H^{\prime }$ either has an inverse syntheme with an internal pair in the complement of the triple $Y\cap H$, yielding a unique $(4,2)$ block, or its inverse syntheme is entirely cross, yielding the unique $(3,3)$ block. These alternatives are exclusive and exhaustive for the same matching-on-two-triples reason.

Every split $r=0,\dots ,5$ therefore gives exactly one containing block. We have constructedAs an independent count, each block contains six five-subsets and $132\cdot 6=792=\left(\genfrac{}{}{0ex}{}{12}{5}\right)$. The case proof establishes uniqueness and existence; the count alone would not have done so.

The complete graph on six points has fifteen edges and fifteen perfect matchings. Its six one-factorizations organize a dual incidence geometry exchanging points with factorizations and edges with matchings. The classical names are duads, synthemes and totals of synthemes. This geometry supports duad-syntheme duality on six points and a construction of the small Witt design.