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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.