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For an $r$-uniform set family $\mathcal{F}$, its lower shadow is the set family of all $(r-1)$-sets contained in a member. Write a positive integer $m$ in its unique binomial representationThe Kruskal-Katona theorem states thatFor $m=0$ the bound is zero. Equality is attained by the first $m$ $r$-sets in colexicographic order. In that order $A$ precedes $B$ when the largest element of their symmetric difference belongs to $B$.

Here is a complete UV-compression proof of the Kruskal-Katona theorem. For disjoint equal-sized sets $U,V$, put $C(A)=(A\setminus V)\cup U$ when $A$ contains every element of $V$ and no element of $U$, and put $C(A)=A$ otherwise. The UV-compression of a set family replaces an eligible $A$ only if its image is absent. It preserves cardinality: a retained collision keeps both the original set and its already present image.

We first prove the required Shadow lemma for UV-compressions. Suppose that, for each $x\in U$, a $y\in V$ can be chosen so the set family is fixed by $C_{U\setminus \{x\},V\setminus \{y\}}$. We claimConsider a lower shadow member obtained by deleting $z$ from a moved set $A^{\prime }=(A\setminus V)\cup U$. If $z\notin U$, it is the compressed image of $A\setminus \{z\}$, an old lower shadow member. If $z=x\in U$, the smaller compression of $A$ produces $A^{\prime }\setminus \{x\}\cup \{y\}$, which is in $\mathcal{F}$ by the assumed invariance. Thus $A^{\prime }\setminus \{x\}$ already belongs to the old lower shadow; it is not eligible to move since it lacks $x$.

It remains to consider a lower shadow member $B$ from a retained original member $A=B\cup \{z\}$. If $B$ is not eligible to move, it stays in the compressed lower shadow. If it is eligible, let $B^{\prime }=(B\setminus V)\cup U$. When $z\notin U$, retaining the eligible $A$ means its image already belongs to $\mathcal{F}$, and that image contains $B^{\prime }$. When $z=x\in U$, the smaller compression of $A$ belongs to $\mathcal{F}$ and equals $B^{\prime }\cup \{y\}$. In either case $B^{\prime }$ is in the old lower shadow too. Consequently the collision rule retains $B$ as well. This proves the claimed containment in every case. Since compression preserves cardinality, it cannot increase lower shadow size.

If $\mathcal{F}$ is not a colexicographic initial segment, an earlier absent set $A$ and a later present set $B$ give $U=A\setminus B$, $V=B\setminus A$ with $\max U<\max V$, and a nontrivial UV-compression. Choose such a compression with $|U|$ smallest. If $|U|>1$, choose $y\in V\setminus \{\max V\}$. For every $x\in U$, the smaller compression still has its largest differing element on the $V$ side and must fix $\mathcal{F}$ by minimality. If $|U|=1$, the smaller compression is the identity. Thus the lower shadow argument applies.

Each nontrivial step strictly decreases the positive integer weightbecause the largest differing coordinate belongs to $V$. Iteration therefore terminates. A terminal set family is a colexicographic initial segment, since an earlier gap and a later member would otherwise provide another nontrivial compression. The final set family has $m$ members and no larger lower shadow.

To compute its lower shadow, take $a_r$ maximal with $\left(\genfrac{}{}{0ex}{}{a_r}{r}\right)\le m$. All $r$-sets of $[a_r]$ occur first. The next block consists of sets containing $a_r+1$ whose remaining $(r-1)$-set runs through the initial segment of length $m-\left(\genfrac{}{}{0ex}{}{a_r}{r}\right)$. By Pascal's identity, the remainder is smaller than $\left(\genfrac{}{}{0ex}{}{a_r}{r-1}\right)$, so its next leading top index is strictly less than $a_r$. Repeating gives the stated unique binomial representation. The first block contributes all $\left(\genfrac{}{}{0ex}{}{a_r}{r-1}\right)$ $(r-1)$-sets in $[a_r]$ to the lower shadow. Deleting $a_r+1$ from the next block contributes only sets already counted; deleting another coordinate contributes $a_r+1$ joined to the lower shadow of the smaller initial segment. This recursive decomposition gives exactly the displayed sum. The base $r=1$ has lower shadow $\{\varnothing \}$ for every nonempty set family. This completes the proof and the sharpness assertion.

Now let $\mathcal{C}$ be the set family of vertex sets of the graph's cliques of order four. Its twice-iterated lower shadow consists of edges of the graph. If there were at least sixteen such copies, choose a sixteen-member subfamily. The relevant binomial representation isThe proved theorem gives at least $\left(\genfrac{}{}{0ex}{}{6}{3}\right)+\left(\genfrac{}{}{0ex}{}{3}{2}\right)=23$ triples in its first lower shadow. Apply it again to any twenty-three of those triples, using $23=\left(\genfrac{}{}{0ex}{}{6}{3}\right)+\left(\genfrac{}{}{0ex}{}{3}{2}\right)$. Their edge lower shadow has at leastmembers, contradicting the fifteen available edges. Conversely, the complete graph $K_6$ has $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ edges and $\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$ copies of $K_4$. Therefore the maximum number of copies is exactly 15.

The Kruskal-Katona theorem states that if $\mathcal{A}\subseteq [n{]}^{(r)}$ has the unique binomial representationthen its lower shadow obeysAn initial segment of colexicographic order has exactly this shadow.

Here is the UV-compression proof of the Kruskal-Katona theorem. For disjoint equal-size sets $U,V$, the UV-compression $C_{U,V}$ replaces the pattern $V$ by $U$ when the image is not already in the family. If the family is not an initial colexicographic segment, choose a changing pair with $\max U<\max V$ and $|U|$ minimal. Minimality ensures that for every $x\in U$ an appropriate smaller compression $C_{U\setminus \{x\},V\setminus \{y\}}$ already fixes the family. The Shadow lemma for UV-compressions then givesMeanwhile the binary weight ${\sum }_{A\in \mathcal{A}}{\sum }_{i\in A}{2}^i$ strictly decreases. Iterating must terminate, and a terminal family is an initial colexicographic segment. Computing that segment's shadow from its binomial representation proves the theorem.

We next classify pairs that decrease the shadow for every uniform family. The answer, including the identity case, isFor $|U|=0$ the operation is the identity. For $|U|=1$, the hypotheses of the Shadow lemma for UV-compressions reduce to stability under the empty compression and therefore hold for every family.

To see failure for larger pairs, relabel freely. If $|U|=|V|=2$, write $U=\{u_1,u_2\}$ and $V=\{v_1,v_2\}$. The familyhas a three-point lower shadow, whereas its compression replaces $v_1{v}_2$ by $u_1{u}_2$ and has a four-point lower shadow. If the common size is $m\ge 3$, takeThe old two shadows overlap once and have size $2m-1$. Compression replaces $V$ by $U$; the two resulting $m$-sets intersect in only $u_1$, so their lower shadows are disjoint and have size $2m$.

For the two specified pairs, the answer is no in both cases, even after assuming the family is left-compressed.

For $(U,V)=(345,126)$, takeThis family is left-compressed. Its compression replaces $126$ by $345$. The old shadow isof size nine, while the new shadow replaces the last two pairs by $34,35,45$ and has size ten.

For $(U,V)=(145,236)$, take the left-compressed familyIts shadow consists of $12,13,23$ and the nine pairs having one element in $\{1,2,3\}$ and one in $\{4,5,6\}$, so it has size twelve. Compression replaces $236$ by $145$; all twelve old shadow pairs remain and $45$ is added. The new shadow therefore has size thirteen.