A set is -rich when every -element subset of has at least common neighbours.
Choose vertices independently and uniformly from , allowing repetitions, and let
By Jensen inequality applied to the convex function ,
Let count the -subsets having fewer than common neighbours in . For each such , the probability that is
so
The assumed inequality gives , so some choice has . Delete one vertex from each bad -subset of . The remaining set has size at least and contains no bad -subset, so it is -rich. This is the basic dependent random choice argument.
Embed the part injectively into the rich set in a graph . List the vertices of as . When embedding , the images of its at most neighbours in have at least common neighbours in : extend that image set to an -subset of if necessary, noting that enlarging a set can only shrink its common neighbourhood. At most vertices have already been used, so one common neighbour remains available for . Choosing it embeds every edge incident with and keeps the map injective. Continuing greedily embeds in . This is the rich-set embedding lemma.
Let and colour the edges of red and blue. One colour, say red, forms a graph with
Apply part (a) with
Its positive term satisfies
while
The difference is at least , so contains a -rich set of size at least .
The hypercube graph is bipartite according to the parity of the sum of its coordinates. Each part has vertices, every vertex has degree , and . Part (b) therefore embeds a red copy of . Every red-blue colouring of has a monochromatic copy, proving

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