Worked solution: A rainbow-tree proof of Ringel's conjecture (2020)

Step 7 of 8: Assembling the pieces: every tree gets a rainbow copy

In plain words

Step 3 guaranteed every large tree falls into Case A, B, or C; Steps 4 and 5 together handle Cases A and B via randomised embedding plus absorption; Step 6 handles Case C by an entirely different deterministic route. Since these cases jointly cover every possible tree shape, no tree with $n$ edges is left unaddressed -- completing the proof that the ND-coloured $K_{2n+1}$ always contains a rainbow copy.

\text{every ND-coloured } K_{2n+1} \text{ has a rainbow copy of every } n\text{-edge tree}

Detailed analysis

Section 2.4 of Montgomery, Pokrovskiy and Sudakov (2021) states the main lemmas -- the Case A and Case B finishing lemmas from Sections 4-5, and the randomised embedding Theorem 2.2/2.5 from Section 6 -- and derives Theorem 2.1 from them for trees in Cases A and B, while Section 7's Method M3 separately covers Case C. Since Lemma 3.5 (Step 3) guarantees every sufficiently large tree lies in at least one of the three cases, combining the case-specific arguments proves Theorem 2.1 in full generality: every ND-coloured $K_{2n+1}$ contains a rainbow copy of every $n$ -edge tree, for $n$ large enough.