Prove or disprove that there exists no more specific match-ups involving these 4 third-placed teams for the round of 16 other than

Question

UEFA EURO 2020 has been just started, still with the format of 24 teams. The top 2 in each groups will proceed to the round of 16 along with the best of 4/6 third-placed finishers. Prove or disprove that there exists no more specific match-ups involving these 4 third-placed teams for the round of 16 other than $$\begin{matrix} \mathsf{1B}\vee \mathsf{3A}/\mathsf{D}/\mathsf{E}/\mathsf{F} & \mathsf{1C}\vee \mathsf{3D}/\mathsf{E}/\mathsf{F}\\ \mathsf{1F}\vee \mathsf{3A}/\mathsf{B}/\mathsf{C} & \mathsf{1E}\vee \mathsf{3A}/\mathsf{B}/\mathsf{C}/\mathsf{D} \end{matrix}$$ FYI, here are the combinations of matches in the round of 16_ in.format,
is it related to a generating function for all the possible scenarios ? Can you give me a hint ? Thanks a real lot !
Edit (07/11/21). I will describe my plan by qubit. I should start with the much easier-to-understood problem:
Hall's Marriage. Given two pairs of men and women whom participate in a couple's show. Realize that Mrs. $\mathsf{A}$ fancies Mr. $\mathsf{C}$ and Mr. $\mathsf{D},$ on the other hand, Mrs. $\mathsf{B}$ only fancies Mr. $\mathsf{D}.$ The best move here is $\mathsf{A}$ chooses $\mathsf{C}$ and $\mathsf{B}$ chooses $\mathsf{D}.$ Well, this is the Mathematics behind:
If $\mathsf{A}$ doesn't care the thinking of $\mathsf{B}$ then $\mathsf{A}$ has more than a possible way, I call it the state $11.$ And $\mathsf{B}$ who has only a possible way, I call it the state $01.$ Assume that $\mathsf{A}$ is full-of-selfish, that means $11\rightarrow 01.$ If $\mathsf{A}$ chooses $\mathsf{D}$ and $\mathsf{B}$ chooses $\mathsf{C},$ who knows ? This is the situation that $\mathsf{B}$ can't forsee, that means $01\rightarrow 10.$ So $10$ is satisfied here, huh ? No. In the logic law, $1\Rightarrow 1, 0\Rightarrow 1, 0\Rightarrow 0$ are true but only $1\Rightarrow 0$ is false, I call the states on purpose (I work with the decimal $\mathsf{base-}10$ and the binary "$\mathsf{base-}10$" too, very frequently). In conclusion, $11\oplus 01\rightarrow\mathsf{unknown},$ maybe $10$ or $01.$ I solve the problem by making $01\rightarrow 00,$ totally true by the logic law, $\mathsf{true}\Rightarrow\mathsf{true}.$ That leads to $11\oplus 01\rightarrow 11\oplus 00\rightarrow 11.$ This is my big picture about that $$\begin{matrix}\mathsf{pairs} & A & B\\ C & & \times\\ D & & \end{matrix}$$ $$\begin{matrix}\mathsf{pairs} & A & B\\ C & & \times\\ D & & \checkmark\end{matrix}$$ $$\begin{matrix}\mathsf{pairs} & A & B\\ C & & \times\\ D & \times & \checkmark\end{matrix}$$ In the beginning, we should limit the choice of $\mathsf{A}$ with only $\mathsf{C},$ that's all.
Combinations of matches. Okay. Let's describe them on the big picture $$\begin{matrix}\mathsf{match-ups} & B & C & E & F\\ A & & \times & & \\ B & \times & \times & & \\ C & \times & \times & & \\ D & & & & \times\\ E & & & \times & \times\\ F & & & \times & \times\end{matrix}$$ There are $14$ positions in the state $11.$ And observing that every rows and every columns has more than $2$ choices, if we make two of rows full, no problem, we always has a possible sequence of choices. Without loss of generality, we consider with only the columns $\mathsf{E}$ and $\mathsf{F}$ permitting the rows $\mathsf{A}$ and $\mathsf{B},$ that's the hardest case $$\begin{matrix}\mathsf{match-ups} & E & F\\ C & & \\ D & & \times\end{matrix}$$ It's already on the top. I think $14$ is also the best constant here. That makes the most suitable states $01$ and $11.$ What's your thinking ? Just let me know.

Answer 1

I don't know if this answers fully your concerns, but it may help. The table below, which is taken directly from the UEFA regulation, paragraph 21.05, shows the different options for the round of 16 pairings, depending on which third-placed teams qualify from the final tournament group matches.

Depending on the quadruple of 3rd-ranked teams (left column), the second column shows which teams will be matched against $1B,1C,1E,1F$ respectively.

$$\begin{matrix} A & B & C & D & | & 3A & 3D & 3B & 3C\\ A & B & C & E & | & 3A & 3E & 3B & 3C\\ A & B & C & F & | & 3A & 3F & 3B & 3C\\ A & B & D & E & | &3D & 3E & 3A & 3B\\ A & B & D & F & | & 3D & 3F & 3A & 3B\\ A & B & E & F & | & 3E & 3F & 3B & 3A\\ A & C & D & E & | & 3E & 3D & 3C & 3A\\ A & C & D & F & | & 3F & 3D & 3C & 3A\\ A & C & E & F & | & 3E & 3F & 3C & 3A\\ A & D & E & F & | & 3E & 3F & 3D & 3A\\ B & C & D & E & | & 3E & 3D & 3B & 3C\\ B & C & D & F & | & 3F & 3D & 3C & 3B\\ B & C & E & F & | & 3F & 3E & 3C & 3B\\ B & D & E & F & | & 3F & 3E & 3D & 3B\\ C & D & E & F & | & 3F & 3E & 3D & 3C\\ \end{matrix}$$

Note that this table is more specific than the one in the question. Indeed, if $3A,3B,3C,3D$ qualify, the original table was allowing also the option $ 3A - 3D - 3C - 3B$ in the first row. So, this is a proof that it was not describing the future matching uniquely.

Instead, the table that describes all possibilities trivially determines uniquely the matchups.