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In this case, it would fail lemma 11 (the triangular staircases in the right corners). Even if you manage to dodge lemma 11 (like repeating the first and the last rows two times each), this still does not prevent the horizontally-oriented dominoes to be rotated to the base of the pyramid leading to a violation of lemma 4.

In this case, it would fail lemma 11 (the triangular staircases in the right corners). Even if you manage to dodge lemma 11 somehow, this still does not prevent the horizontally-oriented dominoes to be rotated to the base of the pyramid leading to a violation of lemma 4.

Proof: Let's start filling the top-left corner of the MUB with a SUD = 2. The two first possibilities are those:

Proof: Let's start filling the top-left corner of a MUB with one dimension sized 3. The two first possibilities are those:

╔═══╦═    ╔═╦═╦═   
╠═══╣ ... ║ ║ ║ ...
╠═══╣     ╠═╩═╣    
╚═══╩═    ╚═══╩═

Considering lemma 4, this can't be a MUB.

Let us try this other configurations:

╔═╦═══╦═══        ╔═╦═╦═╦═══   
║ ╠═══╣    ... -> ║ ║ ║ ║   ...
╠═╩═╦═╩═╗         ╠═╩═╬═╩═╗   
╚═══╩═══╩═        ╚═══╩═══╩═

And, by lemma 4, this is not a MUB either. Having the vertically-oriented domino in the bottom two rows is just a reflection of it in the top two, so this is no way unsolvable.

In fact, lemmas 6, 7 and 8 are here just for the show for it being easier to understand what is going on. We could just proceed to this lemma directly instead.

Proof: Let's start filling the top-left corner of the MUB with a SUD = 2. The only two possibilities of starting are those:

Proof: Let's start filling the top-left corner of a MUB with one dimension sized 3. The only possibilities are those:

╔═══╦═    ╔═╦═╦═    ╔═══╦═    ╔═╦═══╦═══    ╔═══╦═══╦═
╠═══╣ ... ║ ║ ║ ... ╠═╦═╣ ... ║ ╠═══╣   ... ╠═╦═╩═╦═╝ ...
╠═══╣     ╠═╩═╣     ║ ║ ║     ╠═╩═╦═╩═╗     ║ ╠═══╣
╚═══╩═    ╚═══╩═    ╚═╩═╩═    ╚═══╩═══╩═    ╚═╩═══╩═══
  A         B         C           D             E

Considering lemma 4, $A$, $B$ and $C$ aren't MUBs. In $D$ and $E$, with a single rotation, we will also violate that lemma. So, this is no way unsolvable.

In fact, lemmas 6, 7 and 8 are here just for the show for it being easier to understand what is going on. We could just proceed from lemma 4 directly to this lemma instead.

Well, $A$, $B$, $C$, $D$ and $G$ violates lemma 4, so they aren't MUBs.

╔═══╦═      ╔═══╦═ 
╠═══╣       ╠═══╣
╠═╦═╩═╗     ╠═╦═╩═╗
║ ╠═╦═╣     ║ ╠═╦═╩═╗ 
╠═╣ ║ ║ ... ╠═╣ ╠═══╣ ...
║ ╠═╩═╣     ║ ╠═╩═╦═╝
╠═╩═╦═╝     ╠═╩═╦═╝
╚═══╩═      ╚═══╩═
   A           B

╔═══╦═         ╔═══╦═         ╔═══╦═ 
╠═══╣          ╠═══╣          ╠═══╣  
╠═╦═╩═╗        ╠═╦═╬═╗        ╠═══╬═╗
║ ╠═══╣        ║ ║ ║ ║        ╠═══╣ ║ 
╠═╬═══╣ ... -> ╠═╬═╬═╣ ... -> ╠═══╬═╣ ...
║ ╠═══╣        ║ ║ ║ ║        ╠═══╣ ║
╠═╩═╦═╝        ╠═╩═╬═╝        ╠═══╬═╝
╚═══╩═         ╚═══╩═         ╚═══╩═ 

And we've violated lemma 4 again.

Filling it in any way that follows the lines dividing the vertically-oriented block will lead us to violate lemma 4. So, in order to avoid that, we will eventually build a pyramid:

Well, $A$, $B$, $C$, $D$ and $G$ violates lemma 4 (we can split the first two columns off the rest of the board), so they aren't MUBs.

╔═══╦══     ╔═══╦════
╠═══╣       ╠═══╣
╠═╦═╩═╗     ╠═╦═╩═╗
║ ╠═╦═╣     ║ ╠═╦═╩═╗ 
╠═╣ ║ ║ ... ╠═╣ ╠═══╣ ...
║ ╠═╩═╣     ║ ╠═╩═╦═╝
╠═╩═╦═╝     ╠═╩═╦═╝
╚═══╩══     ╚═══╩════
   A           B

╔═══╦══        ╔═══╦══        ╔═══╦══
╠═══╣          ╠═══╣          ╠═══╣  
╠═╦═╩═╗        ╠═╦═╩═╗        ╠═╦═╬═╗
║ ╠═╦═╣        ║ ╠═══╣        ║ ║ ║ ║
╠═╣ ║ ║ ... -> ╠═╬═══╣ ... -> ╠═╬═╬═╣
║ ╠═╩═╣        ║ ╠═══╣        ║ ║ ║ ║
╠═╩═╦═╝        ╠═╩═╦═╝        ╠═╩═╬═╝
╚═══╩══        ╚═══╩══        ╚═══╩══

And we've violated lemma 4 again because we can split the first two columns off the rest of the board.

Filling it in any way that follows the lines dividing the vertically-oriented block will lead us to eventually violate lemma 4. So, in order to avoid that, we will eventually build a pyramid: