Are all polyominoes with even sides tileable by dominoes?

Question

In this paper (Section 8) the author states that it is "trivial" to show that a polyomino with all sides (including the sides of holes in the polyomino) even has a tiling by dominoes.

It is indeed easy to see for polyominoes without holes, but the general case does not seem so obvious to me. One thing that makes it tricky is that the holes can appear anywhere as long as their borders do not overlap or coincide. (This makes it difficult to find a proof that reduces the figure but keeps the even-sides constraint.)

Is there indeed a trivial proof for this?

Note: I asked a earlier question about what the same author calls "even" polyominoes, but those polyominoes are completely different.

Answer 1

After thinking about it all day I discovered a trivial proof.

First note that any figure with all sides even and no holes can be tiled with dominoes so that they are all horizontal, and that each domino shares its long edge with at most one other domino.

Now take any figures with all sides even, possibly with holes (whose sides must also be all even). First tile it as if it had no holes. Then remove all dominoes that lie completely inside a hole. Some dominoes may cross the border of the hole. These dominoes must occur in pairs (since the hole sides have even length), so each pair that overlaps a hole border can be replaced with a single domino outside the hole. The tiling is complete.

Below is the procedure on the figure above.

The proof also works if the outer border has only all horizontal sides of even length and all holes have their vertical sides even length.

Update: The proof also works for figures with sides divisible by $n$ and bars of length $n$ instead of dominoes if the holes are at least $n - 1$ units apart. I think (but am not sure) that we can remove this constraint if the tile set is all $n$-ominoes.