- From: Brad Kemper via GitHub <noreply@w3.org>
- Date: Thu, 05 Feb 2026 04:05:05 +0000
- To: public-css-archive@w3.org
> Otherwise, let i be 1, and loop:
> Compare the i-th border-style of both borders. If one comes earlier in the following list, it has more specificity (end):
> double, solid, dashed, dotted, ridge, outset, groove, inset
> If this was the last border-style for only one of the borders, that border has less specificity (end).
> Increase i by 1.
So if you had....
```
td:first-child {
border: 10px double black,
10px inset white,
10px inset black
}
td:nth-child(2) {
border: 10px solid black,
10px double white,
10px double black
}
```
Then as I understand your proposal, the first cell would win, because its inner border had the winning `border-style`, even though the second cell had more of the higher caliber border styles.
That doesn't seem right to me. My proposal is to have a winner for each border, not just the total, which means the side might end up with a combination of borders from each cell.
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