Elections in which every Condorcet winning set has four candidates

Construct an election in which every Condorcet winning set has at least four candidates, refuting the conjecture that three always suffice.

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Board: where the work stands

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3	3	Written by the site from the issues and claims, never by hand; times are UTC. A claim says who works on what and keeps nobody out: name an issue in your notes to work on it, read the notes of whoever else does, or open an issue for what is not there yet.
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5		-0 open issues, 0 that nobody works on, 0 worked on; 0 open pull requests.
	5	+0 open issues, 0 that nobody works on, 0 worked on; 1 open pull request.
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7	7	`main` at c2dbef5: no checks on it yet.

		@@ -2,6 +2,6 @@
2	2
3	3	Written by the site from the issues and claims, never by hand; times are UTC. A claim says who works on what and keeps nobody out: name an issue in your notes to work on it, read the notes of whoever else does, or open an issue for what is not there yet.
4	4
5		-0 open issues, 0 that nobody works on, 0 worked on; 0 open pull requests.
	5	+0 open issues, 0 that nobody works on, 0 worked on; 1 open pull request.
6	6
7	7	`main` at c2dbef5: no checks on it yet.