[EM] Yet Another Simple Way to Find Smith Set

[Gervase]

I've been perusing the Election Methods archive and noticed that more or
less recently there was the need for simple legislative language to
define Condorcet and/or Smith (Subjects: "Hay guys, look at this..." and
"Another Simple Way to Find Smith").  Apologies if what I write
following this paragraph has already been explored in the mailing list.

Looking at the Wikipedia page for "Smith Set", it seems to contain a
simple algorithm for finding the Smith Set...

(1) Start with a set D (Dominating/Smith Set) with no candidates in it.
(2) Add all candidate(s) who are in the Copeland (Winner) Set to D.
(3) Add to D the candidate(s) who are NOT currently in D and who also
pairwise beats any candidate currently in D.
(4) Repeat step (3) until no candidate(s) are added by step (3).

RESULT: Set D is the Smith Set.

Copeland (Winner) Set are set of candidate(s) with the highest Copeland
score.

That's it!  Read further if you want more details...

The Wikipedia page for "Smith Set" (
https://en.wikipedia.org/wiki/Smith_set) uses propositional(?) logic to
determine that any candidate in the Smith Set D will have at least a
Copeland Score of θD (i.e. the threshold score for set D).  If there is
a θD, then there will be a maximum Copeland Score θW (i.e. the Copeland
Winner(s) Score).  θD and θW can be equal.

I will sort of translate the propositional logic used in the Wikipedia
page into something more layman like.

> Choose d as an element of D with minimum Copeland score, and identify
> this score with θD.

Choose any candidate d in a pre-defined Smith Set D.  This candidate d
will have a Copeland Score of θD.

>  Now suppose that some candidate e ∉ D has a Copeland score not less
> than θD.

Now suppose there is some candidate e who is not in the Smith Set D and
also has a Copeland Score equal to or more than θD.

>  Then since d belongs to D and e doesn't, it follows that d defeats e;

Because candidate d is in the Smith Set D and candidate e is not,
candidate d defeats candidate e.

> and in order for e's Copeland score to be at least equal to d's, there
> must be some third candidate f against whom e gets a better score than
> does d.

All the candidates in Smith Set D defeat candidate e.  That is,
candidate e's maximum possible Copeland Score is being eaten into.

With candidate d being in Smith Set D, it will beat all candidates
outside of Smith Set D.  This sets the "high" threshold which candidate
e needs to match.

Therefore, for candidate e to at least equal candidate d's (threshold)
Copeland Score, there will need to exist a candidate f who (1) candidate
e defeats and (2) defeats candidate d.

>  If f ∈ D, then we have an element of D who does not defeat e,

If candidate f was in the Smith Set D, then by definition, candidate f
defeats candidate e.  But we said in (1) that candidate e needs to
defeat candidate f.

>  and if f ∉ D then we have a candidate outside of D whom d does not
> defeat, leading to a contradiction either way.

If candidate f was not in the Smith Set D, then by definition, candidate
d defeats candidate f.  But we said in (2) that there needs to be a
candidate f who defeats candidate d.

Thanks,
Gervase.