Is Woodall the same as Smith,IRV? (And thus Schwartz-Woodall the same as
Schwartz,IRV?)

Here's my thinking: Suppose that the IRV order is A>B>C>D>E and the
Smith set is {B, C, D}. Woodall's rules say: keep eliminating until only
one of the initial Smith set members is left.

So first eliminate E. The remaining members are {B, C, D}
Then eliminate D: {B, C}
Then eliminate C: {B} - and B is elected.

At each step, we subtract from the Smith set the k last ranked
candidates on IRV's social order (which are those who were eliminated
before round k). We then increase k until the set difference produces a
single candidate.

But this candidate is the Smith set member who is highest ranked on
IRV's elimination order. Hence it is the winner of Smith,IRV.

Sounds about right?

-km

IRV fails ISDA, so removing all non-Smith members can affect the
IRV-ordering of Smith set members.