Webster's definition of his method directly, and in the most obvious
way, puts each party's seats as close as possible to what is called
for by a common ratio beteen seats & votes. This, by itself, is a
strong argument in favor of Webster/Sainte-Lague.

This can't be true of any other method:

But there's another specific statement that can be made about Webster/Sainte-Lague:

If we have a Webster/Sainte-Lague seat allocation, and we take a seat from 1 party & give it to another party, this will always increase the difference between the seats per vote ratios of those 2 parties.

Why that's so:

If I may repeat it again, Webster puts each party's number of seats, S, as close as possible to the result of dividing its votes, V, by the common divisor, D.

S is as close as possible to V/D

Say we divide S & V/D by the same constant number, V:

If S is as close as possible to V/D, then S divided by V is as close as possible to V/D divided by V:

S/V is as close as possible to 1/D

I should clarify that when I say that S is as close as possible to V/D, and that S/V is as close as possible to 1/D, I mean that it's as close as possible, given the constraint that S must be a whole number.

Let's call 1/D "R", for "ratio". R = 1/D

So S/V is as close as possible to R.

In this discussion, when I speak of where a party is, I am referring to where its S/V is.

When one party gives a seat to another, this will of course change the 2 parties S/V in opposite directions.

Therefore, either the 2 parties will move toward eachother, or they'll move away from eachother.

If they move away from each other, the difference between their S/V has been increased, and our case is already proved.

So let's consider the cases where the 2 parties are moved toward eachother:

There are 2 cases:

Case 1. The 2 parties start on the same side of R:

Since we're assuming that the seat transfer moves the 2 parties toward eachother, the party farther from R is moved toward R. But that party, like all the parties, is as close to R as it can be. So if it moves toward R, it must move past R, ending up on the other side of R, and farther from it than it was initially. Not only is it now farther from R, but now R is between it and the other party, unlike before the move. Additionally, the other party has also moved farther from R, moving away from R. Therefore, the parties are farther apart than they were before they moved.

Case 2. The 2 parties start on opposite sides of R:

By assumption, the 2 parties move toward eachother, which means they move toward R. But since they're both as close to R as they can be, this means that they both move past R, ending up farther from it, on the side opposite from the side they started on. They're both now farther from R, and R is still between them. Therefore they're farther from eachother than they were before they moved.

Therefore, starting with a Webster/Sainte-Lague seat allocation, any transfer of a seat from 1 party to another will always put the S/ will always put the S/V of the 2 parties farther apart.

This can't be true of any other method:

Any other method will sometimes give an allocation different from
the Webster/Sainte-Lague allocation. To get this allocation from the
Webster allocation, you'd have to transfer a seat from 1 party to
another, at least once. This puts the involved parties S/V farther
apart. Then, if you transferred the seats back to where they were,
you'd of course *decrease* the difference in those 2 parties' S/V.
You've started with this other allocation and transferred a seat, and
this has decreased the difference in the S/V of the 2 parties involved.
Therefore no other allocation method can have Webster/Sainte-Lague's
exchange property.