Pairwise coalitions and D'Hondt

5-winner example

9 A>F>G>H>I>J
9 B>F>G>H>I>J
9 C>F>G>H>I>J
9 D>F>G>H>I>J
9 E>F>G>H>I>J
8 K
7 L

We can notice a 45-voter coalition of voters backing the candidates A through J. Running D’Hondt versus the largest other coalitions, the 8 K and 7 L voters, this 45-voter coalition takes all 5 seats. So we can then eliminate all candidates not in the coalition, and then by recomputing the Droop quota to factor for the 15 now exhausted K and L ballots, just elect each of the 5 1st choice candidates in the coalition.

Here’s how this concept leads to the Condorcet winner in the single-winner case:
42 M>N>C>K
26 N>C>K>M
15 C>K>N>M
17 K>C>N>M

We can see an anti-M coalition of 58 voters, a majority, so eliminate M from contention (this requires Independence of Smith-dominated Alternatives). Now, the 42 M-top voters’ new 1st choice is N; this gives N 68 1st choice votes, a majority, so N wins (This is the Wikipedia example with Nashville as the CW).

I think this could lead to a simpler way of conceptualizing Condorcet PR; figure out how many seats each coalition is entitled to, then run a sequential PR method eliminating all non-guaranteed candidates at each stage.