A different version of (5), in response to Benign A-Fission, could be a rule that treats Lefty as non-extra and Righty as extra in Split B β maybe for basically the reasons you give for Split B over No Split β, and one of Lefty or Righty as non-extra in Split C. Then youβd choose Split B among the three options.
One incomplete rule that could deliver this result is the following:
If all the splits have non-negative welfare, treat one with the highest welfare as non-extra, and treat the others as extra.
So, No Split gives Anna 80 welfare, Split B, 10+90=100 welfare and Split C, 10+60=70 welfare. Split B is best.
This doesnβt say what to do about splits with negative welfare. Two options:
Treat them all as non-extra.
Treat a worst off one as non-extra, and the rest as extra and ignore them.
We might also consider decreasing marginal value to additional splits beyond the highest positive (and worst negative) welfare one, instead of totally ignoring them. Maybe the highest welfare one gets full weight, the second highest gets weight r,0<r<1, the third highest gets r2, β¦, the n-th highest gets rnβ1. This bounds the weighted sum of welfare in the splits by ββn=0rn=11βr times the welfare of the best off split. They still count for something, but we could avoid Repugnant Fission.
A different version of (5), in response to Benign A-Fission, could be a rule that treats Lefty as non-extra and Righty as extra in Split B β maybe for basically the reasons you give for Split B over No Split β, and one of Lefty or Righty as non-extra in Split C. Then youβd choose Split B among the three options.
One incomplete rule that could deliver this result is the following:
If all the splits have non-negative welfare, treat one with the highest welfare as non-extra, and treat the others as extra.
So, No Split gives Anna 80 welfare, Split B, 10+90=100 welfare and Split C, 10+60=70 welfare. Split B is best.
This doesnβt say what to do about splits with negative welfare. Two options:
Treat them all as non-extra.
Treat a worst off one as non-extra, and the rest as extra and ignore them.
We might also consider decreasing marginal value to additional splits beyond the highest positive (and worst negative) welfare one, instead of totally ignoring them. Maybe the highest welfare one gets full weight, the second highest gets weight r,0<r<1, the third highest gets r2, β¦, the n-th highest gets rnβ1. This bounds the weighted sum of welfare in the splits by ββn=0rn=11βr times the welfare of the best off split. They still count for something, but we could avoid Repugnant Fission.