You can also think of A5 in terms of its contrapositive: For all A,B∈L, if A≻B, then for all 0<p≤1A≻pA+(1−p)B
This is basically just the strict version of A4. I probably should have written it that way instead. I wanted to use ⪰ instead of ≻, because it is closer to the base definition, but that is not how I was natively thinking about it, and I probably should have written it the way I think about it.
You have the inequality backwards. You can’t apply A5 when the mixture is better than the endpoint, only when the mixture is worse than the endpoint.
Got it, thanks!
You can also think of A5 in terms of its contrapositive: For all A,B∈L, if A≻B, then for all 0<p≤1A≻pA+(1−p)B
This is basically just the strict version of A4. I probably should have written it that way instead. I wanted to use ⪰ instead of ≻, because it is closer to the base definition, but that is not how I was natively thinking about it, and I probably should have written it the way I think about it.