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So to actually do reflexives I've used two solutions. One is to use an extra predicate like Lojban {du}, so that "X loves X" gets rendered as "X loves Y & X du Y". This, incidentally, is the strategy I use for so-called 2D-Livagian notation of predicate logic, where the notation can't handle reflexivity.
The other solution is to use special inflections that augment a stem to reflexivize it; these inflections would augment the inflection that shows the predicate's adicity and how many ESAPs it has. In 2014 Livagian I use this latter method, at least for predicates of maximal tetradic adicity (including event arg, which can't be inflectionally marked as reflexive).
As for additive merger, the particular case of "brothers" is unlikely to arise, because rather than having symmetric predicates like "X is brother of Y", I tend to use the "X joi Y are brothers" method. But something like "mother and child (are well)" (i.e. "X is mother of Y, X joi Y are well") can arise, and iirc it works by making every item on the vlist visible to an additively-merged incoming ESAP.
There are still more complications in that you can get cycles of alternate additive and intersective merger involving the same incoming ESAP. For example, "xorxes joi And discuss" might have a word order "discuss, xorxes, And", such that -- ignoring event arguments -- there is a vlist <Discuss1, xorxes1> and so first And1 additively merges with xorxes1, and then the result, And1&xorxes1 intersectively merges with Discuss1. And if the phrase were "Obama and discussants xorxes and And", with word order "Obama, discuss, xorxes, And", (And1&xorxes1)+Discuss1 would additively merge with Obama1, yielding ((And1&xorxes1)+Discuss1)&Obama1. I currently require that the counting for this iterative process be successive, so that each new merger, of the same ESAP, involves counting back from the vlist position of the previous merger, not from the end of the list all over again. This is the one respect in which predicate order is not free, because predicates must occur in an order that matches iterated merging. As I write this paragraph I realize that in principle (tho improbable), you could have two distinct iterated merger sequences involving the same predicates but each requiring the predicates to be in a different sequence -- which means I am going to have to rethink my solution to this bit.