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Re: [lojban] What gets into 'lo broda'
On the other hand, using C-sets makes it hard to get back to inidividuals, which may be buried a large number of set boundaries away by the time we need them. So, let's stick with L-sets or pluralities and seek a solution elsewhere. The next attempt is to look at the notion of satisfaction as it applies in logic simpliciter and as it works in the present case. Clearly the men who lifted the piano did not lift the piano in the same way as the trios who lifted the piano did. So let's save "satisfaction" for what the trios did (the basic metalanguage notion) and use "realization" for the broader concept. Then we can say that lo broda are those who realize 'broda' and everything seems to work out right. A realizes 'broda' just in case for every individual, x, among A there are y that 1)are among A, 2) x is among, and 3) satisfy 'broda' (this is in plural refernce and plural quantification language, I hope). The only problem that I see immediately
is when the "individuals" are themselves sets of some sort, which then must be C-sets to keep them together, rather than floating out into the great plurality. [btw how to read 'broda cimei' (or whichever mVV i mean) "A triad of things that broda": is it the things or the triads that broda?]
----- Original Message ----
From: John E Clifford <kali9putra@yahoo.com>
To: lojban@googlegroups.com
Sent: Fri, April 23, 2010 5:36:13 PM
Subject: Re: [lojban] What gets into 'lo broda'
So it now appears that the problem is that, while A, B, C, say get into lo broda because they collectively broda, once they are in lo broda, there entry pass is lost. Within lo broda (say it is A, B, C, D, E,) A,B,C is no different from A, D, E which might no broda collectively (nor individually neither), So to solve how 'lo broda broda, we need to keep that collection intact. Neither plural reference nor L-sets will do this, but C-sets will, so perhaps we should fall back to them (with some regret). But they give the "wrong answer" to the reward question. Or do they? In the reward question as set up, the players were indistinguishable, each participated in the same number of lifts. Suppose, to simplify, there are only three players, none of whom can do the lift alone, but AB can and BC, while AC tries a few times without success. Now it seems less right that each gets a hundred dollars. Giving each pair 100 dollars results, assuming even
splits, in A and C each getting 50, while B, who did more successful lifting, gets 100 (the fact that this say a c-note, we will not mention). There are sure to be a number of problems with this solution, too, but it has the advantage of using a single type of reference and of quantification and of answering sensibly both the reference and the satisfaction question. To be sure, there is a lot of unmarked indivual/collective ambiguity -- or just not mentioned -- going on, but that is for fine-turing.
----- Original Message ----
From: John E Clifford <kali9putra@yahoo.com>
To: lojban@googlegroups.com
Sent: Fri, April 23, 2010 11:04:27 AM
Subject: [lojban] What gets into 'lo broda'
In these kinds of discussions it is often a good idea to go back to the basics, in this case L-set theory. And it turns out, not surprisingly, that xorxes intuitions are quite right, however unclear his explanations may have been. In a set of L-sets, the internal boundaries dissolve and we get back to a union (actually the ancestral of a union): all goes back to "individuals" (L-sets without subsidiary L-sets). So the set that contained all those triples and quads that actually lifted the piano (L-sets all) turns out to be just the members of those quads and triples, as xorxes said.
This leaves the question of how 'lo broda cu broda' is true and the answer appears to be that from the things that broda, we can extract a collection that simply is lo broda (lo broda are among the things that broda, in pluralist terms). This still has some problem, since members of disjoint sets that satisfied 'broda' before can now appear to form another set within the brodaers which does not, in fact, satisfy broda. I have no idea how the semantics will puzzle this bit out, but at least the initial question about how things get into lo broda has a solid answer.
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