Wouldn’t It be Cool if These Were Equivalent?
in [Logic]
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From: Charlie-Boo on 1 Jul 2010 11:22 On Jul 1, 10:27 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Frederick Williams <frederick.willia...(a)tesco.net> writes: > > Charlie-Boo wrote: > > >> (Hmmm . . . If ZFC can't have a universal set, how can it have even > >> quantifiers??) > > > First order languages "have" quantifiers. Universes of set theory > > "have" sets (i.e. the sets are individuals in the universe). Are you > > muddling up these two kinds of "having"? The quantifiers refer to > > those individuals just as they do in any first order theory. > > I'm too tired to write anything sensible about this, but > > web.mit.edu/arayo/www/Introduction.pdf > > makes for an interesting read, for those who find these sort of things > interesting. Those who don't might not find it the least bit > interesting. (The collection _Absolute Generality_ also contains some > rather insightful essays, such as the one by Shapiro and Wright.) Again, > based on Charlie-Boo's earlier musings about universal sets my > considered opinion is that these sensible questions and philosophical > ruminations are entirely unrelated to his inane rambling. Yet another sheepish mea culpa! In CBL we have, P(x) + EQ(I,J) => P(I) P(I) + TRUE(x) => P(x) If you can list a set and check for equality, you can decide the set. If you can decide a set and list the universal set, you can list the set. Now, there are some things you can't list but can decide. So if you have a universal set TRUE(a) then be careful you aren't violating the rules on what can and can't be listed. In other words, it's Hilbert all over again. But more to the point. Does ZFC have a universal set? Does it have quantifiers? Do quantifiers need a universal set? (This is called Discovery in the US judicial system - feeling out what differences there are between the two opposing versions of reality.) C-B > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, dar ber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Charlie-Boo on 1 Jul 2010 11:24 On Jul 1, 10:33 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > On Jul 1, 9:49 am, Frederick Williams <frederick.willia...(a)tesco.net> > > wrote: > > > Charlie-Boo wrote: > > > > > In the meantime, if I just said that the universal set is N (which was > > > > my intention, ... > > > > Look, > > > > (allX) |- P(X) > > > > is hard to understand _whatever_ the domain of quantification (is that > > > what you mean by 'universal set'?). Does it mean something like > > > > whatever term is substituted for the free variable X in P, > > > the result is provable > > > > or what? > > > I thought I said it, but does this work: Read it left-to-right, > > reading substrings that form familiar mathematical concepts. In this > > case we have "(allX)" which is . . . etc. > > > Yes, of course it means that. How would you express it as a formal > > wff? > > Wff of _what_? If we are talking about PA then there is a provability > predicate to express |-. Let's call it prov, then > > prov(n) iff |-phi > > where n is the numeral for the G"odel number of phi. Yes, good. I was wondering if anyone would point that out. I was going to figure out the exact relationship between that approach and mine. Which is? C-B > -- > I can't go on, I'll go on.- Hide quoted text - > > - Show quoted text -
From: Charlie-Boo on 1 Jul 2010 15:03 On Jul 1, 11:42 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > ... I was > > going to figure out the exact relationship between that approach and > > mine. > > Go on then. Well, let's see. I call this system ABC because I represent wffs using letters where A = |- B = ~ C = (all X) Notice that P is actually free. The empty string [] represents P. So we have e.g. A = |-P B = ~P C = (allX)P(X) AA = |- |- P AB = |- ~P AC = |- (allX)P(X) BA = ~|-P BB = ~~P BC = ~(allX)P(X) CA = (aA)|-P(A) CB = (aA)~P(A) CC = (allX)(allY)P(X,Y) etc. Then the task is to (1) List a bunch of ABC wffs. (2) Take the ones that aren't in Logic. (3) See how to represent them using the provability predicate. I just did (1). If I do (2), would you do (3)? I'll also do (3) but with no committment to degree. (Life goes on.) Actually, then (4) List wffs using the provability predicate and (5) see how to represent them in ABC! C-B > -- > I can't go on, I'll go on.
From: Charlie-Boo on 3 Jul 2010 13:36
On Jul 3, 5:30 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > Well, let's see. I call this system ABC because I represent wffs > > using letters where > > > A = |- > > B = ~ > > C = (all X) > > > Notice that P is actually free. The empty string [] represents P. So > > we have e.g. > > > A = |-P > > B = ~P > > C = (allX)P(X) > > AA = |- |- P > > AB = |- ~P > > AC = |- (allX)P(X) > > BA = ~|-P > > BB = ~~P > > BC = ~(allX)P(X) > > CA = (aA)|-P(A) > > CB = (aA)~P(A) > > CC = (allX)(allY)P(X,Y) > > etc. > > Is this alphabet soup of interest to anyone other than you? Are you saying that you don't understand it? Do you know what I'm doing? I'm listing the first few wffs. Their representation is any string of alphabet {A,B,C} so it's real easy to list wffs. Then the idea is to see how each would be represented using the provability predicate, to compare the two approaches. Does that help? C-B > -- > I can't go on, I'll go on.- Hide quoted text - > > - Show quoted text - |