The End of an Era - That of The Natural Numbers
in [Math]
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From: Jesse F. Hughes on 18 Jun 2010 11:28 Marshall <marshall.spight(a)gmail.com> writes: > On Jun 18, 4:29 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> >> Your capability of misreading is peerless. > > Do you think he does it on purpose, at least some of the time? > I mean, he even gets really simple and obvious things wrong. > (Which is not to say that all of the things he gets wrong are > simple and obvious.) > > I'm beginning to wonder if he isn't more of a troll than > a crank. I'm sure I have no idea whether it's on purpose or not. -- Jesse F. Hughes | "There's no other star but one star | and you want it to make light, | but it's not making light." | -- A blues tune by Quincy P. Hughes
From: Nam Nguyen on 18 Jun 2010 23:51 Alan Smaill wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Jesse F. Hughes wrote: >>> If a theory has constants, then its >>> structures must be non-empty. >> No (model) structure within FOL= can be the empty _set_. It has at least >> these 2 elements: >> >> <'A',U> and <'=',{all 2-tuples of some form} | empty set> >> >> _by definition_. > > > The issue is whether the *universe* of the model can be non-empty. > > Since constants correspond to elements of the universe > (0-ary functions in Shoenfiled's terminology), A constant symbol is a language symbol. An element is an element of a universe U which is just a set (which could be empty). A correspondence is _just a meta mapping_ between a symbol to an element. When there's no element because U is empty, you could just change the meta correspondence (mapping): you map the element to "no element" which, if you use the unformalized set as the meta language, is usually designated by the empty set {}. > if the language has > constants, the *universe* cannot be empty. Again, a language doesn't have a power to make a set be non-empty or not. All there really is here is just a meta (mental) mapping, using a meta language as a mapping description vehicle.
From: Marshall on 19 Jun 2010 00:17 On Jun 18, 8:51 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > When there's no element because U is empty, > you could just change the meta correspondence > (mapping): you map the element to "no element" You cannot do that, and still say that you have a model of the language. If the language has a constant, then in order for a model to qualify as a model *of that language* it must designate a value from the carrier set to correspond to that constant. If it doesn't, then it doesn't model the constant, and hence doesn't model the language. It may in fact be a model, but it's sure not a model of that language. Marshall
From: Nam Nguyen on 19 Jun 2010 00:18 Marshall wrote: > On Jun 17, 11:32 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> Since there are _no n-tuples_ _in_ an empty >> predicate and since there's no non-empty predicate in an empty U, >> no formula can be true in the degenerated structure where U is empty. > > Here is the heart of your mistake. Without justification (and in fact > incorrectly,) you make the jump from speaking of predicates > to speaking of all formulas. Oh I didn't jump Marshall: you're just not knowing what you are uttering about (as usually is the case). I had a long post on June 12 where I gave some technical details mostly about the relationship amongst set membership, predicate, formula truth evaluation. Most of those are within "Point 1" - "Point 4" where I mentioned about and PIC (predicate interpretation condition), etc... And that wasn't the only post I had technical explanation for my argument. You're the one who is clueless and jumping around and could NOT utter a _technical_ definition for a general formula (open or closed) being true in a model. Even as we speak. > You are absolutely right that in a structure with an empty carrier > set, > every predicate must be empty. > > You are absolutely wrong in thinking that in a structure in which > every predicate must evaluate to false, that every *formula* > must also evaluate to false. Where did I state something like "every predicate must evaluate to false"? What does it mean for a predicate to evaluate to false, can you elaborate? > > Formulas are more than just predicates. To be precise: formulas are different from predicates. But where did I say the two are equal? (Actually your utterance here is nonsensical: "more than" in what respect?)
From: Nam Nguyen on 19 Jun 2010 00:30
Marshall wrote: > On Jun 18, 8:51 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> When there's no element because U is empty, >> you could just change the meta correspondence >> (mapping): you map the element to "no element" > > You cannot do that, and still say that you have > a model of the language. If the language has > a constant, then in order for a model to qualify > as a model *of that language* it must > designate a value from the carrier set to correspond > to that constant. > If it doesn't, then it doesn't model > the constant, and hence doesn't model the language. A model is obligated to model a truth value (true or false) for a formula, not a constant: a constant isn't true or false, a formula is. That's why there's no 0-ary _predicate_. > > It may in fact be a model, but it's sure not a model > of that language. It's a degenerated language model, but a language model. Because for each n-ary predicate symbol (n > 0) there are complementary predicate sets (even though empty) to evaluate truth value for formulas, based on Tscot (Tarski's Semantic Concept of Truth). |