The End of an Era - That of The Natural Numbers
in [Math]
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From: Marshall on 19 Jun 2010 00:40 On Jun 18, 9:30 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > 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 Wrong. A model is obligated to supply an appropriate interpretation of every member of the signature of the language. > > 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. Wrong. If that were correct, then we might as well just say that anything is a model of any language. This banana on the counter over here is a model of the language of basic arithmetic. It doesn't have a member of the domain that models the constant zero, but it *is* high in potassium. Marshall
From: Nam Nguyen on 19 Jun 2010 01:12 Jesse F. Hughes wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> So the closed formula A is in L(fancyA). So x=x is not in L(fancyA), >> an extension of L, in the sense that x=x is already in L. > The fuller of what I said was: > That's actually not true, in closer inspection. What might make > one easily confused on the issue is he actually referring to 2 > different formulas: the original L in question, and the L(fancyA). > Then he said (pg. 19): "We shall now define a truth value fancyA(A) > for each closed formula in L(fancyA)"! > > So the closed formula A is in L(fancyA). So x=x is not in L(fancyA), > an extension of L, in the sense that x=x is already in L. I had a typo there: "referring to 2 different formulas" should have been "referring to 2 different languages", as it was clear in the next sentence I was referring to 2 different languages: L and L(fancyA). > L is a subset of L(fancyA), and so every formula of L is also a formula > of L(fancyA). But didn't I mention "in the sense that x=x is already in L" as a caveat? Especially L(fancyA) extends L in a contrived way so as to have more individual symbols _for naming purposes only_! I was actually explaining to you (which you missed) that _you_ got so confused between a closed A and the open x=x Shoenfield was referring to and repeatedly but incorrectly claimed he wasn't defining being true (valid) for an open formula such as x=x. My explanation here was that by a closed formula A, he meant A must necessarily be in L(fancyA) while you should only care that the _open_ x=x is in L as it's intended for the purpose of truth evaluation: the fact it's also an expression in a contrived extended L(fancyA) is immaterial and ignorable, as as truth evaluation Shoenfield was doing is concerned. You got to read people's caveat more carefully, before jumping to conclusion and .... > > Lordy, but you're incapable of reading a relatively clear and simple > text. and before committing yourself at kind of pathetic dialog, as above. As far as I'm concerned, you're in such attack mode whenever you want to create a smokescreen to hide your technical mistakes.
From: Nam Nguyen on 19 Jun 2010 01:25 Marshall wrote: > On Jun 18, 9:30 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> 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 > > Wrong. A model is obligated to supply an appropriate > interpretation of every member of the signature of the > language. Wrong? Are you now abandoning the idea x=x is true, it being wrong to say "A model is obligated to model a truth value (true or false) for a formula"? >>> 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. > > Wrong. If that were correct, then we might as well just say > that anything is a model of any language. How would you get to that conclusion, given each language would have its own degenerated language model? > > This banana on the counter over here is a model of the > language of basic arithmetic. It doesn't have a member > of the domain that models the constant zero, but it *is* > high in potassium. But the banana doesn't have this ordered pair <'S',{}> for example. So that's not even the degenerated model of L(PA). Give it up Marshall. Once you're technically wrong you're technically wrong. Just admit it!
From: Marshall on 19 Jun 2010 01:40 On Jun 18, 10:25 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Marshall wrote: > > On Jun 18, 9:30 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > >> 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 > > > Wrong. A model is obligated to supply an appropriate > > interpretation of every member of the signature of the > > language. > > Wrong? Are you now abandoning the idea x=x is true, it being wrong > to say "A model is obligated to model a truth value (true or false) > for a formula"? > > >>> 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. > > > Wrong. If that were correct, then we might as well just say > > that anything is a model of any language. > > How would you get to that conclusion, given each language > would have its own degenerated language model? > > > > > This banana on the counter over here is a model of the > > language of basic arithmetic. It doesn't have a member > > of the domain that models the constant zero, but it *is* > > high in potassium. > > But the banana doesn't have this ordered pair <'S',{}> for > example. So that's not even the degenerated model of L(PA). > > Give it up Marshall. Once you're technically wrong you're technically > wrong. Just admit it! I admit that I was wrong to even bother to try to speak with such a dipshit as you. Marshall
From: Nam Nguyen on 19 Jun 2010 03:04
Marshall wrote: > On Jun 18, 10:25 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> Marshall wrote: >>> On Jun 18, 9:30 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >>>> 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 >>> Wrong. A model is obligated to supply an appropriate >>> interpretation of every member of the signature of the >>> language. >> Wrong? Are you now abandoning the idea x=x is true, it being wrong >> to say "A model is obligated to model a truth value (true or false) >> for a formula"? >> >>>>> 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. >>> Wrong. If that were correct, then we might as well just say >>> that anything is a model of any language. >> How would you get to that conclusion, given each language >> would have its own degenerated language model? >> >> >> >>> This banana on the counter over here is a model of the >>> language of basic arithmetic. It doesn't have a member >>> of the domain that models the constant zero, but it *is* >>> high in potassium. >> But the banana doesn't have this ordered pair <'S',{}> for >> example. So that's not even the degenerated model of L(PA). >> >> Give it up Marshall. Once you're technically wrong you're technically >> wrong. Just admit it! > > I admit that I was wrong to even bother to try to speak with > such a dipshit as you. The "dipshit" is those who fiercely defend the idea x=x is true in all contexts of FOL reasoning, and yet couldn't utter the definition of x=x being true - post after post. Like you. *** Any rate, consider the context of an inconsistent theory T, is x=x true in in such a T? Or is it that one could assign the formula x=x to true or false as one wishes, relatively speaking? Your side has lost the argument on more than one account already. There's no point to continue anyway. |