The End of an Era - That of The Natural Numbers
in [Logic]
|
Prev: What are deliberately flawed & fallacious Arguments? Sophistry!
Next: sci.lang is not meant for advertising
From: Nam Nguyen on 27 May 2010 23:58 Alan Smaill wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Alan Smaill wrote: >>> Fine, take U = {0,1,2}, and take everything else as above. >> Well, so far you've only spelled out U (and in effect <'A',U>). >> You've not spelled out the mapping (ordered pair) <'blue',p_blue> >> where p_blue is an actual _set_. Iow, if R is p_blue, can you >> spell out the predicate-set R? > > The set is the extension of the relation R: > > { x in U | R(x) } = {1} Now then, let's extend the language L(t4) to L(T4b) so it has another 1-ary symbol 'non-blue', and extend T4 into T4b so it has another axiom: non-blue(c1) <-> ~blue(c1). Can we keep the model M4 for T4b? If not what can we keep, and what should we add? > Are you claiming that your notion of model is equivalent to > Shoenfield's? Of course I do. And I'm still in the process of doing the explanation so I hope you don't mind answering the new question above. > >>> Do you agree that it follows from his definition that a constant >>> is interpreted as an element of the domain, >> Suppose you have a theory T5 = {Ax[~(x=e)]}, which element of your >> "domain" U (whatever U might be) would get interpreted as e? > > Could be any object in U. > But whatever it is, the structure is not a model for > T5, i.e. it will not satisfy the statement Ax[~(x=e)]. > Just follow Tarski's definition. Not a true model of course. But there's a false model for it. But let's settle the other issue above first and we'll come back to address the existing of the false model for an inconsistent theory. > >>> and that therefore >>> the domain is not empty whenever there is a constant in the >>> language? >> In his definition, as I had before, his structure is a non-empty >> set of ordered pairs, in each of which the 2nd component _is a set_ >> (un-formalized kind of set that should be taken a priori but which >> nonetheless could be empty). > > That's for predicates, not for constants. > Look at his treatment of constants (& function symbols). _Everything is n-ary_ in his treatment: including 0-ary (constant) symbol, 0-tuple, 0-ary function value for a constant itself. So if an _n-ary predicate is a set of n-tuples_, then the predicate is always a set (even if it's empty)!. So the 2nd component _is a set_ in all cases.
From: Nam Nguyen on 28 May 2010 00:49 Jesse F. Hughes wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Jesse F. Hughes wrote: >>> Nam Nguyen <namducnguyen(a)shaw.ca> writes: >>> >>>> Aatu Koskensilta wrote: >>>>> Nam Nguyen <namducnguyen(a)shaw.ca> writes: >>>>> >>>>>> Other than that, I'm afraid any conversation I might have with you >>>>>> would be fruitless. >>>>> Well, yes, as already noted I've at long last concluded it's totally >>>>> pointless to try to discuss logic with you, my contributions thus >>>>> reduced to general observations and cheap pot-shots. >>>>> >>>> I guess cheap shots will be flying around then. >>> What else is possible, when the topic is logic, but one of the >>> conversants is a blowhard incapable of realizing that, whenever P is >>> false in a structure, ~P is true in that same structure? >> Why not, if this is the _degenerated structure_ (the false structure) >> of a language? > > Because, by the definition of truth in a structure, if P is false in > M, then ~P is true. > > That's why not. You're wrong of course. What you've said resembles boolean algebra where ~F = T. But this isn't. This is Tarski's concept of truth a la set membership: if all predicates in the degenerated (false) structure are empty, then non-membership will occur and, by definition, all formulas will be interpreted as (assigned to) "false". Do you know what _NON_ membership mean? Does it mean "true" that something is in an empty predicate-set? > > This definition is clearly stated on p. 19 of Shoenfield, but you're > too addle-brained to understand it. Look who is talking! > >> Are you surprised that a tautology and a contradiction are >> equivalent in a _degenerated formal system_ that's called >> inconsistent? Surely you're not incapable of understanding that, >> are you? > > Of course. In an inconsistent theory, ~P <-> P is provable. That > observation does *not* imply that both ~P and P are false in some > mathematical structure. > > (Hint: inconsistent theories have no models. Duh.) That's where you misunderstanding is: when we say "no model" we implicitly mean no _true_ model. Model is a structure and therefore is a set that's _always_ non-empty, even though its universe U and hence its predicates might be empty! Have a good grip on that in your brain. > > I'm not sure why I'm actually in this discussion with Nam, who has > shown an infuriating unwillingness to learn from others and a very sad > inability to understand the authors he reads. I suppose I'll let it > go at this time. Well, yeah: if you don't have an ability to remove your filters and misunderstanding, then you should have let it go, way back.
From: Nam Nguyen on 28 May 2010 01:25 Daryl McCullough wrote: > Nam Nguyen says... > >> Jesse F. Hughes wrote: > >>> What else is possible, when the topic is logic, but one of the >>> conversants is a blowhard incapable of realizing that, whenever P is >>> false in a structure, ~P is true in that same structure? >> Why not, if this is the _degenerated structure_ (the false structure) >> of a language? Are you surprised that a tautology and a contradiction >> are equivalent in a _degenerated formal system_ that's called inconsistent? > > You are confusing two different things: truth in a structure, > and provability in a theory. No. It's you who are confusing between what's an analogy and what isn't. > If S is a structure for a language > L, then truth in a structure is defined in such a way that > every closed sentence of L (a sentence without free variables) > is assigned a value "true" or "false", and the set of true sentences > is disjoint from the set of false sentences. Yes, but you're referring only to the non-degenerated structures! > > In the case of the empty structure (no elements), this assignment > is pretty simple: > > (1) Every sentence of the form "Ax Phi(x)" is assigned "true". > (2) Every sentence of the form "Ex Phi(x)" is assigned "false". > (3) There are no quantifier-free closed sentences. You're stipulating _one_ mapping between formulas and a set of 2 binary values. Is it the only way that make sense? And more to the point, would this way conform with Tarski's concept of a being true and being false, using set membership? > > What about sentences that are *not* closed? Well, to interpret > open sentences (ones with free variables), we also have to have > an assignment function (one that assigns an element of the domain > to each variable). In the case of the empty domain, there are > no assignment functions, so the empty domain cannot be extended > to give a truth value to open sentences. > > There is a subtle distinction between "true" and "valid" for > a structure. If Phi is an open sentence, then it is considered > "valid" for a structure if every assignment results in a sentence > that is true. Under this definition, since there are no assignments > for the empty structure, it follows vacuously that *every* formula > (and its negation) is valid for the empty structure. > > So, the set of *true* formulas for the empty domain is a consistent > set. The set of *valid* formulas for the empty domain is inconsistent. Except that a) it's supposed to be a set of binary (2) values that all formulas would get mapped to, not quaternary (4) values; and b) an inconsistent theory would have _all_ the formulas as its theorems: the purportedly "true" as well as the "false" ones! > > In any case, the open formula "x=x" is valid in every structure, > including the empty structure. The open formula "~(x=x)" is only > valid in the empty structure. The key question is how would we go from 4 to 2 and still make sense in term of set-membership (to satisfy Tarski's)? The answer is we can't. The way that would make sense the most is going from 4 to 1: because that'd would be consistent with the logic of non-membership of the empty set.
From: William Hughes on 28 May 2010 07:41 On May 28, 1:49 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Do you know what _NON_ membership mean? Does it mean "true" that > something is in an empty predicate-set? No it means "true" that nothing is in an empty predicate set. In any model (empty universe or not) one of There exists an x such that blue(x) and There does not exist an x such that blue(x) must be true. - William Hughes
From: Marshall on 28 May 2010 09:13
On May 26, 4:32 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > I'm not sure why I'm actually in this discussion with Nam, who has > shown an infuriating unwillingness to learn from others and a very sad > inability to understand the authors he reads. There's a minor typo here: you said "authors" when in fact he just quotes endlessly from the one book by the one guy. Marshall |