How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: MoeBlee on 28 Jun 2010 12:14 On Jun 27, 10:49 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 28, 2:11 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > On Jun 27, 1:25 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > On Jun 28, 12:44 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > On Jun 26, 6:38 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > The proof in the theory ZF-Inf > > > +~Inf that infinite sets do not exist encodes (among other things) > > > that models of PA cannot exist. > > > PROVE IT. > > Here is one possible approach. Consider the theory F that I had > defined in another post, where And what is the meta-theory you are using for defining and proving things about this theory F? > F = ZF - Inf + D=0, > > Where 0 is the null set and D is defined as > > D = {x: An(x not in P_n(0))} Prove (and please state the formal language, logic, and axioms in which you conduct this proof) that there exists a D such that for all x, we have x in D iff An x not in P_n(0). Unless you can do that, then your 'D' is not well defined. So, instead, I'd state your axiom as: AxEn x in P_n(0) > Here P_n(0) is power set operation iterated n times on 0, and P_0(0) = > P(0), P_1(0)=P(P(0)), etc. Clearly only hereditarily finite sets can > exist in models of F. Prove it. (And please do not waste my time by typing a bunch of stuff that does not conform to some ordinary technical definition of 'structure (model) for a language' and 'model of a theory'.) > I assume that the theory F can be interpreted in PA. Assume this has > been done. Why do you assume that? It might be true, but still it deserves to be shown. > I have a question for you: > > Do you think that the sentence D=0 should map to a PA-sentence that is > equivalent to ~Con(PA) (since D=0 basically asserts that "infinite > sets do not exist")? (1) As I noted, your 'D' is not yet demonstrated to be properly defined. So I take the axiom as: AxEn x in P_n(0) (2) What do you mean by "map a sentence to"? What PARTICULAR FUNCTION do you have in mind? But wait, actually don't bother. I'm not interested in performing exercises for you. If you have a proof of something, then just state your formal system, rules, and axioms, and show me the proof. (3) I'm not even interested in whatever sense you might mean "basically states that" "infinite sets do not exist". If you have a proof of something, let's see it. MoeBlee
From: MoeBlee on 28 Jun 2010 12:24 On Jun 28, 7:19 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > Which post by MoeBlee says anything about how to prove PA consistent > in ZFC? About a couple of years ago, I posted to you about it. As to this thread, like Alan, I recall recently posting on it but can't find the post in this thread (mabye it was in another recent thread?). In any case, you wrote to Aatu something to the effect that ZFC (by the way, Z-R is more than adequate) provides for N and then proves the axioms of PA about N (or something like that; I'm not representing that I'm accurately capturing your remark; but merely trying to remind you of the post). And that is essentially how Z-R proves the consistency of PA: In Z-R we define '0' (the empty set), 'w' (omega), 'S' (successor operation for omega), '+' (addition operation for omega), '*' (multiplication operation for omega). Then we prove, as theorems of Z-R, the axioms of PA. Thus <w 0 S + *> is a model of PA. And it's easy enough to see that if a theory has a model then that theory is consistent. And, of course, since Z-R is a subtheory of ZFC, perforce, anything Z- R proves also ZFC proves. MoeBlee
From: R. Srinivasan on 28 Jun 2010 12:54 On Jun 28, 8:53 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 27, 4:32 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > On Jun 28, 2:11 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > On Jun 27, 1:25 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > On Jun 28, 12:44 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > > On Jun 26, 6:38 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > The proof in the theory ZF-Inf > > > > +~Inf that infinite sets do not exist encodes (among other things) > > > > that models of PA cannot exist. > > > > PROVE IT. > > > Trivial. > > If it's trivial, then it shouldn't take you but a minute to prove it. > Please do. > > > > Look, in ordinary set theory such as Z set theory, we may prove that > > > there is a certain object that is a certain theory and we define the > > > rubric 'PA' to be that theory. > > > Sure. If we want a metatheory of PA in which it *is* possible to > > construct a model of PA (like Z set theory) then I do agree that we > > need all the machinery you describe. > > You're not listening. Theory Y (AF-I+~I) does not provide for > constructing the THEORY PA ITSELF. > So what? Are you so obtuse that you cannot understand the simple arguments that I presented to you? > > Why do you keep skipping this?: Y proves there are no infinite sets. > PA is an infinite set of sentences, moreover axiomatized by an > infinite set of axioms, moreover with a language that has an infinite > set of variables. So Y proves THERE IS NO SUCH theory. > You are utterly off target here. There is no way, it seems, to drill this very simple point into your head: "The theory PA exists" IS NOT EVEN A PROPOSITION in the language of Y. Then how can Y PROVE that "THERE IS NO SUCH theory" as you claim? However, the theory Y, being equivalent to PA, does encode sentences equivalent to the arithmetical propositions Con(PA) and ~Con(PA). Do you understand this? Assuming that you do, let us proceed. I am saying that the proof in Y that "infinite sets do not exist" is logically equivalent to the sentence ~Con(PA), WHICH IS A LEGITIMATE PROPOSITION OF Y (under suitable coding). Do you dispute this? If so present your arguments. If Y does not know what PA is, how is it possible that Y can encode sentences like Con(PA)? > > Please do not continue to argue PAST that very simple state of > affairs. > You are a simpleton. Try to get out of the box and do some thinking for a change. > > > >But in ZF-Inf +~Inf we may prove that > > > there does NOT exist any such object that meets the description of > > > being a first order theory, let alone one having the specifics > > > properties that, in Z, we ascribe to PA. So, in ZF-Inf +~Inf we cannot > > > even define the rubric 'PA' in a way that PA is a theory. > > > But now we are looking at a metatheory that only needs to *deny* that > > models of PA exist. > > Same as I just wrote above. > Same rebuttal. > > > So it does not need all the machinery that you > > describe to accomplish this. It is enough for this metatheory to deny > > the existence of infinite sets for it to rule out the existence of > > objects that can be interpreted as models of PA. > > Same as I just wrote above. > Same rebuttal > > > > > That is all I need to calll ZF-Inf > > > > +~Inf as a metatheory of PA. > > > > It can't be a very "meaningful" metatheory for PA since it proves that > > > there IS NO theory that meets the description we give, in such as Z, > > > to'PA'. > > > This and what you state below is again your own confusion. > > You've shown no confusion on my part. Saying "your confusion" and > "trivial" when asked for proofs of things is not substantive argument. > > Anyway, just go back to this: > > Y proves there are no infinite sets. PA is an infinite set of > sentences, moreover axiomatized by an infinite set of axioms, moreover > with a language that has an infinite set of variables. So Y proves > THERE IS NO SUCH theory. Y proves not only is there such a theory, but > there is not even such a LANGUAGE for such a theory. > Complete rubbish RS
From: MoeBlee on 28 Jun 2010 13:57 On Jun 28, 11:54 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 28, 8:53 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > On Jun 27, 4:32 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > On Jun 28, 2:11 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > On Jun 27, 1:25 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > > On Jun 28, 12:44 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > > > On Jun 26, 6:38 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > The proof in the theory ZF-Inf > > > > > +~Inf that infinite sets do not exist encodes (among other things) > > > > > that models of PA cannot exist. > > > > > PROVE IT. > > > > Trivial. > > > If it's trivial, then it shouldn't take you but a minute to prove it. > > Please do. NO ANSWER HERE. > > > > Look, in ordinary set theory such as Z set theory, we may prove that > > > > there is a certain object that is a certain theory and we define the > > > > rubric 'PA' to be that theory. > > > > Sure. If we want a metatheory of PA in which it *is* possible to > > > construct a model of PA (like Z set theory) then I do agree that we > > > need all the machinery you describe. > > > You're not listening. Theory Y (AF-I+~I) does not provide for > > constructing the THEORY PA ITSELF. > > So what? I EXPLAINED "so what". > Are you so obtuse that you cannot understand the simple > arguments that I presented to you? Oh goodie, now the 'obtuse' card. I DO understand that what you've presented is ill-premised, as I've EXPLAINED to you. > > Why do you keep skipping this?: Y proves there are no infinite sets. > > PA is an infinite set of sentences, moreover axiomatized by an > > infinite set of axioms, moreover with a language that has an infinite > > set of variables. So Y proves THERE IS NO SUCH theory. > > You are utterly off target here. Not at all. Deal with it: Y proves there are no infinite sets. A theory is an infinite set of sentences. So Y proves there are no theories. > There is no way, it seems, to drill > this very simple point into your head: > > "The theory PA exists" OF COURSE PA exists, as we prove formally in Z set theory, but NOT in Y. In Z set theory we may formally define 'first order theory', and all kinds of things, and then PROVE that there is a certain first order theory with certain properties, and we call that particular first order theory 'PA'. That happens in Z set theory or some other sufficiently strong formal theory (if we formalize) or just in informal set theoretic mathematical discussion (if we don't formalize). But it does NOT happen in Y. > IS NOT EVEN A PROPOSITION in the language of Y. YES IT IS, you ignoramus. The language of Y is the language of set theory (i.e., the 2-place relation symbol 'e' and the logical symbols). And in that language (here I'm only giving an informal English rendering), we may define 'PA' as "the set of sentences closed in the language with '0', '+', 'S', and '*', under the [such and such] a set of axioms'. > Then how can Y PROVE > that "THERE IS NO SUCH theory" as you claim? I TOLD YOU, oh obtuse one. Y proves there are no infinite sets. But a theory is an infinite set. So Y proves there are no theories. > However, the theory Y, > being equivalent to PA, They are equivalent in the sense that they can be interpreted one into the other. > does encode sentences equivalent to the > arithmetical propositions Con(PA) and ~Con(PA). Do you understand > this? I understand it a thousand times better than you. To be exact, there is a sentence S (we may call it "Con(PA)") in the language of PA such that S is true in the standard model for the language of PA iff PA is consistent. Notice all of the above is formulated in a META-theory about PA, presumably (or usally) a formal or informal set theory such as Z. > Assuming that you do, let us proceed. > > I am saying that the proof in Y that "infinite sets do not exist" is > logically equivalent to the sentence ~Con(PA), That's not even coherent. You're saying a PROOF is logically equivalent to a sentence. Probably what you mean is that one sentence (in the language of Y) "infinite sets do not exist" is LOGICALLY equivalent to another sentence (in the language of PA) "~Con(PA)". Then PROVE it. And, by the way, show a notion of LOGICAL equivalence of sentences from different languages (I think you could work up such a definition, but I'd like you to provide just for clarity.) And note (which I take you do understand, but I mention just in case), "PA" does not occur in the language of PA, so "~Con(PA)" is a NICKNAME for some actual sentence in the language of PA and thus "PA" does not occur in the ACTUAL sentence that "~Con(PA)" stands for. Moreover, WHATEVER you devise along these lines, it still would not refute the simple fact: Y proves "there does not exist an infinite set", so Y proves "there does not exist a set of sentences that is infinite" (since Y can express 'set of sentences'), and (and it can be proven IN Y) every theory is an infinite set of sentences, therefore Y proves there does not exist a theory. Just STOP for a minute right there. Whatever else about your "coding", the above argument stands. You cannot refute it. > WHICH IS A LEGITIMATE > PROPOSITION OF Y (under suitable coding). > Do you dispute this? If so present your arguments. If Y does not know > what PA is, how is it possible that Y can encode sentences like > Con(PA)? You've not shown a contradiction between what I've said and the theorem (proven in Z set theory, not proven in Y) that there is a sentence S of the language of PA (that can be interepreted in Y) such that S is true in the standard model of PA iff PA is consistent. If you claim there is a contradiction between what I said and the above then STATE the exact contradiction. State the exact theorem P of Z set theory such that what I've said is ~P. I'll lay it out for you ONE MORE TIME: Y is a formal theory (or you can even consider an informal counterpart of Y). Y proves there is no thing that that is not 1-1 with some natural number (i.e., there are no infinite sets). And the language of Y is the same language as ZFC (the language of set theory). And in this language we can define predicates such as 'is a theory', etc. and prove that if T is a theory then T is infinite (a certain kind of infinite set of sentences). So Y proves there is no theory (i.e. Y can define the PREDICATE 'is a theory' but also Y proves there is no object that has that predicate), so, perforce, Y proves there is no theory such that [...] where for [...] we list the properties used to define (IN Z) 'PA' (and note: Z defines 'PA' but Y does not, and that does not contradict that the PROPERTIES may be discussed in both theories while Z proves there is an object having such properties and Y proves there is no object having such properties). In Z set theory, whether formal or informal, (or some other suitable formal or informal theory, but not Y, because Y cannot even show there EXISTS a theory having the properties that we show (IN Z) PA to have) we prove that there is a sentence S in the language of PA (and which can be interpreted in Z) such that S is true in the standard model of PA iff PA is consistent (and so the negation of S is true in the standard model of PA iff PA is inconsistent). Now, if there is a contradiction in what I just said, then you need to state the system in which the claimed contradiction occurs and you would do well to show the actual contradiction as a formula and its negation. > > Please do not continue to argue PAST that very simple state of > > affairs. > > You are a simpleton. Try to get out of the box and do some thinking > for a change. Yes, that really proves your point. Actually it demonstrates that AGAIN you moved PAST the simple fact that Y proves there are no theories. > > > >But in ZF-Inf +~Inf we may prove that > > > > there does NOT exist any such object that meets the description of > > > > being a first order theory, let alone one having the specifics > > > > properties that, in Z, we ascribe to PA. So, in ZF-Inf +~Inf we cannot > > > > even define the rubric 'PA' in a way that PA is a theory. > > > > But now we are looking at a metatheory that only needs to *deny* that > > > models of PA exist. > > > Same as I just wrote above. > > Same rebuttal. No, you did NOT rebut me. If you just keep arguing by CLAIM and not by proof, then we're done. > > > So it does not need all the machinery that you > > > describe to accomplish this. It is enough for this metatheory to deny > > > the existence of infinite sets for it to rule out the existence of > > > objects that can be interpreted as models of PA. > > > Same as I just wrote above. > > Same rebuttal No, you did NOT rebut me. If you just keep arguing by CLAIM and not by proof, then we're done. > > > > > That is all I need to calll ZF-Inf > > > > > +~Inf as a metatheory of PA. > > > > > It can't be a very "meaningful" metatheory for PA since it proves that > > > > there IS NO theory that meets the description we give, in such as Z, > > > > to'PA'. > > > > This and what you state below is again your own confusion. > > > You've shown no confusion on my part. Saying "your confusion" and > > "trivial" when asked for proofs of things is not substantive argument. > > > Anyway, just go back to this: > > > Y proves there are no infinite sets. PA is an infinite set of > > sentences, moreover axiomatized by an infinite set of axioms, moreover > > with a language that has an infinite set of variables. So Y proves > > THERE IS NO SUCH theory. Y proves not only is there such a theory, but > > there is not even such a LANGUAGE for such a theory. > > Complete rubbish That is NOT a rebuttal. If what I wrote is "rubbish" then state which sentences are not correct: Y proves there are no infinite sets. PA is an infinite set of sentences, moreover axiomatized by an infinite set of axioms, moreove with a language that has an infinite set of variables. So Y proves there is no such theory. It's PLAIN: ~Ex x is infinite If x has [without mentioning 'PA', just fill in the PROPERTIES we would use to define 'PA'] then x is infinite Therefore, ~Ex x has [without mentioning 'PA', just fill in the PROPERTIES we would use to define 'PA'] And all those properties ARE expressible in Y, since they are only predicates, and all predicates expressible in Z are expressible in Y (since Z and Y have the same language, and definition of PREDICATES relies only on language, not on axioms), as, in particular, such things as 'first order language', 'theory', etc. ARE definable in the language of set theory (which is the language of Y and is the language of Z). It's only that in Z we CAN prove there is a unique (if we are specific enough about the symbols) object (a set) having those properties and in Y we prove there is NOT an object having those properties; so in Z we can define 'PA' as that unique object and in Y we cannot define 'PA' in that way because Y proves there is no such object having those properties. Now, don't be a pouting little baby and cry "rubbish" et. al. Instead, look at EXACTLY what I've written, then read a good textbook in mathematical logic. MoeBlee
From: R. Srinivasan on 28 Jun 2010 15:36
On Jun 28, 10:57 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 28, 11:54 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > On Jun 28, 8:53 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 27, 4:32 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > On Jun 28, 2:11 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > > On Jun 27, 1:25 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > > > On Jun 28, 12:44 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > > > > On Jun 26, 6:38 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > > The proof in the theory ZF-Inf > > > > > > +~Inf that infinite sets do not exist encodes (among other things) > > > > > > that models of PA cannot exist. > > > > > > PROVE IT. > > > > > Trivial. > > > > If it's trivial, then it shouldn't take you but a minute to prove it. > > > Please do. > > NO ANSWER HERE. > > > > > > Look, in ordinary set theory such as Z set theory, we may prove that > > > > > there is a certain object that is a certain theory and we define the > > > > > rubric 'PA' to be that theory. > > > > > Sure. If we want a metatheory of PA in which it *is* possible to > > > > construct a model of PA (like Z set theory) then I do agree that we > > > > need all the machinery you describe. > > > > You're not listening. Theory Y (AF-I+~I) does not provide for > > > constructing the THEORY PA ITSELF. > > > So what? > > I EXPLAINED "so what". > > > Are you so obtuse that you cannot understand the simple > > arguments that I presented to you? > > Oh goodie, now the 'obtuse' card. > > I DO understand that what you've presented is ill-premised, as I've > EXPLAINED to you. > > > > Why do you keep skipping this?: Y proves there are no infinite sets. > > > PA is an infinite set of sentences, moreover axiomatized by an > > > infinite set of axioms, moreover with a language that has an infinite > > > set of variables. So Y proves THERE IS NO SUCH theory. > > > You are utterly off target here. > > Not at all. Deal with it: Y proves there are no infinite sets. A > theory is an infinite set of sentences. So Y proves there are no > theories. > > > There is no way, it seems, to drill > > this very simple point into your head: > > > "The theory PA exists" > > OF COURSE PA exists, as we prove formally in Z set theory, but NOT in > Y. In Z set theory we may formally define 'first order theory', and > all kinds of things, and then PROVE that there is a certain first > order theory with certain properties, and we call that particular > first order theory 'PA'. That happens in Z set theory or some other > sufficiently strong formal theory (if we formalize) or just in > informal set theoretic mathematical discussion (if we don't > formalize). But it does NOT happen in Y. > > > IS NOT EVEN A PROPOSITION in the language of Y. > > YES IT IS, you ignoramus. > > The language of Y is the language of set theory (i.e., the 2-place > relation symbol 'e' and the logical symbols). And in that language > (here I'm only giving an informal English rendering), we may define > 'PA' as "the set of sentences closed in the language with '0', '+', > 'S', and '*', under the [such and such] a set of axioms'. > OK. I think I can restrict the language of Y such that no explicit references to any infinite sets are allowed. So I want the language of Y to only contain references to finite sets and the axiom that "Infinite sets do not exist" is stated in the form you mentioned: Ax En x in P_n(0) without any explicit reference to infinite sets (By the way, I think that the axiom D=0 which I wanted to add is very interesting in its own right, but for the time being, let us use your definition). If I do that you will not be able to define any object called 'PA' in the language of Y. The propositions in this language can only contain explicit references to finite sets. But I do not even need to do this. Let me play along with you. Suppose that Y proves that there is no infinite object called PA. It still does not undermine my argument in any way, as I will argue below. > > > Then how can Y PROVE > > that "THERE IS NO SUCH theory" as you claim? > > I TOLD YOU, oh obtuse one. Y proves there are no infinite sets. But a > theory is an infinite set. So Y proves there are no theories. > > > However, the theory Y, > > being equivalent to PA, > > They are equivalent in the sense that they can be interpreted one into > the other. > > > does encode sentences equivalent to the > > arithmetical propositions Con(PA) and ~Con(PA). Do you understand > > this? > > I understand it a thousand times better than you. > > To be exact, there is a sentence S (we may call it "Con(PA)") in the > language of PA such that S is true in the standard model for the > language of PA iff PA is consistent. > > Notice all of the above is formulated in a META-theory about PA, > presumably (or usally) a formal or informal set theory such as Z. > > > Assuming that you do, let us proceed. > > > I am saying that the proof in Y that "infinite sets do not exist" is > > logically equivalent to the sentence ~Con(PA), > > That's not even coherent. > > You're saying a PROOF is logically equivalent to a sentence. > > Probably what you mean is that one sentence (in the language of Y) > "infinite sets do not exist" is LOGICALLY equivalent to another > sentence (in the language of PA) "~Con(PA)". > There is a sentence in the language of Y that encodes ~Con(PA). What I am saying is that the sentence "infinite sets do not exist" is equivalent to this sentence. Yes, there was a typo in what I said above (although I do believe truth is provability, but for the time being, let me stay within classical thinking). > > Then PROVE it. And, by the way, show a notion of LOGICAL equivalence > of sentences from different languages (I think you could work up such > a definition, but I'd like you to provide just for clarity.) > > And note (which I take you do understand, but I mention just in case), > "PA" does not occur in the language of PA, so "~Con(PA)" is a NICKNAME > for some actual sentence in the language of PA and thus "PA" does not > occur in the ACTUAL sentence that "~Con(PA)" stands for. > > Moreover, WHATEVER you devise along these lines, it still would not > refute the simple fact: > > Y proves "there does not exist an infinite set", so Y proves "there > does not exist a set of sentences that is infinite" (since Y can > express 'set of sentences'), and (and it can be proven IN Y) every > theory is an infinite set of sentences, therefore Y proves there does > not exist a theory. > > Just STOP for a minute right there. Whatever else about your "coding", > the above argument stands. You cannot refute it. > OK. So what? It does not undermine the point I want to make. As I said, let me play along with you. Here is what I want to drill into your head: 1. The theory PA can be interpreted in Y. 2. Y formalizes sentences like Con(PA) (via suitable encoding). 3. Y denies the existence of infinite sets. Basically these three points are to be understood as follows. Y allows all the arithmetic operations of PA, Y understands what it means for PA to be consistent (or for a model of PA to exist) and Y denies the existence of infinite sets. These three facts together mean that Y proves ~Con(PA) in a meaningful way, for Y denies that models of PA can exist. Now let us get to your point. Y does not recognize any explicit object that stands for the theory 'PA' and even denies its existence. Perfectly in tune with the fact that Y proves ~Con(PA). For if PA is inconsistent, infinite sets cannot exist and indeed there cannot exist any infinite objects like the theory 'PA' from Y's perspective. So what? In what way does this undermine my argument above based on the 3 points noted above? The point is that Y understands what Con(PA) is via encoding. It does not need to recognize any explicit object called 'PA' to prove ~Con(PA) (or its encoding in the language of Y). Kindly keep your focus on the 3 points I mentioned above and avoid nitpicking. > > Now, don't be a pouting little baby and cry "rubbish" et. al. Instead, > look at EXACTLY what I've written, then read a good textbook in > mathematical logic. > > This is your problem. You want to go by the book, but here you need to think out of the box. I certainly do not want to get dragged into the same mess that you have fallen into. RS |