Tristan L 3 Posted February 6 Share Posted February 6 (edited) To me, axiom-systems seem to basically be ownships (properties). For instance, the group-axiom-system is basically the ownship of being an ordered pair \((G, *)\) such that \(G\) is a set and \(*\) is a function from \(G\times G\) to \(G\) such that \(*\) is associative and has an identity element and each member of \(G\) has an inverse element with regard to \(*\). Just as the axiom-system itself is an ownship, so are what are called “propositions in the language/speech of the system” actually properties. For instance, when we say: “The proposition that the sum of the inner angles of a triangle is always 180° follows from the Euclidean axioms“, we actually mean that for every structure E, if E has the Euclidean axiom-system as a property, then E has the property that every triangle in it has an inner-angle-sum of 180°. Some axiom-systems, such as the group-axiom-system and the field-axiom-system, are had by several structures of which not all are isomorphic to each other. In other words, such axiom-systems have at least two models which aren’t isomorphic to each other. Let’s call these axiom-systems “not-characterizing”. Others, such as the Dedekind-Peano-axiom-system (called “DP” henceforth) , are only had by one structure up to isomorphy – they have a model, and all their models are isomorphic to each other. Let’s call these axiom-systems “characterizing”. The only model of DP up to isomorphy (and indeed up to unique isomorphy), th.i. the only entity, up to (unique) isomorphy, which has DP as a property, is the structure of the natural numbers. The German mathematician Richard Dedekind showed this in 1888 with his Theorem of Isomorphy. Now, there are two ways in which a ‘proposition’ \(P\) (in reality a property, see above) can be undecidable (neither provable nor disprovable) in an axiom-system \(AS\). One way is that \(AS\) has several non-isomorphic models of which some have \(P\) and others don’t. For instance, being unendly (infinite) isn’t decidable from the field-axioms since there are finite (endly) fields as well as unendly ones. This underkind of undecidability is, I think, obviously not very interesting. The Continuum-Hypothesis (CH) is one example, for it’s true in some models of ZFC and false in others. Here, the lack of our knowledge stems from the fact (deedsake) that ZFC doesn’t contain all information about the set-universe to start with. The second way in which \(P\) can be undecidable in \(AS\) is that it either holds in all models of \(AS\) or its negation holds in all models of \(AS\), and yet we still can derive neither one from \(AS\) because our logical (witcrafty) tools are too weak. This reflects a true unableness on our part to get info out of \(AS\) which nevertheless is there. This is always the case if \(AS\) is characterizing (has only one model up to isomorphy). Such an axiom-system contains all info about its model, so undecidability in it means inability to get info which is nonetheless there. So while undecidableness in ZFC need not be interesting, undecidableness in DP always is. Since characterizing axiom-systems have just one model up to isomorphy, what we call “propositions in their speech” – in truth ownships – can actually be regarded as propositions, namely the propositions resulting from predicating those properties of the system’s unique model. Now, from Gödel’s Incompleteness Theorems, we know that there are undecidable propositions in DP, that is, not every statement about the naturals can be shown or disproven. However, my question is this: Is there an individual proposition about the naturals of which we know that we can neither prove it nor disprove it in DP? Edited February 6 by Tristan L had to change LaTeX-syntax since display-environment not known 1 Link to post Share on other sites

joigus 370 Posted February 6 Share Posted February 6 (edited) Cohen's proof that it is undecidable whether there is a cardinality (number of elements) intermediate between that of \( \mathbb{N} \) and that of \( \mathbb{R} \). https://en.wikipedia.org/wiki/Continuum_hypothesis "Undecidable" is perhaps my colourful way of saying it. People seem to prefer the wording: Quote The continuum hypothesis was advanced by Georg Cantor in 1878, and establishing its truth or falsehood is the first of Hilbert's 23 problems presented in 1900. The answer to this problem is independent of ZFC, so that either the continuum hypothesis or its negation can be added as an axiom to ZFC set theory, with the resulting theory being consistent if and only if ZFC is consistent. Edited February 6 by joigus Link to post Share on other sites

Tristan L 3 Posted February 6 Author Share Posted February 6 (edited) 1 hour ago, joigus said: Cohen's proof that it is undecidable whether there is a cardinality (number of elements) intermediate between that of N and that of R . https://en.wikipedia.org/wiki/Continuum_hypothesis Yes, CH is undecidable in / independent of ZFC, but that's because ZFC has several not-isomorphic models: 4 hours ago, Tristan L said: One way is that AS has several non-isomorphic models of which some have P and others don’t. For instance, being unendly (infinite) isn’t decidable from the field-axioms since there are finite (endly) fields as well as unendly ones. This underkind of undecidability is, I think, obviously not very interesting. The Continuum-Hypothesis (CH) is one example, for it’s true in some models of ZFC and false in others. Here, the lack of our knowledge stems from the fact (deedsake) that ZFC doesn’t contain all information about the set-universe to start with. I'm asking whether there is a proposition which is known to be independent of an axiom-system with only one model (up to isomorphy), such as the Dedekind-Peano-axioms. Edited February 6 by Tristan L Link to post Share on other sites

joigus 370 Posted February 6 Share Posted February 6 25 minutes ago, Tristan L said: Yes, CH is undecidable in / independent of ZFC, but that's because ZFC has several not-isomorphic models: I'm asking whether there is a proposition which is known to be independent of an axiom-system with only one model (up to isomorphy), such as the Dedekind-Peano-axioms. Oh, boy. I'm sorry. I should have read you out. You are much more knowledgeable on this than I am. But your comments reinforce my impression that Gödel's theorem does not provide much in the way of a constructive process, even though the proof itself is constructive. I think what you mean is something like theorems in number theory, propositions on prime numbers, and the like, right? I simply don't know. I've never heard of any of those. That is a very good question. Maybe @wtf has something interesting to say about it. 1 Link to post Share on other sites

Tristan L 3 Posted February 6 Author Share Posted February 6 16 minutes ago, joigus said: I think what you mean is something like theorems in number theory, propositions on prime numbers, and the like, right? Yes, exactly. However, the conjecture that there are odd perfect numbers is not what I seek, for since we know that (if it is true, we can show that it's true), we know that (if it's undecidable, it is false), so if we knew that it is undecidable, we'd know that it's false, making it decidable after all and thus leading to a contradiction. I seek a proposition \(A\) 24 minutes ago, joigus said: in number theory, propositions on prime numbers, and the like such that we know that we can't know whether \(A\) is true or false. Link to post Share on other sites

wtf 152 Posted February 6 Share Posted February 6 (edited) 8 hours ago, joigus said: Maybe @wtf has something interesting to say about it. I'll take a run at this. I think the key issue is that there is no completeness theorem for second-order logic. In general the OP's post doesn't mention the key distinctions between first and second order logic. OP should give this a read, perhaps there is some insight to be had. https://plato.stanford.edu/entries/logic-higher-order/ In particular, note section 5.2: "We shall now see that there is no hope of a Completeness Theorem for second-order logic." 13 hours ago, Tristan L said: To me, axiom-systems seem to basically be ownships (properties). For instance, the group-axiom-system is basically the ownship of being an ordered pair (G,∗) such that G is a set and ∗ is a function from G×G to G such that ∗ is associative and has an identity element and each member of G has an inverse element with regard to ∗ . Ok, that's a funny way of putting it but I know what you mean. 13 hours ago, Tristan L said: Just as the axiom-system itself is an ownship, so are what are called “propositions in the language/speech of the system” actually properties. For instance, when we say: “The proposition that the sum of the inner angles of a triangle is always 180° follows from the Euclidean axioms“, we actually mean that for every structure E, if E has the Euclidean axiom-system as a property, then E has the property that every triangle in it has an inner-angle-sum of 180°. Ok. But you are conflating syntax (axiom systems) and semantics (models). That is, the 180 degree theorem follows syntactically from the axioms of Euclidean geometry. That's a purely syntactic fact. And it's also the case that the theorem is true in any model of those axioms. That's a syntactic fact. Two different things. In first-order logic, a theorem is provable (syntax) if and only if it's true in every model of the axioms (semantics). This is Gödel's completeness theorem. But beware, as the SEP article I linked above indicates, there is no completeness theorem for second-order logic. I confess to not knowing much about the fine points of second-order logic, but I suspect the OP's questions relate to this fact, that second order logic does not have a completeness theorem. 13 hours ago, Tristan L said: Some axiom-systems, such as the group-axiom-system and the field-axiom-system, are had by several structures of which not all are isomorphic to each other. In other words, such axiom-systems have at least two models which aren’t isomorphic to each other. Let’s call these axiom-systems “not-characterizing”. Others, such as the Dedekind-Peano-axiom-system (called “DP” henceforth) , are only had by one structure up to isomorphy – they have a model, and all their models are isomorphic to each other. Let’s call these axiom-systems “characterizing”. The only model of DP up to isomorphy (and indeed up to unique isomorphy), th.i. the only entity, up to (unique) isomorphy, which has DP as a property, is the structure of the natural numbers. The German mathematician Richard Dedekind showed this in 1888 with his Theorem of Isomorphy. The terminology is that an axiom system with a unique model up to isomorphism is categorical. And one that has non-isomorphic models is non-categorical. OP, please read this. https://en.wikipedia.org/wiki/Categorical_theory Now the first-order theory of Peano arithmetic most definitely does have nonstandard models. For example the hyperintegers of the hyperreal numbers are a nonstandard model of PA. They include the usual finite natural numbers 0, 1, 2, 3, ... as well as all the infinite ones. However the second-order theory of PA is categorical, and perhaps that's what you are referring to as Dedekind's theorem. So this is an example of where you need to distinguish between first and second order theories. There ARE non-categorical models of first-order PA; but none of second-order PA. This I believe is the crux of your concern, if I'm understanding your post. 13 hours ago, Tristan L said: Now, there are two ways in which a ‘proposition’ P (in reality a property, see above) can be undecidable (neither provable nor disprovable) in an axiom-system AS . One way is that AS has several non-isomorphic models of which some have P and others don’t. For instance, being unendly (infinite) isn’t decidable from the field-axioms since there are finite (endly) fields as well as unendly ones. This underkind of undecidability is, I think, obviously not very interesting. The Continuum-Hypothesis (CH) is one example, for it’s true in some models of ZFC and false in others. Here, the lack of our knowledge stems from the fact (deedsake) that ZFC doesn’t contain all information about the set-universe to start with. Yes, but note that the real numbers axiomatized as an infinite, complete ordered field is categorical. That's because completeness is a second-order property. It quantifies over subsets of the real numbers, not just individuals. That is, the least upper bound property says that all nonempty subsets of the reals that are bounded above have a least upper bound. Since we have quantified over subsets, that's a second-order property, and then it turns out that the second-order reals are categorical. 13 hours ago, Tristan L said: The second way in which P can be undecidable in AS is that it either holds in all models of AS or its negation holds in all models of AS , and yet we still can derive neither one from AS because our logical (witcrafty) tools are too weak. This reflects a true unableness on our part to get info out of AS which nevertheless is there. This is always the case if AS is characterizing (has only one model up to isomorphy). Such an axiom-system contains all info about its model, so undecidability in it means inability to get info which is nonetheless there. So while undecidableness in ZFC need not be interesting, undecidableness in DP always is. Since characterizing axiom-systems have just one model up to isomorphy, what we call “propositions in their speech” – in truth ownships – can actually be regarded as propositions, namely the propositions resulting from predicating those properties of the system’s unique model. This is where my knowledge ends. Since there's no completeness theorem for second-order logic, there must be some axiom system in which something is a theorem that's not true in all models, or true in all models but not a theorem. I don't know any specific examples or anything more about it. EDIT -- this isn't right, see below 13 hours ago, Tristan L said: Now, from Gödel’s Incompleteness Theorems, we know that there are undecidable propositions in DP, that is, not every statement about the naturals can be shown or disproven. However, my question is this: Is there an individual proposition about the naturals of which we know that we can neither prove it nor disprove it in DP? You're asking if there are undecidable statements in the categorical theory of second-order PA. This I do not know. A related question of interest is whether there are "natural" statements of arithematic that are undecidable (presumably in first-order PA). Harvey Friedman has been searching for examples of such. Here's an overview. https://plus.maths.org/content/picking-holes-mathematics Well that's what I know about it, hope something in here was helpful. The key is that first-order theories are never categorical and second-order theories sometimes are. But there's no completeness theorem for second-order theories, leading to all the aspects of this that I don't know, and to the question you're asking. ps -- Aha, a clue. I was perusing the Wiki article on categorical theories at https://en.wikipedia.org/wiki/Categorical_theory and it says: Every categorical theory is complete. However, the converse does not hold So this tells us that IF a theory is categorical (has only one model up to isomorphism) THEN every theorem provable from the axioms is true in every model of those axioms. But if the converse fails, then there is an axiomatic system with a statement that is true in all models of the axioms, but that is not provable from the axioms. Well, "today I learned!" pps -- The Wiki footnote leads to Carl Mummert's answer to this math.SE question: https://math.stackexchange.com/questions/933466/difference-between-completeness-and-categoricity/933632#933632 Mummert gives a concrete example of a comple, noncategorical theory. This doesn't seem to be the same thing as what we are looking for, a categorical, non-complete theory. There may well be clues on this page though. Reading a little more of Mummert's response, he's using "complete" in a different sense, that every statement is either provable or its negation is. That's not the same kind of completeness as in Gödel's completeness theorem, which says that a statement is provable if and only if it's true in every model. These two subtly different meanings of complete are yet another confusing aspect of all of this. So I was wrong about what it means for a theory to be categorical but not complete. Nevermind that part. These are deep waters and I got in a little over my head. From the Wiki article on completeness: https://en.wikipedia.org/wiki/Complete_theory Quote In mathematical logic, a theory is complete if, for every closed formula in the theory's language, that formula or its negation is demonstrable. Recursively axiomatizable first-order theories that are consistent and rich enough to allow general mathematical reasoning to be formulated cannot be complete, as demonstrated by Gödel's first incompleteness theorem. This sense of complete is distinct from the notion of a complete logic, which asserts that for every theory that can be formulated in the logic, all semantically valid statements are provable theorems (for an appropriate sense of "semantically valid"). Gödel's completeness theorem is about this latter kind of completeness. Edited February 6 by wtf 2 Link to post Share on other sites

studiot 2146 Posted February 7 Share Posted February 7 13 hours ago, wtf said: I'll take a run at this. I think the key issue is that there is no completeness theorem for second-order logic. In general the OP's post doesn't mention the key distinctions between first and second order logic. Thank you so much for such a comprehensive and well thought out addition to the discussion , including the links. +1 This will take much thoughtful chewing over and investigating. Link to post Share on other sites

Tristan L 3 Posted February 7 Author Share Posted February 7 Thank you very much for your detailed and informative (informatul) answer! 👍 18 hours ago, wtf said: Ok. But you are conflating syntax (axiom systems) and semantics (models). Or rather, I was only talking about semantics. I like to think of an axiom-system as standing for an ownship, and of being a model of that axiom-system as having that ownship. 18 hours ago, wtf said: Two different things. Right. 18 hours ago, wtf said: I suspect the OP's questions relate to this fact, that second order logic does not have a completeness theorem. Yes, it does. 18 hours ago, wtf said: However the second-order theory of PA is categorical, and perhaps that's what you are referring to as Dedekind's theorem. Yes, and to avoid confusion with the much too weak first-order theory, I called the second-order theory "DP". Dedekind's Isomorphy Theorem indeed shows the categoricity of DP. 18 hours ago, wtf said: This I believe is the crux of your concern, if I'm understanding your post. True. 18 hours ago, wtf said: Yes, but note that the real numbers axiomatized as an infinite, complete ordered field is categorical. You're right; instead of taking not-categorical first-order ZFC as our basis, we take categorical second-order analysis as our basis, with the true set-theory (about which ZFC doesn't give enough info) as the meta-logical (over-witcrafty) framework of second-step logic. Then CH is indeed a proposition of the kind that I was searching for. 19 hours ago, wtf said: Every categorical theory is complete. However, the converse does not hold As I understand it, the word "complete" as brooked (used) here means what I'll call "semantic completeness": every sentence expressible in the speech is either true in all models or untrue in all models, as opposed to what I'll call "syntactic completeness": every sentence expressible in the speech is either derivable from the axioms with the logical calculus or its negation is. 19 hours ago, wtf said: every theorem provable from the axioms is true in every model of those axioms. Well, that has to be so if the logical calculus is right, which it should. I think that semantic completeness rather than syntactic completeness is meant, though, which makes sense: If the axiom-system has only one model, every sentence is either true in all models or it's false in all models. 19 hours ago, wtf said: Well that's what I know about it, hope something in here was helpful. Most certainly your whole answer is! 😃 19 hours ago, wtf said: You're asking if there are undecidable statements in the categorical theory of second-order PA. This I do not know. Yes, and more broadly, I was searching for statements which are semantically decidable in (either true in all models of or untrue in all models of) some axiom-system, but not syntactically decidable in the axiom-system. Am I right in thinking that these are the ones that show true lack of being able to know on our part? Link to post Share on other sites

Col Not Colin 9 Posted February 7 Share Posted February 7 Hi everyone. Firstly, thanks to @wtf who has provided plenty of reading material. It'll take me a few days to browse through that lot. Admitting that I haven't read much of those links, what follows is just an initial reaction or speculation: On 2/6/2021 at 10:18 AM, Tristan L said: However, my question is this: Is there an individual proposition about the naturals of which we know that we can neither prove it nor disprove it in DP? What do you mean by a proposition about the naturals? Does it ONLY have to involve the naturals, or can the proposition just involve one natural number somewhere? For example: Take a typical undecideable statement (like something examined by Godel), let's call that proposition P, and just add an obviously provable statement like Q = "2+2 = 4". Then, using ^ as the logical connective for AND, the statement P_{ }_{^} Q would be a new statement that is undecideable and involves natural numbers. Link to post Share on other sites

wtf 152 Posted February 7 Share Posted February 7 (edited) 4 hours ago, Tristan L said: 23 hours ago, wtf said: I suspect the OP's questions relate to this fact, that second order logic does not have a completeness theorem. Yes, it does. I'm not sure what you mean by this. I linked and quoted a SEP article saying there is no completeness theorem for second-order logic. Do you have a reference to the contrary? I admit I'm no specialist on this subject. Or is "yes it does" referring to what your question relates to? Edited February 7 by wtf Link to post Share on other sites

studiot 2146 Posted February 7 Share Posted February 7 8 minutes ago, wtf said: I'm not sure what you mean by this. I linked and quoted a SEP article saying there is no completeness theorem for second-order logic. Do you have a reference to the contrary? I admit I'm no specialist on this subject. Or is "yes it does" referring to what your question relates to? Actually page 45 of the stanford link you provided (I have already thanked you for the link, paid my $10 and downloaded the pdf ) says rather more, particularly on page 45 et seq. As always the devil is in the detail. Link to post Share on other sites

wtf 152 Posted February 8 Share Posted February 8 (edited) 1 hour ago, studiot said: Actually page 45 of the stanford link you provided (I have already thanked you for the link, paid my $10 and downloaded the pdf ) says rather more, particularly on page 45 et seq. As always the devil is in the detail. The SEP article doesn't have page numbers. Can you please point me to what you're referring to? https://plato.stanford.edu/entries/logic-higher-order/ Early on it says, "The situation changed somewhat when Henkin proved the Completeness Theorem for second-order logic with respect to so-called general models (§9). Now it became possible to use second-order logic in the same way as first order logic, if only one keeps in mind that the semantics is based on general models." Whatever a "general model" is. Maybe I'll read it later. Edited February 8 by wtf Link to post Share on other sites

Tristan L 3 Posted February 8 Author Share Posted February 8 10 hours ago, wtf said: I'm not sure what you mean by this. Oh, only now do I realize that what I said was ambiguous. 10 hours ago, wtf said: Or is "yes it does" referring to what your question relates to? Yes, it does. If I wanted to contradict you regarding second-step logic, I'd have said: "No, it does". The problem is that the English speech doesn't have an equivalent of German "doch", French "si", and Arabic "بَلَى" (and in fact, none of these three European languages has an equivalent of Arabic "كَلَّا"). So I guess that my question has been half-answered: Since second-order witcraft is incomplete, there are undecidable statements in it, of which CH is an example. However, while we know that second-order logic will always stay incomplete (if only finite proofs are allowed), we might some day find a further axiom that applies to the true set universe with which to decide CH. 14 hours ago, Col Not Colin said: Then, using ^ as the logical connective for AND, the statement P_{ }_{^} Q would be a new statement that is undecideable and involves natural numbers. That's right. I should have been more specific: a statement expressible in the speech of the Dedekind-Peano axioms, th.i. second-step or some higher-step logic. 14 hours ago, Col Not Colin said: What do you mean by a proposition about the naturals? Does it ONLY have to involve the naturals Yes, that's what I mean, though sets of naturals and sets of sets of naturals and so on are also allowed. Link to post Share on other sites

studiot 2146 Posted February 8 Share Posted February 8 10 hours ago, wtf said: The SEP article doesn't have page numbers. Can you please point me to what you're referring to? I suppose that depends upon which presentation you download did you download the HTML version ? I said I downloaded the pdf presentation, which comes in two flavours. US and A4 paper sizes. I chose A4, which is clearly paginated, as are all pdf documents as far as I know. Anyway, since I can't copy and paste without loosing the the symbols from the pdf, here are a couple of screenshots, note the pagination at the bottom corners. Thank you again for the reference excellent and seems bang up to date as practicable. Link to post Share on other sites

Col Not Colin 9 Posted February 8 Share Posted February 8 Puzzlement: This is extremely unusual. The language used by Tristan L is sometimes extra-ordinarily precise but it is frequently unusual. Some examples: "ownships" used in the OP instead of properties; A hyperlink to witcrafty that seems almost irrelevant. On brief inspection witcrafty seems to be a website about the formation of words. Brief inspection of some other posts Tristan has created or contributed to include one about entropy where an alternative way to patition microstates was presented that seemed to involve runes and the ability to communicate with runes. Other recent posts on this forum include one about notation studies (not by Tristan but by another new member, whose background is unknown). Speculation: @Tristan L are you more interested in notation than anything else? Are you trialing software that analyses and then utilises notation, perhaps software developed in some ML environment? What was the witcrafty hyperlink about? I can't understand how to turn off tracking cookies and disable other features of that website, so I haven't examined anything in it too deeply. None of this essential, it's probably just me overthinking the situation. Best wishes to you. Bye for now. Link to post Share on other sites

wtf 152 Posted February 8 Share Posted February 8 (edited) 8 hours ago, studiot said: I suppose that depends upon which presentation you download did you download the HTML version ? You paid ten dollars to download a pdf? When you click on the article it comes up in your browser for free. You don't need to download anything. I don't even see a way to download a pdf, let alone pay for one. I am confused. Anyway thanks for the link. I did go back and read section 9 of the SEP article and did not understand a word of it. It seems to be fairly advanced mathematical logic. But there does seem to be a completeness theorem for "general" models, which is apparently a technical term that lets you get a completeness theorem but has some drawbacks. Edited February 8 by wtf Link to post Share on other sites

studiot 2146 Posted February 8 Share Posted February 8 44 minutes ago, wtf said: You paid ten dollars to download a pdf? I like to have my own copy of things. But I also thought you would understand more. I don't know where you found that article, but it is part of the Stanford Encyclopedia, which contains many many articles, with a choice of formats. Paid members may download any of these. Like our member with the cigar, I support Wikipedia with a few dollars a year and thought I would like to support the Standford project with a few dollars more. Neither are greedy unlike some sites. 2 hours ago, Col Not Colin said: "ownships" used in the OP instead of properties; yes, I was going to ask about this. But I don't see any conspiracy background, just a member whose first language is not English struggling with a translator. 🙂 51 minutes ago, wtf said: I did go back and read section 9 of the SEP article and did not understand a word of it. It seems to be fairly advanced mathematical logic. But there does seem to be a completeness theorem for "general" models, which is apparently a technical term that lets you get a completeness theorem but has some drawbacks. Yes that is also my understanding so far. Link to post Share on other sites

wtf 152 Posted February 8 Share Posted February 8 (edited) 44 minutes ago, studiot said: I support Wikipedia with a few dollars a year and thought I would like to support the Standford project with a few dollars more. Most praiseworthy, and you put my freeloading self to shame. I should mention though that according to Google, Stanford University's endowment is 27.7 billion USD. Now 27.7B and ten dollars 😉 Also I went to Cal, and therefore am required to consider Stanford my mortal enemy, at least one day a year when they play football in a rivalry that goes back to 1892. https://en.wikipedia.org/wiki/Big_Game_(American_football) Edited February 8 by wtf Link to post Share on other sites

Tristan L 3 Posted February 9 Author Share Posted February 9 20 hours ago, Col Not Colin said: are you more interested in notation than anything else? No, it's just a side-hobby of mine to help free speeches broadly, and the English tongue in particular, from the effects of linguistic imperialism, which I do by brooking (using) truly English words instead of ones that came into this language by way of speech-imperialism in actual natural situations, such as forum talks (mutatis mutandis for other victims of language-imperialism). Speech in general interests me, too, but not more that what it is brooked to stand for (which, spellbindingly, includes speech itself). 20 hours ago, Col Not Colin said: What was the witcrafty hyperlink about? The website to which the link leads is the English Wordbook. It is part of the Anglish Moot, which is all about liberating English from the effects of linguisic imperialism. The English Wordbook gives the right English equivalents to un-English words that came in through language-imperialism, to a big part, but not only, due to the Norman Conquest. For many proper English words which have not yet become widely used, when I brook them, I link their first instances in a text to the Wordbook so as to back my brooking of those words up, and I also write their foreign-derived equivalents in brackets after them (or vice versa). Brooking truly English words has the futher boot that they are usually shorter and thus more efficient that their foreign counterparts. For instance, "boot" is only a third as long (in terms of syllables) and therefore thrice as efficient as foreign-derived "advantage", and "often" has only two syllables while "frequently" has three. "Witcraft" is onefoldly (simply) the English synonym for "logic", the former being drawn from "wit" (from OE "witt", from PIE "*weyd-") and "craft" (from OE "cræft", from PG "*kraftuz"), while the latter is derived from OG "logikḗ" (from PIE "*leǵ-"). Another properly English word for witcraft is "flitecraft", which stems from the Old English word "flītcræft" for witcraft. For more truly English techical language (craftspeak), see the Anglish Moot's leaf on Craftspeak. 21 hours ago, Col Not Colin said: "ownships" used in the OP instead of properties Yes, and after my first use of that word in my post, a fayed (added) "(properties)" to clarify for those not yet/anymore familiar with "ownship". This word is the exact English equivalent of the German (Þeech) word "Eigenschaft" for ownship; "own" and "eigen" both come from Ortheedish (Proto-Germanic) "*aiganaz", and "-ship" and "-schaft" both come from Orþeedish "*-skapiz", which is closely related (beteed) to the forebear of English "shape". 21 hours ago, Col Not Colin said: Other recent posts on this forum include one about notation studies (not by Tristan but by another new member, whose background is unknown). And who bears no beteeing (relationship) to me (although the flitecrafta/logician would remind us that strictly speaking, everything bears some beteeing to everything else, e.g. the relation of standing to each other in such a way that 1=1) 😉. 21 hours ago, Col Not Colin said: Brief inspection of some other posts Tristan has created or contributed to include one about entropy where an alternative way to patition microstates was presented that seemed to involve runes and the ability to communicate with runes. Huh?! That post is about the fact (deedsake) that for any current microstate, you can partition phase space in such a way that the current entropy with regard to that partition (for entropy is always relative to a partition of phase space) is low. Hence, I argue there (and please set me right if I'm wrong), there's almost always something interesting going on (often perhaps even life), though it is likely that the lifeforms of one partition can only detect what is interesting with resprect to their own partition of phase-space. I just used particle-configurations that look like runes as an example; I could just as well have chosen tennis-rackets or Chinese characters or chess-pieces or whatever has structure. This shows us how weighty it is to read (or listen) and understand before one judges 😉. 21 hours ago, Col Not Colin said: I can't understand how to turn off tracking cookies and disable other features of that website, so I haven't examined anything in it too deeply. It's part of the great platform https://www.fandom.com/, formerly called "Wikia", which has been co-made by the co-maker Jimmy Wales of Wikipedia. So what's the big deal? 20 hours ago, studiot said: But I don't see any conspiracy background, Those who thought that there was any conspiracy, be careful ⚠️ lest you become conspiracy-theorists. 20 hours ago, studiot said: just a member whose first language is not English struggling with a translator. 🤣 Judging the canship (ability) or knowledge of another as wanting can result from one's own want of canship or knowledge, as seems to be the case here. Dost ðou even know ðy own speech? For instance, ðou shouldst not address an individual as "you", but as "ðou", for "you" is the plural accusative and dative (many-tale whon-case and whom-case) form of ðe English second personal pronoun, akin to Þeech "euch", while "ðou" is its singular nominative (one-tale who-case) shape, akin to German "du". Ðe sentence "Alice, can you please help me?" translates into Þeech literally as "Adelheid, kann euch mir bitte helfen?", which is so bad German ðat it would not be easy to even understand. Didst ðou know ðat? Sadly, the fair English speech has been messed up quite a bit, but we should at least try to tidy it up again. However, all that has little to do with the topic at hand, and I'd rather discuss it in a speechlore (linguistics) forum rather that a mathematics one. Link to post Share on other sites

wtf 152 Posted February 9 Share Posted February 9 (edited) 6 hours ago, Tristan L said: lest you become conspiracy-theorists. "I'm not a conspiracy theorist. I'm a conspiracy analyst." -- Gore Vidal 6 hours ago, Tristan L said: However, all that has little to do with the topic at hand, Yes, so how did you get from linguistic imperialism to the fine points of the second-order completeness theorem for Henkin models? Thanks for explaining witcraft. Most English-speaking readers probably took that as witchcraft, which made no sense in context. Edited February 9 by wtf Link to post Share on other sites

Col Not Colin 9 Posted February 9 Share Posted February 9 (edited) On 2/6/2021 at 10:18 AM, Tristan L said: Now, from Gödel’s Incompleteness Theorems, we know that there are undecidable propositions in DP, that is, not every statement about the naturals can be shown or disproven. However, my question is this: Is there an individual proposition about the naturals of which we know that we can neither prove it nor disprove it in DP? Hi again. Godel has been mentioned everywhere so I assume you are aware of Godel's second incompleteness theorem. I'll para-phrase it here: Where P is the Peano arithemetic system, we have the result that the consistency of that system cannot be proved (unless P is actually inconsistent). This provides a fairly simple statement, ¬Prov("0=1"), in the Naturals which (almost) seems to meet your criteria. If that's all you wanted. See the references given earlier by wtf, I've tried to use the same notation. Why do I say ALMOST meets your criteria? It does not establish that P is consistent, we simply believe that it is. If P is consistent, then it satisfies your requirement. So the answer to your question is YES - we have a statement in the Naturals that cannot be proved or disproved. Conversely, if P is inconsistent, then we can prove anything, so the answer to your question would be NO - every statement can be proved. Additionally I'm not entirely sure what you mean by DP. You refer to the Dedekind-Peano-Axiom System which seems to suggest you are extending the Naturals to the entire Real Numbers. If this is the case, then the Godel numeralisation technique used in his original proof of the incompleteness theorems may not apply. As I understand it, we are required to find a unique numeralisation for every object (including numbers, field operations etc.) under consideration. This is not difficult for the Naturals as constructed by Peano's axioms since we need only numeralise the object 0 and the successor. However, finding a suitable numeralisation for the reals is obviously much harder. In particular we seek a 1-to-1 correspondance from R (the set of reals) into a subset of N (the Naturals), which defies what we understand about the cardinality of these sets. You may be correct in your original post but I have to say that you have me at a disadvantage: I was NOT aware that Godel's incompleteness theorems provide an example of an undecideable proposition in this case (if you have a link to this result, I'd be grateful for that) Hmmmm...... I'm aware this is becoming confusing perhaps I could make my point this way - Did you mean to write the following: On 2/6/2021 at 10:18 AM, Tristan L said: Now, from Gödel’s Incompleteness Theorems, we know that there are undecidable propositions in P, that is, not every statement about the naturals can be shown or disproven. However, my question is this: Is there an individual proposition about the naturals of which we know that we can neither prove it nor disprove it in DP? If that's the sort of question you intended to ask, then it's difficult. We know that sometimes a Natural number can arise as the result of operations on irrationals, for example [math] \sqrt{2} \times \sqrt{2} [/math] is a perfectly good natural number. So we cannot exclude formula involving irrationals and just restrict our attention to statements provable or unprovable in P and the Godel numeralisation technique does not seem to generalise to the entire reals. Based on earlier replies you won't accept the easy way-out by exploiting the general weakness of most logic systems: For example, informally constructing a statement that just mentions Natural numbers in passing (like "this statement cannot be proved from the axioms AND 2+1 = 3") doesn't seem to be sufficient. Instead you seem to be putting some addditional requirments on the statement: Is it that the proposition "this statement cannot be proved from the axioms" requires formal construction within the logical language, or is it that you have very exacting requirements on the symbols that can be used? (For example only natural numbers and field operations like +, x ?). Apologies for length and lack of coherence, I am tired and off to sleep. Best wishes to you. Edited February 9 by Col Not Colin Too tired to make sense. Link to post Share on other sites

joigus 370 Posted February 9 Share Posted February 9 7 hours ago, Tristan L said: No, it's just a side-hobby of mine to help free speeches broadly, and the English tongue in particular, from the effects of linguistic imperialism, which I do by brooking (using) truly English words instead of ones that came into this language by way of speech-imperialism in actual natural situations, such as forum talks (mutatis mutandis for other victims of language-imperialism). Speech in general interests me, too, but not more that what it is brooked to stand for (which, spellbindingly, includes speech itself). Your point is well taken, @Tristan L, but I think the topic is difficult enough in and of itself that facilitating communication to as wider readership as possible overrides every other need. You can't speak a language that only you understand. I was totally thrown off by "ownship" and "witcraft". Link to post Share on other sites

StringJunky 2302 Posted February 10 Share Posted February 10 1 hour ago, joigus said: Your point is well taken, @Tristan L, but I think the topic is difficult enough in and of itself that facilitating communication to as wider readership as possible overrides every other need. You can't speak a language that only you understand. I was totally thrown off by "ownship" and "witcraft". If one wants to be understood, best to speak in the language of the listener. Link to post Share on other sites

Tristan L 3 Posted February 10 Author Share Posted February 10 (edited) 11 hours ago, wtf said: Yes, so how did you get from linguistic imperialism to the fine points of the second-order completeness theorem for Henkin models? Firstly, I brook (use) right English words in discussions broadly without wanting to talk about that deedsake (fact) in those discussions. I just brought up the topic here because I was asked about my use of proper (or at least more proper) English. Secondly, I don't have much interest in Henkin semantics since I want soothfast (real), full-blown second-order (even better: higher-order) flitecraft. 11 hours ago, wtf said: Thanks for explaining witcraft. Most English-speaking readers probably took that as witchcraft, which made no sense in context. You're welcome. Since the reader knows that "witchcraft" wouldn't make much sense in the context, don't they come up with the thought to read again more carefully so as to check what actually stands written there? 11 hours ago, Col Not Colin said: Hi again. Hi again, too 👋. 11 hours ago, Col Not Colin said: Where P is the Peano arithemetic system 11 hours ago, Col Not Colin said: Additionally I'm not entirely sure what you mean by DP. I'm not talking about puny 😉 first-order Peano arithmetic, but about the higher-order Dedekind-Peano axioms (with full-fledged induction axiom): On 2/6/2021 at 12:18 PM, Tristan L said: [...] the Dedekind-Peano-axiom-system (called “DP” henceforth) [...] 11 hours ago, Col Not Colin said: You refer to the Dedekind-Peano-Axiom System which seems to suggest you are extending the Naturals to the entire Real Numbers. No, by "DP", I mean the higher-step (second-order will do) Dedekind-Peano axioms with the full induction axiom. 11 hours ago, Col Not Colin said: For example, informally constructing a statement that just mentions Natural numbers in passing (like "this statement cannot be proved from the axioms AND 2+1 = 3") doesn't seem to be sufficient. Instead you seem to be putting some addditional requirments on the statement: Is it that the proposition "this statement cannot be proved from the axioms" requires formal construction within the logical language [...]? Yes, for we cannot simply accept any self-referential statement, like "This statement is false", as meaningful. If we can formalize "this statement cannot be proved from the second-order DP axioms AND 2+1 = 3" in the speech of second-order (or any higher-step) flitecraft (logic), I'll accept it. 11 hours ago, Col Not Colin said: Apologies for length and lack of coherence, I am tired and off to sleep. Best wishes to you. Not at all (regarding the first part), you've given interesting food for thought. Best wishes to you, too! 10 hours ago, joigus said: Your point is well taken, @Tristan L, but I think the topic is difficult enough in and of itself that facilitating communication to as wider readership as possible overrides every other need. You're right in that one shouldn't overdo it, that is, the freeing of English should be done slowly enough to be comfortable. That's better for both the talks about other stuff like flitecraft (logic) and the liberation undertaking itself (of which I didn't very consciously think in this thread until my style was brought up by others, by the way, since it has become so natural for me to try to speak and write proper English). 10 hours ago, joigus said: I was totally thrown off by "ownship" and "witcraft". I didn't mean to do that, which is why I usually include the improper, foreign-derived (unright, outlandish*-drawn) word along with the as-of-yet uncommon proper English one. See e.g. On 2/6/2021 at 12:18 PM, Tristan L said: To me, axiom-systems seem to basically be ownships (properties). and On 2/6/2021 at 12:18 PM, Tristan L said: [...] and yet we still can derive neither one from AS because our logical (witcrafty) tools are too weak. *Remark: I don't seek to replace words on account of being foreign, but on account of being due to speech-imperialism. 10 hours ago, joigus said: You can't speak a language that only you understand. True, but firsly, I'm far from the only one who tries to free English; there are others who put much more work into it, such as those behind the Anglish Moot, and many (likely most) of them already speak far righer English than I do. Secondly, a start has to be made somewhere, and while right English isn't widely spoken as of yet, it will hopefully become widespread in the forthtide (future). 9 hours ago, StringJunky said: If one wants to be understood, best to speak in the language of the listener. Indeed, but what if I am speaking in the laguage of the listener, as is soothly (really) the case, yet the listener doesn't truly know their own speech rightly? Moreover, I always give the improper-but-still-common equivalents of not-yet-widespread proper English terms, so the listener/reader shouldn't have problems. X-posted with joigus (just learned the term "x-post" ). Edited February 10 by Tristan L Link to post Share on other sites

joigus 370 Posted February 10 Share Posted February 10 (edited) 9 hours ago, StringJunky said: If one wants to be understood, best to speak in the language of the listener. I couldn't agree more. X-posted with Tristan L. 3 minutes ago, Tristan L said: True, but firsly, I'm far from the only one who tries to free English; there are others who put much more work into it, such as those behind the Anglish Moot, and many (likely most) of them already speak far righer English than I do. Secondly, a start has to be made somewhere, and while right English isn't widely spoken as of yet, it will hopefully become widespread in the forthtide (future). OK. Thanks for drawing my attention to right English. I'm always eager to learn about language. But how many of those Anglish speakers are knowledgeable enough to have a meaningful conversation about second-order logic? See my point? I'm sure a compromise is possible. Mathematics has nothing to do with empire-building, does it? Edited February 10 by joigus -1 Link to post Share on other sites

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