The Definition of Points
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From: cbrown on 24 Apr 2007 12:51 On Apr 23, 10:18 am, Tony Orlow <t...(a)lightlink.com> wrote: > cbr...(a)cbrownsystems.com wrote: > > <snip> > > So the "tree" criss-crosses like a chain link fence. This type of > > partial order is usually called a lattice. > > Yes, it becomes a sort of a lattice-looking thing from one level to the > next. It's actually the set of vertices of a |S|-dimensional cube, if > the same subset may only occur once. More is required: to consider the ordering, we also need to include the edges /between/ the vertices; and they must be /directed/ edges; and there must be no cycles. That's a very "special" kind of cube; not just /any/ cube will do. > If you allow the redundancies, so > that S-[x,y] appears both as a child of S-{x} and of S-{y}, then you get > a tree, but not every element is unique. And that's why most would not consider it a tree, but instead a lattice. Similarly, most would not consider a pentagon to be a cube; even though one can (by replicating points to create "redundancies"), label the points of a cube variously with the five points of a pentagon. > I guess on each level n, where > S is on level 0, one gets each unique subset n times, and the number of > unique elements generated at each level is 2^n-n? Something like that. > I'm not sure why you want a tree to be a proper representation of a lattice. I was simply pointing out that it is completely at odds with the definition of a tree to consider the partial ordering by inclusion as being a tree when we consider vertices as representing distinct elements, and the directed edges to represent the "<=" relation (which seems perectly natural to me), > > > > In order to get a "tree", we need to make sure that we don't do this > > kind of criss-crossing; which is more than you've stated in 1. and 2. > > > It's easy to see how to do this is S is finite. It's harder to see how > > to do this if S is not finite, without some sort of choice function. > > And then we get to the whole axiom of choice implies well-ordering > > thing. > > I was assuming redundancy. > To me, that means you were (implicitly) assuming, e.g., x = {a,b} and y = {a,b}, but not x = y. Seems a bit convoluted, don't you think? <snip> > > A neN E meN : m>n > > > doesn't imply that m is the /smallest/ m satisfying m > n. Yes, we > > know from the properties of the naturals that there /is/ a smallest > > such m which would then be the m that "comes right after" n (thus > > satisfying your allusion to a successor); but that isn't always the > > case. > > Right. IF we define '>' to mean "successor" we get an inductive set, and > if we define "successor" to mean "increment", then it seems to me we > really get the naturals. > Again, that is simply not enough. Yes, the naturals have the property that 1 = 0 + 1, 2 = 1 + 1, 3 = 2 + 1 and so on. But that is not a /definition/ of the naturals; it is an observation /about/ the naturals. This is just like your statement "Order is defined by x<y ^ y<z -> x < z". Order is /not/ defined that way; you need more. And you need more to define the naturals. > > > > I'll formalize "m is the smallest natural which is larger than the > > natural n" by: > > > (m,n in N) and (m > n) and (for all s in N\{m}, if s > n, then s > m) > > > But if we switch to the rationals, > > > (m,n in Q) and (m > n) and (for all s in Q\{m}, if s > n, then s > m) > > > then there is no such m for any n in Q. > > Well, once you have ordered the rationals so as to appear countable, ... "Appear"? 6 doesn't "appear" to be even; it /is/ even. Similarly, the rationals can be bijected with naturals. Therefore they don't simply "appear" countable, they /are/ countable. > ... then there is such an m for any n, in that order. That's changing the > meaning of '>' from the normal quantitative interpretation of a rational > to one of a different order. I wonder why, then, most people think that 3/2 > 5/6. What on earth do you think they mean? > Personally, I think comparing subsets of > the reals out of quantitative order is one of the methods that lead to > screwy results. I mean, what you just demonstrated is that, in the > normal quantitative order, N is sparse and Q dense, which to most naive > intuitions says |Q|>|N|. It all depends on what "most naive intuitions" mean by "|Q| > |N|". By some definitions of ">" and "|X|" this would be true, and by others, it would be false. This is similar to "3/2 > 5/6" being true for some definitions of ">", and false by other definitions. <snip> > >>>>> Let's order the subsets of {a,b,c} by inclusion. Then: {a,b} < > >>>>> {a,b,c}. {a,c} < {a,b,c}. {a} < {a,b}. {a} < {a,c}. But not ({a,b} < > >>>>> {a,c}); and not ({a,c} < {a,b}). Does that mean that {a,b} = {a,c} > >>>>> "for the purposes of that order"? > >>>> Where b<c, {a,b}<{a,c}. > >>> Since "<" indicates subset inclusion, it is the case that not b < c. > >> If a, b and c are members of a set such that xeS ^ yeS -> x<y v y<x v > >> x=y, then either b<c v b<c v b=c. > > > Yes, if a set is totally ordered (i.e., the ordered pair (S, <) > > defines a /total order/ on the /elements/ of S), then (trivially) it > > can be totally ordered. > > > But I was talking about the ordered pair (P(S), <) where < is subset > > inclusion; which defines a /partial/ order on the /subsets/ of S. > > > In that case there are distinct /subsets/ X, Y of S with not (X < Y or > > Y < X). Which is to say that there are distinct /elements/ X, Y of > > P(S) with not (X < Y or Y < X) > > Yes, in that partial order. But, if you partition a set... what do you mean by "partition a set"? > ... such that there > is a total order on the subsets, and a total order within each subset, > then you can establish a total order on the entire set... Trivially, if you establish a total order on the subsets of S, then you have a total order on the subsets of S. On the other hand, if you select some limited class of subsets of S upon which we have a total order ">", then it says nothing about subsets which are /not/ in that limited class. For example, if you establish a total order on the finite subsets of some infinite set S, then it doesn't say /anything/ about how that ordering should be extended to /infinite/ subsets of S. > ... using something > like a choice function. I just don't see how an uncountable set can be > partitioned into a countable set of subsets, each countable, to produce > a well ordering. > Equally, I find it difficult to imagine how you could have a countably infinite collection of non-empty subsets {X_1, X_2, ..., X_n, ...} and yet somehow /not/ be able to select one element from each of these sets. Thus the joke: "The axiom of choice is obviously true, the well- ordering principle is obviously false, and who can say about Zorn's lemma?". <snip> > Consider Some countable set S, and a set of binary strings representing > subsets, with one bit for each element of S, which is a 1 if that subset > contains that element and 0 otherwise. Don't we have the set of all > binary reals in [0,1)? Given any two, can't we determine which is greater? > No (at least not in the way you state); because there are two different subsets of S which correspond to the same real number. Therefore, by representing the same real number, it is not the case that one is "greater" than the other using R's usual ordering. They are different subsets; but we have identified them with the same real number; so we cannot determine which is "greater" in this way. <snip> > > Imposing the ordering of R as you propose is a /pre/-order (or a pre- > > total-order) on the sequences you describe. > > Huh? Isn't that a total ordering on P(S)? > No. In a pre-order, it is not required that x <= y and y <= x implies y = x. The other rules regarding a partial order /do/ apply: x <= x x <= y and y <= z implies x <= z and that is why it is called a "pre-order". > > > > I think you would agree that the "sequence of bits": > > > 0000111... > > > is not the /same/ sequence of bits as: > > > 0001000... > > No, but of course you consider them the same value. I'm not sure I do, > outside of standard reals. > We start with subsets of N. We apply a /bijective/ function that maps each subset of N to a sequence of "bits". Next, we map a sequence of "bits" to lim sum (b_n*2^-n) of those bits ("the quantitative order of the reals in [0,1)"). It turns out that the latter function is not a bijection. "Outside the standard reals" a sequence of bits could mean anything you wish it to. What did you intend? > > > > Equivalently, the set X = {n : n > 3} and Y = {3} are not equal. > > > And yet, with the ordering you describe, (0000111... <= 0001000...) > > and (0001000... <= 0000111...); which is to say X <= Y and Y <= X and > > yet, not (X = Y). > > If they are ordered such that x<y iff the first bit where they differ > has a 0 in x and a 1 in y, then 00001111.. < 0001000... And if they are ordered in such a way X <= Y such that every bit which is a 1 in X is also a 1 in Y, then we get get partial order by inclusion. We can define many different orderings to sequences of bits, depending on our interests. So what is your point? > That is, if the > elements of the set are well ordered, then there can be a total order on > the subsets. But the order you describe is /not/ a well-order on the elements of P(S). Do you recall the definition of a well-order? > I think you are equating those two strings according to the > standard conception of the reals, but those are two different subsets, > one of which includes an element which comes before any element in the > other, in the well ordering on S, so that one comes first. > You are the one who said: > >> if We have > >> created a binary string, and if we consider all bits to be to the right > >> of the binary point, with the first being the first bit, > >> then we have > >> ordered the subsets of N according to the quantitative order of the > >> reals in [0,1]. If instead of the quantitative order of the /reals/, you meant some other thing, then perhaps your statement is true; or perhaps not. I can't say. As we agreed in the section I snip, it is surely the case that there is a total order on the set of all bit sequences (lexicographic order). <snip> > > But I wasn't talking about ordering the /elements/ of P(N). I was > > talking about ordering the /subsets/ of P(N); which is to say ordering > > the /elements/ of P(P(N)). E.g., one element of P(P(N)) is the set X > > of /all/ subsets of N having prime order; and another is the set Y of / > > all/ subsets of N consisting of a finite number of primes. How should > > I proceed to determine whether X <= Y or Y <= X in such a way that <= > > is a total order on P(P(N))? > > Well, if you have established the total order on P(N), then for > xeP(P(N)) and yeP(P(N)), x<y iff the first bit in x which is different > from y is a 0. Think: What is meant by "the /first/ bit in x" in this context? Yes, we have /totally/ ordered P(N) using the usual total order (yea; even, well-order) of N. But we have /not/ thereby put a /well-order/ on P(N); which is to say, we have not (yet) defined an ordering on the /elements/ of P(N) such that every /subset/ of P(N) has a least element. So we have not defined what the "first" or smallest "bit" of each possible /subset/ x of P(N), which is to say element x of P(P(X)), might be. > That is, if you can determine which subset is the most > significant, or first in their union which is not in their intersection, > the element containing that subset precedes the other. > This is the very nut of the idea behind the definition of "well- ordered". Can we /always/ do this, given that we start with an agreed upon well-ordering on N? Maybe, and maybe not. Consider the fact that this question already has a name. That implies that other people, besides you, have already wondered about this. > Can I apply this to your example? Let's see. No, because of the infinity > of bits in P(N), one cannot use the normal ordering of binary reals to > accomplish this, because that order is not a well order suitable for > addressing the bits. Yes. Exactly! See, on one level, you /do/ get it. > One might try using the order of the binary > naturals, but for infinite sets, we would have trouble ordering, say, > ...1010 and ...011. No; that we /can/ do (lexiographic ordering). What is in question is whether we can relatively totally order /sets/ of bit sequences; not whether we can relatively totally order /bit sequences/ (the latter we can easily do by lexicographic ordering of N x {0,1}). A better analogy would be: Suppose (S, <=) is a well-order on S. Then there is an ordering (P(S), <=') such that <=' is a total order. But as we have just seen, it is not /obviously/ true that (without outright assuming something like AoC, which makes the first clause unnecessary): Suppose (S, <=) is a total order on S. Then there is an ordering (P(S), <=') such that <=' is a total order. > So, perhaps that doesn't work for P(P(N)), but what > was the point about a total order on P(P(N)) again? > I brought all of this up to highlight the problems with your assertion: > >>>>>> Defining equality > >>>>>> where there is no relative order doesn't make sense. It is not obvious what the "right" definition, or even "a" definition, of ordering creates a total order (let alone a well-order) on P(P(N)). However, it is quite sensible how we "should" define what /equality/ means for two elements x, y of P(P(N)). To whit, the same as is usual for sets: x = y iff for all z, z in x iff z in y. Cheers - Chas
From: cbrown on 24 Apr 2007 12:55 On Apr 23, 10:18 am, Tony Orlow <t...(a)lightlink.com> wrote: > cbr...(a)cbrownsystems.com wrote: > > <snip> > > So the "tree" criss-crosses like a chain link fence. This type of > > partial order is usually called a lattice. > > Yes, it becomes a sort of a lattice-looking thing from one level to the > next. It's actually the set of vertices of a |S|-dimensional cube, if > the same subset may only occur once. More is required: to consider the ordering, we also need to include the edges /between/ the vertices; and they must be /directed/ edges; and there must be no cycles. That's a very "special" kind of cube; not just /any/ cube will do. > If you allow the redundancies, so > that S-[x,y] appears both as a child of S-{x} and of S-{y}, then you get > a tree, but not every element is unique. And that's why most would not consider it a tree, but instead a lattice. Similarly, most would not consider a pentagon to be a cube; even though one can (by replicating points to create "redundancies"), label the points of a cube variously with the five points of a pentagon. > I guess on each level n, where > S is on level 0, one gets each unique subset n times, and the number of > unique elements generated at each level is 2^n-n? Something like that. > I'm not sure why you want a tree to be a proper representation of a lattice. I was simply pointing out that it is completely at odds with the definition of a tree to consider the partial ordering by inclusion as being a tree when we consider vertices as representing distinct elements, and the directed edges to represent the "<=" relation (which seems perectly natural to me), > > > > In order to get a "tree", we need to make sure that we don't do this > > kind of criss-crossing; which is more than you've stated in 1. and 2. > > > It's easy to see how to do this is S is finite. It's harder to see how > > to do this if S is not finite, without some sort of choice function. > > And then we get to the whole axiom of choice implies well-ordering > > thing. > > I was assuming redundancy. > To me, that means you were (implicitly) assuming, e.g., x = {a,b} and y = {a,b}, but not x = y. Seems a bit convoluted, don't you think? <snip> > > A neN E meN : m>n > > > doesn't imply that m is the /smallest/ m satisfying m > n. Yes, we > > know from the properties of the naturals that there /is/ a smallest > > such m which would then be the m that "comes right after" n (thus > > satisfying your allusion to a successor); but that isn't always the > > case. > > Right. IF we define '>' to mean "successor" we get an inductive set, and > if we define "successor" to mean "increment", then it seems to me we > really get the naturals. > Again, that is simply not enough. Yes, the naturals have the property that 1 = 0 + 1, 2 = 1 + 1, 3 = 2 + 1 and so on. But that is not a /definition/ of the naturals; it is an observation /about/ the naturals. This is just like your statement "Order is defined by x<y ^ y<z -> x < z". Order is /not/ defined that way; you need more. And you need more to define the naturals. > > > > I'll formalize "m is the smallest natural which is larger than the > > natural n" by: > > > (m,n in N) and (m > n) and (for all s in N\{m}, if s > n, then s > m) > > > But if we switch to the rationals, > > > (m,n in Q) and (m > n) and (for all s in Q\{m}, if s > n, then s > m) > > > then there is no such m for any n in Q. > > Well, once you have ordered the rationals so as to appear countable, ... "Appear"? 6 doesn't "appear" to be even; it /is/ even. Similarly, the rationals can be bijected with naturals. Therefore they don't simply "appear" countable, they /are/ countable. > ... then there is such an m for any n, in that order. That's changing the > meaning of '>' from the normal quantitative interpretation of a rational > to one of a different order. I wonder why, then, most people think that 3/2 > 5/6. What on earth do you think they mean? > Personally, I think comparing subsets of > the reals out of quantitative order is one of the methods that lead to > screwy results. I mean, what you just demonstrated is that, in the > normal quantitative order, N is sparse and Q dense, which to most naive > intuitions says |Q|>|N|. It all depends on what "most naive intuitions" mean by "|Q| > |N|". By some definitions of ">" and "|X|" this would be true, and by others, it would be false. This is similar to "3/2 > 5/6" being true for some definitions of ">", and false by other definitions. <snip> > >>>>> Let's order the subsets of {a,b,c} by inclusion. Then: {a,b} < > >>>>> {a,b,c}. {a,c} < {a,b,c}. {a} < {a,b}. {a} < {a,c}. But not ({a,b} < > >>>>> {a,c}); and not ({a,c} < {a,b}). Does that mean that {a,b} = {a,c} > >>>>> "for the purposes of that order"? > >>>> Where b<c, {a,b}<{a,c}. > >>> Since "<" indicates subset inclusion, it is the case that not b < c. > >> If a, b and c are members of a set such that xeS ^ yeS -> x<y v y<x v > >> x=y, then either b<c v b<c v b=c. > > > Yes, if a set is totally ordered (i.e., the ordered pair (S, <) > > defines a /total order/ on the /elements/ of S), then (trivially) it > > can be totally ordered. > > > But I was talking about the ordered pair (P(S), <) where < is subset > > inclusion; which defines a /partial/ order on the /subsets/ of S. > > > In that case there are distinct /subsets/ X, Y of S with not (X < Y or > > Y < X). Which is to say that there are distinct /elements/ X, Y of > > P(S) with not (X < Y or Y < X) > > Yes, in that partial order. But, if you partition a set... what do you mean by "partition a set"? > ... such that there > is a total order on the subsets, and a total order within each subset, > then you can establish a total order on the entire set... Trivially, if you establish a total order on the subsets of S, then you have a total order on the subsets of S. On the other hand, if you select some limited class of subsets of S upon which we have a total order ">", then it says nothing about subsets which are /not/ in that limited class. For example, if you establish a total order on the finite subsets of some infinite set S, then it doesn't say /anything/ about how that ordering should be extended to /infinite/ subsets of S. > ... using something > like a choice function. I just don't see how an uncountable set can be > partitioned into a countable set of subsets, each countable, to produce > a well ordering. > Equally, I find it difficult to imagine how you could have a countably infinite collection of non-empty subsets {X_1, X_2, ..., X_n, ...} and yet somehow /not/ be able to select one element from each of these sets. Thus the joke: "The axiom of choice is obviously true, the well- ordering principle is obviously false, and who can say about Zorn's lemma?". <snip> > Consider Some countable set S, and a set of binary strings representing > subsets, with one bit for each element of S, which is a 1 if that subset > contains that element and 0 otherwise. Don't we have the set of all > binary reals in [0,1)? Given any two, can't we determine which is greater? > No (at least not in the way you state); because there are two different subsets of S which correspond to the same real number. Therefore, by representing the same real number, it is not the case that one is "greater" than the other using R's usual ordering. They are different subsets; but we have identified them with the same real number; so we cannot determine which is "greater" in this way. <snip> > > Imposing the ordering of R as you propose is a /pre/-order (or a pre- > > total-order) on the sequences you describe. > > Huh? Isn't that a total ordering on P(S)? > No. In a pre-order, it is not required that x <= y and y <= x implies y = x. The other rules regarding a partial order /do/ apply: x <= x x <= y and y <= z implies x <= z and that is why it is called a "pre-order". > > > > I think you would agree that the "sequence of bits": > > > 0000111... > > > is not the /same/ sequence of bits as: > > > 0001000... > > No, but of course you consider them the same value. I'm not sure I do, > outside of standard reals. > We start with subsets of N. We apply a /bijective/ function that maps each subset of N to a sequence of "bits". Next, we map a sequence of "bits" to lim sum (b_n*2^-n) of those bits ("the quantitative order of the reals in [0,1)"). It turns out that the latter function is not a bijection. "Outside the standard reals" a sequence of bits could mean anything you wish it to. What did you intend? > > > > Equivalently, the set X = {n : n > 3} and Y = {3} are not equal. > > > And yet, with the ordering you describe, (0000111... <= 0001000...) > > and (0001000... <= 0000111...); which is to say X <= Y and Y <= X and > > yet, not (X = Y). > > If they are ordered such that x<y iff the first bit where they differ > has a 0 in x and a 1 in y, then 00001111.. < 0001000... And if they are ordered in such a way X <= Y such that every bit which is a 1 in X is also a 1 in Y, then we get get partial order by inclusion. We can define many different orderings to sequences of bits, depending on our interests. So what is your point? > That is, if the > elements of the set are well ordered, then there can be a total order on > the subsets. But the order you describe is /not/ a well-order on the elements of P(S). Do you recall the definition of a well-order? > I think you are equating those two strings according to the > standard conception of the reals, but those are two different subsets, > one of which includes an element which comes before any element in the > other, in the well ordering on S, so that one comes first. > You are the one who said: > >> if We have > >> created a binary string, and if we consider all bits to be to the right > >> of the binary point, with the first being the first bit, > >> then we have > >> ordered the subsets of N according to the quantitative order of the > >> reals in [0,1]. If instead of the quantitative order of the /reals/, you meant some other thing, then perhaps your statement is true; or perhaps not. I can't say. As we agreed in the section I snip, it is surely the case that there is a total order on the set of all bit sequences (lexicographic order). <snip> > > But I wasn't talking about ordering the /elements/ of P(N). I was > > talking about ordering the /subsets/ of P(N); which is to say ordering > > the /elements/ of P(P(N)). E.g., one element of P(P(N)) is the set X > > of /all/ subsets of N having prime order; and another is the set Y of / > > all/ subsets of N consisting of a finite number of primes. How should > > I proceed to determine whether X <= Y or Y <= X in such a way that <= > > is a total order on P(P(N))? > > Well, if you have established the total order on P(N), then for > xeP(P(N)) and yeP(P(N)), x<y iff the first bit in x which is different > from y is a 0. Think: What is meant by "the /first/ bit in x" in this context? Yes, we have /totally/ ordered P(N) using the usual total order (yea; even, well-order) of N. But we have /not/ thereby put a /well-order/ on P(N); which is to say, we have not (yet) defined an ordering on the /elements/ of P(N) such that every /subset/ of P(N) has a least element. So we have not defined what the "first" or smallest "bit" of each possible /subset/ x of P(N), which is to say element x of P(P(X)), might be. > That is, if you can determine which subset is the most > significant, or first in their union which is not in their intersection, > the element containing that subset precedes the other. > This is the very nut of the idea behind the definition of "well- ordered". Can we /always/ do this, given that we start with an agreed upon well-ordering on N? Maybe, and maybe not. Consider the fact that this question already has a name. That implies that other people, besides you, have already wondered about this. > Can I apply this to your example? Let's see. No, because of the infinity > of bits in P(N), one cannot use the normal ordering of binary reals to > accomplish this, because that order is not a well order suitable for > addressing the bits. Yes. Exactly! See, on one level, you /do/ get it. > One might try using the order of the binary > naturals, but for infinite sets, we would have trouble ordering, say, > ...1010 and ...011. No; that we /can/ do (lexiographic ordering). What is in question is whether we can relatively totally order /sets/ of bit sequences; not whether we can relatively totally order /bit sequences/ (the latter we can easily do by lexicographic ordering of N x {0,1}). A better analogy would be: Suppose (S, <=) is a well-order on S. Then there is an ordering (P(S), <=') such that <=' is a total order. But as we have just seen, it is not /obviously/ true that (without outright assuming something like AoC, which makes the first clause unnecessary): Suppose (S, <=) is a total order on S. Then there is an ordering (P(S), <=') such that <=' is a total order. > So, perhaps that doesn't work for P(P(N)), but what > was the point about a total order on P(P(N)) again? > I brought all of this up to highlight the problems with your assertion: > >>>>>> Defining equality > >>>>>> where there is no relative order doesn't make sense. It is not obvious what the "right" definition, or even "a" definition, of ordering creates a total order (let alone a well-order) on P(P(N)). However, it is quite sensible how we "should" define what /equality/ means for two elements x, y of P(P(N)). To whit, the same as is usual for sets: x = y iff for all z, z in x iff z in y. Cheers - Chas
From: Lester Zick on 24 Apr 2007 14:20 On Mon, 23 Apr 2007 21:27:22 -0400, Bob Kolker <nowhere(a)nowhere.com> wrote: >Ben newsam wrote: >> >> Here we go, tap dancing in porridge again. > >I thought it was word stew. There you go thinking again, Bob, or what passes for it in your case. Occasionally you get off a decent observation which is about the only differentia between you and the Gang of Four Horses Asses of the Apocalypse. ~v~~
From: Lester Zick on 24 Apr 2007 14:31 On Tue, 24 Apr 2007 00:43:24 +0100, Ben newsam <ben.newsam.remove.this(a)gmail.com> wrote: >On Mon, 23 Apr 2007 11:54:05 -0400, Tony Orlow <tony(a)lightlink.com> >wrote: > >>If you want to talk about the truth values of individual facts used in >>deduction, by all means, go for it. > >I would counsel you seriously not to attempt it, Lester will only >obfuscate the "discussion", and then will start to hurl personal >insults at you. Or you could bore us to death with the experimental confirmation of spatial dimentionality. Is there nothing you can contribute to the advancement of science besides gratuitous meretricious ad hominem attacks on moi? ~v~~
From: Lester Zick on 24 Apr 2007 14:44
On Mon, 23 Apr 2007 21:47:51 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >>> Truth tables and logical statements involving variables are >>> just that. If I say, 3x+3=15, is that true? No, we say that IF that's >>> true, THEN we can deduce that x=4. >> >> But here you're just appealing to syllogistic inference and truisms >> because your statement is incomplete. You can't say what the "truth" >> of the statements is or isn't until x is specified. So you abate the >> issue until x is specified and denote the statement as problematic. > >I don't consider it a problem, but a state of affairs. If one doesn't >know what the state of affairs IS, they can say, IF it's THIS then THIS, >and IF THAT then THAT. I don't know if THIS or THAT, but once I know, >then I'll know. That how logic is. This comes back to axioms. Surely (or >probably, or maybe) you see that. Well certainly it's a problematic state of affairs, Tony, but look up the meaning of "problematic". It means "questionable" not necessarily causing problems. If one doesn't know certain factors circumstances are certainly problematic regarding the truth of the proposition. ~v~~ |