Resolution of Russell's and Cantor's paradoxes

[DanTeplitskiy]

I need your help, guys.

The link below is to the paper, that is, in my opinion, contains resolution of one of the most interesting problems of math logic and set theory – Russell’s paradox. As Cantor’s and Russell’s paradoxes is a “paired” problem you will find resolution of both in it.

The paper is now in pre-print phase. I need the folowing kind of help on it: your comments and questions.

The paper is written in a clear organized way – I do not think it will take more than 35-40 minutes from a person knowing the very basics of set theory to get it all.

https://docs.google.com/viewer?a=v&...YTZlMy00NDJhLWJjN2MtMDAzNDUzOWQ2Y2Ew&hl=en_US

The paper is in English (according to a professional mathematician having position in USA - quite readable English), though, of course, it is not English of an English speaking person.

Thanks a lot in advance!

Dan

 

[Citan Uzuki]

You really didn't need anywhere near that much verbiage. You could have summarized the entire paper by saying "All Russel's paradox shows is that the set $\{x:x\not\in x\}$ cannot exist, but it says nothing about the set of all sets." And this is actually true! The contradiction only comes from the generally accepted principle in naive set theory (which is formalized as an axiom schema in ZFC) that given any set x, and any predicate φ, there exists a set consisting of all the elements of x satisfying that predicate. In ZFC, this allows you to show that if there was a set of all sets, then the set $\{x:x\not\in x\}$ would exist, which is plainly absurd, so in ZFC there can be no universal set. However, in a more general setting, you can keep the universal set and instead weaken the axiom of specification so that it only covers certain classes of formulas, which leads you to set theories like new foundations instead.

The problem with NF and related theories which keeps them from being more widely used is that because they do rely on restricting specification, you have to constantly keep checking every time you want to make a new set whether the set you're making can be specified by one of the allowed types of formula. In contrast, ZFC allows you to use any formula you like, even nice diagonalizing ones that would have the power to create paradoxes if applied to a universal set, provided only that you first verify that there is some set that contains the set you wish to create. This is much more flexible and easier to work with, and this is generally more important than keeping a universal set around.

 

[DanTeplitskiy]

Dear Citan Uzuki,

Thanks a lot for reply!

Sorry, it seems to me that you missed the point of the paper. It is not on keeping the set of all sets. Nor on any alternative formal system keeping it some way or another (like NF and the like).

It is on resolution of the paradoxes within non-axiomatic approach. And on the methods of using the classical logic (classical logic itself, not in the context of some formal system) to complete the task.

Could you please read the part of the paper named "analysis of contradictory equivalence" and further to the very end of the paper? I hope you will be able to get the point. I would be very obliged if you do and write some comments then.

Thanks a lot in advance, Citan!

Yours,

Dan

 

[micromass]

You are working with some kind of "replica" of the set of sets, but a replica that apparently doesn't equal the set of sets.

Can you show that such a replica in fact exists?
What do you mean with "replica" anyways, can you give a definition of it?

 

[micromass]

If R includes itself, R is not that R = {x: x∉x}. What I mean is that under the assumption “R ∈ R” R includes a member that is included in itself (R itself is such a member). Thus under this assumption R is not a set of all sets that are not members of themselves - there is no logical ground to conclude that R (as an element) is not included in R (as a set).

There is no difference what so ever between R as an element and R as a set. It is the same R.
And if we say $R\in R$, then it is always the same R. R will still always equal $\{x~\vert~x\notin x\}$. You will have to give a convincing logical argument as to why these R's are different. Defenitions in mathematics are not dynamic!

We could state that a set that includes itself is not included in R if R were like {x: x∉x} but under the assumption “R ∈ R” R ≠ {x: x∉x} so, under this assumption, we cannot state that a particular set that includes itself is not included in R. That is under the assumption R ∈ R we cannot conclude that R ∉ R.

As before, R is defined as $\{x~\vert~x\notin x\}$ so R wil always equal that set. Whether $R\in R$ or not, is of no importance.

You assume somehow that definitions change one way or the other. But they don't.

 

[DanTeplitskiy]

Dear Micromass,

Sorry, but the definition of the word 'replica' is given right where it is used.
From the very definition it is clear that I do not have to prove that it exists.
The whole point is that if use the diagonal argument "some way" we can prove that some set is bigger than itself. Please, try to read the whole paper (or at least the whole part of it ).

As for "dynamic" definitions there is an explanation below that you might want to undestand.
The explanation starts with "Or, if you prefer seeing..."

Anyway thanks a lot for even trying to help! I really mean it, pal.

Yours,

Dan

 

[micromass]

Dear Micromass,

Sorry, but the definition of the word 'replica' is given right where it is used.
From the very definition it is clear that I do not have to prove that it exists.
The whole point is that if use the diagonal argument "some way" we can prove that some set is bigger than itself. Please, try to read the whole paper (or at least the whole part of it ).

I did not see any definition. From what I can read, you use the word replica just too distinguish between the uses. So the replica of the set of sets is exactly the set of sets. Or am I wrong??
But this does not make any sense, since later you go on about how there is no bijection between the replica and the set of sets. It's extremely confusing. What are you really trying to say?

As for "dynamic" definitions there is an explanation below that you might want to undestand.
The explanation starts with "Or, if you prefer seeing

Yeah, that parts makes no sense at all. You say that R becomes like $R\neq \{x~\vert~x\notin x\}$, which makes no sense. And then you say that we have to see the formula dynamically. Uh well, I hate to break it to you, but "dynamic formula's" are not part of mathematics. I don't even know what a dynamic formula is supposed to mean...

 

[DanTeplitskiy]

I did not see any definition. From what I can read, you use the word replica just too distinguish between the uses. So the replica of the set of sets is exactly the set of sets. Or am I wrong??
But this does not make any sense, since later you go on about how there is no bijection between the replica and the set of sets. It's extremely confusing. What are you really trying to say?



Yeah, that parts makes no sense at all. You say that R becomes like $R\neq \{x~\vert~x\notin x\}$, which makes no sense. And then you say that we have to see the formula dynamically. Uh well, I hate to break it to you, but "dynamic formula's" are not part of mathematics. I don't even know what a dynamic formula is supposed to mean...

1. Yes, replica is a copy. What I am getting at is that if we use the diagonal method like in Cantor's paradox we finally get even bigger absurd - the set of all sets is bigger than itself.

2. The point is we eigher consider R as both R = {x: x∉x} and, at the same time R≠ {x: x∉x} which is absurd OR consider the formula dynamically: that is R≠ {x: x∉x} "replaces" R = {x: x∉x}.

 

[micromass]

1. Yes, replica is a copy. What I am getting at is that if we use the diagonal method like in Cantor's paradox we finally get even bigger absurd - the set of all sets is bigger than itself.

Well, what is the definition of a copy?? And why does such a copy exist?? The same points remain...

2. The point is we eigher consider R as both R = {x: x∉x} and, at the same time R≠ {x: x∉x} which is absurd OR consider the formula dynamically: that is R≠ {x: x∉x} "replaces" R = {x: x∉x}.

No, $R=\{x~\vert~x\notin x\}$ always. No exception.
Dynamic formula's is not part of mathematics.

 

[pwsnafu]

Is it just me, or does this read like a philosophy paper to anyone?

 

[micromass]

Is it just me, or does this read like a philosophy paper to anyone?

It's not just you.

 

[Citan Uzuki]

Okay, I've read this paper a second time -- I think you're right, I did miss your point. You spent a lot of time trying to resolve the paradoxes of Cantor and Russel, so naturally I assumed that that was what the paper was about.

Assuming that I've understood you correctly this time, you're arguing for a non-monotone approach to mathematical reasoning, whereby naive arguments are accepted (even knowing that they might lead to a contradiction), up until a paradox actually occurs, at which point the paradox is analyzed to determine where a case of invalid set formation actually occurred, and the contradictory conclusion then disposed of. Your argument is that since such an instance can always be found, we will gain a better understanding of mathematics by finding such paradoxes and analyzing them than by trying to confine our arguments to a set of rules where such paradoxes cannot occur. Have I understood you correctly this time?

 

[DanTeplitskiy]

Okay, I've read this paper a second time -- I think you're right, I did miss your point. You spent a lot of time trying to resolve the paradoxes of Cantor and Russel, so naturally I assumed that that was what the paper was about.

Assuming that I've understood you correctly this time, you're arguing for a non-monotone approach to mathematical reasoning, whereby naive arguments are accepted (even knowing that they might lead to a contradiction), up until a paradox actually occurs, at which point the paradox is analyzed to determine where a case of invalid set formation actually occurred, and the contradictory conclusion then disposed of. Your argument is that since such an instance can always be found, we will gain a better understanding of mathematics by finding such paradoxes and analyzing them than by trying to confine our arguments to a set of rules where such paradoxes cannot occur. Have I understood you correctly this time?

Dear Citan Uzuki,

Thanks a lot for reply!

I do not state that the instance can always be found, sorry.

By the way, we usually find some approach useful if it gives us something at all - we do not need everything to be resolved by some approach in some area to find it useful. That is "...since such an instance can always be found, we will gain a better understanding..." does not seem to be logically grounded to me.

In my paper I determine where and how a case of invalid paradox formation actually occurred really - that is, metamath over the standard form of Russell's paradox. Then, separately but connected to, we get a proof that some collection of objects does not exist.

I hope the above will be helpful to you to get my point more correctly .

Thanks a lot in advance, Citan!

Yours,

Dan

 

[DanTeplitskiy]

Is it just me, or does this read like a philosophy paper to anyone?

Dear Pwsnafu,

Thanks a lot for your reading my paper and comments!

In order to make it simpler a great amount of words used. Maybe (I mean maybe) because of this my paper might be perceived as "some philosophy". Actually it is all logics on math objects (sets, diagonal argument and the theorem usually called "Russell's paradox").

Yours,

Dan

 

[pwsnafu]

In order to make it simpler a great amount of words used. Maybe (I mean maybe) because of this my paper might be perceived as "some philosophy". Actually it is all logics on math objects (sets, diagonal argument and the theorem usually called "Russell's paradox").

No, I mean the arguments and techniques that you use makes this a philosophy of mathematics paper, not a mathematics paper.

 

[DanTeplitskiy]

No, I mean the arguments and techniques that you use makes this a philosophy of mathematics paper, not a mathematics paper.

Dear Pwsnafu,

As far as I can see, if it is logic over math objects the logic may be either correct or not.

Yours,

Dan

 

[DanTeplitskiy]

Okay, I've read this paper a second time -- I think you're right, I did miss your point. You spent a lot of time trying to resolve the paradoxes of Cantor and Russel, so naturally I assumed that that was what the paper was about.

Assuming that I've understood you correctly this time, you're arguing for a non-monotone approach to mathematical reasoning, whereby naive arguments are accepted (even knowing that they might lead to a contradiction), up until a paradox actually occurs, at which point the paradox is analyzed to determine where a case of invalid set formation actually occurred, and the contradictory conclusion then disposed of. Your argument is that since such an instance can always be found, we will gain a better understanding of mathematics by finding such paradoxes and analyzing them than by trying to confine our arguments to a set of rules where such paradoxes cannot occur. Have I understood you correctly this time?

Dear Citan Uzuki,

One more thing (just in case): the contradictory conclusion is disposed of not because we prove some non-paradoxical way that the set does not exist but from analyzing the Russell's paradox itself. Maybe you understood this - though I decided to point it out here in case you did not. Meant no disrespect by this.

Yours,

Dan

 

[Citan Uzuki]

Okay, I think I understand what you're asserting well enough to offer some constructive advice now. It's a long post, so grab yourself some tea before you start reading.

Let's start with a few organizational points. Since the point of your paper is to argue for the utility of a non-axiomatic approach, you should state that fact up front. Ideally in the abstract. Something like "We analyze the paradoxes of Cantor and Russel from the standpoint of naive set theory and discuss possible resolutions. We then argue that these paradoxes enhance, rather than diminish, the utility of a non-axiomatic approach to set theory."

Second, actually give an argument for why these paradoxes are okay, and make it the bulk of the paper. This is after all, your main point, and the only one that is specific to your paper. Right now the only mention of this point is a few sentences buried in the conclusion, and it's easy to miss. Conversely, spend a lot less time talking about the paradoxes themselves -- analyses of why the paradoxes of Cantor and Russel exist are commonplace, and will be boringly familiar to anyone familiar with the paradoxes themselves. Also, you should avoid saying that you "resolve" these paradoxes -- such language is usually used by people trying to claim, in plain contradiction of the obvious facts, that the arguments are invalid or otherwise do not constitute paradoxes in the first place, and this will set off every crank detector that gets anywhere near your paper.

Next some linguistic points. If you don't intend to assert the mathematical existence of two distinct "copies" of the same set, you should avoid talking about a replica of the set. I know that you're trying to make the distinction between the role of the domain and the codomain clearer, but you actually just end up confusing people (cf. micromass's post). Likewise, referring to the set used in the proof as the "proof" set (quotes in original) is also confusing to English readers. It would be better to just give the set a name, and refer to it by that name.

Next an academic point. In American universities, Wikipedia is not considered a reliable source, and should NEVER be cited in an academic paper, unless the subject of the paper IS the Wikipedia article in question (e.g. how accurate it is). I would be very surprised to learn that Russian universities work any differently. Further, if you must cite Wikipedia, you should link to a stable version of the page, so that if it is revised in the meantime others can still access the version of the article you were working from. The current stable link to the Russel's paradox article is http://en.wikipedia.org/w/index.php?title=Russell's_paradox&oldid=440736446

All right, now on to the logic of the paper itself. I don't think your argument in the section "analysis of contradictory equivalence" actually holds. As micromass stated, there is no such thing as a dynamic formula in mathematics. Once a definition such as R={x:x∉x} is established, it does not "become" anything else. We consider different possible assumptions about what relationship R might have to itself, but we are always reasoning about the same object {x:x∉x}. It is precisely because we do this that we can turn it into a proof by contradiction that the set {x:x∉x} doesn't actually exist. Using a fixed symbol to refer to an impossible object in order to derive a contradiction is utterly commonplace in mathematics, and completely logically valid. My favorite example is this proof that there is no greatest integer: "Let n be the greatest integer. Then n+1 is greater than the greatest integer. Q.E.A." The very first thing we do is assign n to be an impossible object! Another example is the proof that 2 has no rational square root. Actually, I'd like to emphasise this point, because you say in your paper:

Though just in case: reasoning on square root of 2 is about some property of the object, not about the existence of the object.

Actually, the proof that √2 is not rational is about the existence of a mathematical object -- namely, integers n and m such that (n/m)^2 = 2. You can prove that 2 has no rational square root even if you have not yet proven that √2 exists at all (a useful thing, if you haven't yet done the construction of the real numbers). It's exactly the same as the proof that {x:x∉x} cannot exist, so why is one a paradox needing "resolution" and the other just a straight proof by contradiction?

Finally, I would like to say a few words in defense of the axiomatic method itself. You seem to be under the impression that the axiomatic approach is used only because it's "good enough" and out of tradition. This is not true. Mathematics is a unique discipline amongst the sciences. In literally every other scientific discipline, the accumulation of large amounts of empirical evidence is considered more than enough reason to accept something as scientific truth. Only in mathematics do we do something silly like insisting on a proof of, say, the Riemann hypothesis, when it has already been confirmed numerically up to the first ten trillion zeroes. But in these other disciplines, our models are only approximations to the underlying physical reality. We live with an unavoidable possibility of error. So we learn very quickly not to do anything that might magnify these errors to the point where they have practical consequences. We use the simplest principles possible and usually in straightforward ways. And when we do construct something theoretically using complicated arguments, we build an actual physical model and field-test it against reality before we do anything critical with it. And very frequently, we find that http://news.yahoo.com/contact-lost-hypersonic-glider-launch-163016325.html" due to small unexpected problems.

Mathematics is different, because the mathematical objects we're dealing with are either creations of our own mind or living in some platonic ideal realm (depending on your philosophy), and in either case are completely specified by the properties we define them to have. This gives us the possibility of error-free reasoning, which we regularly exploit to give http://planetmath.org/?method=l2h&f...remAndThatEAndPiAreTranscendental2&op=getobj", which take the tiniest of subtle differences and slowly build up their consequences until at last you are able to derive an absurdity, thus showing your initial assumption wrong. But the same arguments that amplify subtle differences between objects also amplify subtle mistakes, so if you try an argument like that one with a physical object, the difference between theory and reality will get blown up to the point where there is no obvious resemblance between reality and your conclusions. Thus not only does mathematics make use of error-free reasoning, but it depends on it. The smallest of mistakes, the most subtle confusions, will eventually be magnified to the point where they ruin your whole conclusion.

It was ultimately this necessity, combined with the illustration given by Russel's paradox of just how easily mathematicians could get themselves into trouble, that led to the development of axiomatic treatments of set theory. By formalizing logic and confining our arguments to those that could be derived using rules of inference known to be sound, we could reduce any possibility of error to a small number of places, which could then be very thoroughly inspected to ensure that there were no fallacies lurking about. We wanted to go even further and prove using purely finitistic methods the consistency of set theory, but unfortunately Godel showed that to be impossible. So we're left with having to take some set of axioms on faith. Still, reasoning from the axioms of ZFC or some other similar system is the closest we can get to "there is no possibility whatsoever of error". That is why we accept the axiomatic method. A system in which paradoxes are accepted would in every reasonably complicated argument (which is most of them nowdays) leave us wondering "Is the theorem I proved really true, or did I just magnify a subtle paradox to the point of absurdity?" Unless you can find a way to resolve that fear -- and I don't think that you can -- your suggested approach, while perhaps a nice exercise for philosophers, is inadequate to the task of actually doing mathematics.

 

[DanTeplitskiy]

Dear Citan Uzuki,

Thanks a lot!

I know there should not be such a thing like dynamic formula! But Russell's formula actually is like that - I only write about it! I write that we have to consider the formula dinamically as the other choice is to consider R (unconsciously or not - after reading my paper) as both R ≠ {x: x∉x} and R = {x: x∉x} whithin one reasoning (I mean within standard formulation of Russell's paradox) - that is breaking the basic laws of logic. Maybe the word "dinamically" or the word "becomes" confuses people. Maybe.

It seems to me (maybe I am wrong) that you missed an important point: Analysis of contradictory equivalence R ∈ R ↔ R ∉ R (which I consider to be the resolution of the paradox) is not about whether the R exists or not. Not at all! It is quite a separate thing.

By the way do you understand what I mean by this: "If R includes itself, R is not that R = {x: x∉x}. What I mean is that under the assumption “R ∈ R” R
includes a member that is included in itself (R itself is such a member)."? Just in case: I am not asking if you agree or not.

Thanks a lot in advance!

Yours,

Dan

 

[micromass]

Dear Pwsnafu,

As far as I can see, if it is logic over math objects the logic may be either correct or not.

Yours,

Dan

And in this case I'm afraid that it's not correct.

 

[Citan Uzuki]

Dear Citan Uzuki,

Thanks a lot!

I know there should not be such a thing like dynamic formula! But Russell's formula actually is like that - I only write about it! I write that we have to consider the formula dinamically as the other choice is to consider R (unconsciously or not - after reading my paper) as both R ≠ {x: x∉x} and R = {x: x∉x} whithin one reasoning (I mean within standard formulation of Russell's paradox) - that is breaking the basic laws of logic. Maybe the word "dinamically" or the word "becomes" confuses people. Maybe.

Certainly after considering Russel's paradox we are led to the conclusion that R ≠ {x:x∉x} in the same line of reasoning that began with R = {x:x∉x}, but this doesn't mean the laws of logic are broken, it simply means that we have reached a contradiction. It's a peculiar feature of the human mind that we are able to envision actual contradictions while understanding that they are impossible, and it's why I brought up the example of "Let n be the greatest integer...," because the fact that we can understand that proof at all illustrates that we can imagine logical impossibilities. Thus, there is no problem in saying that when we grasp Russel's paradox we have understood that R = {x: x∉x} throughout.

It seems to me (maybe I am wrong) that you missed an important point: Analysis of contradictory equivalence R ∈ R ↔ R ∉ R (which I consider to be the resolution of the paradox) is not about whether the R exists or not. Not at all! It is quite a separate thing.

Yes, I understood that these are separate things. I was simply illustrating that the issues with envisioning a logical contradiction also arise in more traditional nonexistence proofs, and thus that there is nothing special about Russel's paradox in this regard.

By the way do you understand what I mean by this: "If R includes itself, R is not that R = {x: x∉x}. What I mean is that under the assumption “R ∈ R” R
includes a member that is included in itself (R itself is such a member)."? Just in case: I am not asking if you agree or not.

Yes, you're saying that when we introduce the assumption that R∈R into our line of reasoning that the symbol R can no longer be taken to refer to {x: x∉x}. I understand this, I just disagree.

 

[DanTeplitskiy]

Certainly after considering Russel's paradox we are led to the conclusion that R ≠ {x:x∉x} in the same line of reasoning that began with R = {x:x∉x}, but this doesn't mean the laws of logic are broken, it simply means that we have reached a contradiction. It's a peculiar feature of the human mind that we are able to envision actual contradictions while understanding that they are impossible, and it's why I brought up the example of "Let n be the greatest integer...," because the fact that we can understand that proof at all illustrates that we can imagine logical impossibilities. Thus, there is no problem in saying that when we grasp Russel's paradox we have understood that R = {x: x∉x} throughout.

Dear Citan Uzuki,

The laws of logic are broken not because we are led to the ultimate conclusion that R ≠ {x:x∉x} (that is there is no such collection of objects)

The laws of logic are broken because under each assumption separately R ≠ {x:x∉x} but
we still treat R like R = {x:x∉x} doing this R ∈ R -> R ∉ R And this R ∉ R -> R ∈ R. That is, in the standard formulation of Russell's paradox, after we really get R ≠ {x:x∉x} (under each assumption separately) we conclude something based on R = {x:x∉x} within the same reasoning (under each assumption separately).

Yours,

Dan

 

[Citan Uzuki]

Now you seem to be missing my point. My point is that there is no logical problem with treating R as {x: x∉x} throughout, even though R≠{x: x∉x}.

 

[DanTeplitskiy]

Now you seem to be missing my point. My point is that there is no logical problem with treating R as {x: x∉x} throughout, even though R≠{x: x∉x}.

Dear Citan Uzuki and other guys interested,

Please see new version of my paper, improved a lot.

https://docs.google.com/viewer?a=v&...YTZlMy00NDJhLWJjN2MtMDAzNDUzOWQ2Y2Ew&hl=en_US

The root part is on pages 12-14, that is, questions/comments on this part of my paper are mostly welcome!

Other parts of the paper have been improved as well.

Thanks a lot in advance!

Yours,

Dan

P. S. Citan, it seems to me that I got your point. Could you please look through pages 12-14? Maybe you will better get mine .

 

[DanTeplitskiy]

Now you seem to be missing my point. My point is that there is no logical problem with treating R as {x: x∉x} throughout, even though R≠{x: x∉x}.

Dear Citan,

In the following sentence:

“Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.”

Can you confirm that for you there is no logical problem with treating 'Dan' legless and conclude "...he should be taken to the nearest hospital for legless people.” even though has at least one leg (if he has an ankle)?

Please do not take it for a personal attack :) It is just to be sure I got your point quite correctly.

Thanks a lot in advance!

Yours,

Dan

 

[Citan Uzuki]

Dear Citan,

In the following sentence:

“Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.”

Can you confirm that for you there is no logical problem with treating 'Dan' legless and conclude "...he should be taken to the nearest hospital for legless people.” even though has at least one leg (if he has an ankle)?

Please do not take it for a personal attack :) It is just to be sure I got your point quite correctly.

Thanks a lot in advance!

Yours,

Dan

Yes, that is correct.

 

[DanTeplitskiy]

Yes, that is correct.

Dear Citan,

As far as I understand it is the classical logic that is the basis of the whole science (not only math).

In classical logic we can not conclude "...he should be taken to the nearest hospital for legless people.” if "...his right ankle is bleeding"

Yours,

Dan

 

[Citan Uzuki]

In classical logic we can not conclude "...he should be taken to the nearest hospital for legless people.” if "...his right ankle is bleeding"

Actually, we can. In classical logic, the material implication $A \rightarrow B$ is the same thing as $(\lnot A) \lor B$. Now if a man has no legs, then the antecedent "his right ankle is bleeding" is false, which means the implication "if his right ankle is bleeding, then he should be taken to the nearest hospital for legless people" is true.

 

[DanTeplitskiy]

Actually, we can. In classical logic, the material implication $A \rightarrow B$ is the same thing as $(\lnot A) \lor B$. Now if a man has no legs, then the antecedent "his right ankle is bleeding" is false, which means the implication "if his right ankle is bleeding, then he should be taken to the nearest hospital for legless people" is true.

Dear Citan,

Sorry, could you clarify the above for me? I can not see your point.

Yours,

Dan

 

[micromass]

Dear Citan,

Sorry, could you clarify the above for me? I can not see your point.

Yours,

Dan

His point is that if we take the implication $p\rightarrow q$ and if p is false, then $p\rightarrow q$ is always true. It doesn't matter what q is.

So if we have a legless man then p="his right ankle is bleeding" is always false. So $p\rightarrow q$= "if his right ankle is bleeding, then we take him to the hospital" is true.

 

[DanTeplitskiy]

His point is that if we take the implication $p\rightarrow q$ and if p is false, then $p\rightarrow q$ is always true. It doesn't matter what q is.

So if we have a legless man then p="his right ankle is bleeding" is always false. So $p\rightarrow q$= "if his right ankle is bleeding, then we take him to the hospital" is true.

Dear Micromass,

If we are given this: “Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.” - we do not know what is actually true (whether he is actually legless or not) as all we are given is contradiction.

Yours,

Dan

 

[micromass]

Dear Micromass,

If we are given this: “Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.” - we do not know what is actually true (whether he is actually legless or not) as all we are given is contradiction.

Yours,

Dan

If we are given that the man is legless then, p="his right ankle is bleeding" is false. Right??

And thus $p\rightarrow q$ is always true. This are the basic laws of logic.

 

[DanTeplitskiy]

If we are given that the man is legless then, p="his right ankle is bleeding" is false. Right??

And thus $p\rightarrow q$ is always true. This are the basic laws of logic.

Dear Micromass,

We are not given that the man is legless.

We are given this: “Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.” which is logically equivalent to "IF (Dan is a completely legless man And his right ankle is bleeding) THEN he should be taken to the nearest hospital for legless people."

Do you consider the conclusion "...he should be taken to the nearest hospital for legless people" can be drawn from “Dan is a completely legless man. If his right ankle is bleeding..."?

Yours,

Dan

 

[micromass]

Dear Micromass,

We are not given that the man is legless.

We are given this: “Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.” which is logically equivalent to "IF (Dan is a completely legless man And his right ankle is bleeding) THEN he should be taken to the nearest hospital for legless people."

Do you consider the conclusion "...he should be taken to the nearest hospital for legless people" can be drawn from “Dan is a completely legless man. If his right ankle is bleeding..."?

Yours,

Dan

Ah, but the same conclusion holds. Let p="Dan is completely legless and his right ankle is bleeding", then p is false. Thus the implication $p\rightarrow q$ holds true never the less.
Why is p false? Well, p is the conjuction p="Dan is completely legless" AND "His right ankle is bleeding". And this conjunction is clearly false.

 

[DanTeplitskiy]

Ah, but the same conclusion holds. Let p="Dan is completely legless and his right ankle is bleeding", then p is false. Thus the implication $p\rightarrow q$ holds true never the less.
Why is p false? Well, p is the conjuction p= "Dan is completely legless" AND "His right ankle is bleeding". And this conjunction is clearly false.

Dear Micromass,

If we can derive anything from a contradiction will the thing we derive this way be of any value for us? Will this thing be logically grounded if this could be anything?

For example, we can derive "he should be taken to the nearest hospital for legless people And he should not be taken to the nearest hospital for legless people" from any contradiction (including the above- "Dan is completely legless" AND "His right ankle is bleeding"). Right?

Yours,

Dan