**Moderator:** Flannel Jesus

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The absolute Russell set exists.

Proof.

Lemma. If a statement is false, then it materially implies some contradiction exists.

Proof of Lemma. It is given that a statement is false. So, the hypothesis of the material implication “the statement materially implies some contradiction exists” is false. By the truth table for material implication, any material implication with a false hypothesis is true. So, the statement materially implies some contradiction exists. This concludes the proof of the lemma.

Assume the absolute Russell set exists. As I’ve proved in my post at http://www.ilovephilosophy.com/viewtopic.php?p=2699066#p2699066, the absolute Russell set both is and is not an element of itself. For that reason, some contradiction exists. It follows by ex contradictione quodlibet that no contradiction exists. Discharge the assumption. Thus, by implication introduction, it is materially true that if the absolute Russell set exists, no contradiction exists. So, as a step of the proof that may be questionable, it is not materially true that if the absolute Russell set exists, some contradiction exists. In other words, the absolute Russell set exists does not materially imply some contradiction exists. Using modus tollens with the lemma and previous statement, the statement “the absolute Russell set exists” is not false. Thus, the statement is true. Therefore, the absolute Russell set exists. This concludes the proof.

A possible question with the proof is whether “the absolute Russell set exists materially implies no contradiction exists” strictly implies “the absolute Russell set exists does not materially imply some contradiction exists.” It's fairly intuitive that if a statement implies no contradiction, then it does not imply some contradiction. “The absolute Russell set exists materially implies no contradiction exists” does not tautologically imply “the absolute Russell set exists does not materially imply some contradiction exists;" the four corner entries in the bottommost two rows and the rightmost two columns of the following truth table reveal that in the case where the absolute Russell set does not exist, the former statement is true, but the latter is false.

Truth Table.

p = “The absolute Russell set exists.”

q = “Some contradiction exists.”

p.....q..|..¬q.....p → q.....p → ¬q.....¬(p → q)

T.....T...|...F.........T............F...............F

T.....F...|...T.........F............T...............T

F.....T...|...F.........T............T...............F

F.....F...|...T.........T............T...............F

This concludes the truth table.

The question of interest, however, regards strict implication, not tautological implication. Both statements describe the epistemically possible case in which the absolute Russell set does exist. So, the question can be reconsidered as whether, in the epistemically possible case in which the absolute Russell set exists, “no contradiction exists” strictly implies “it is not true that some contradiction exists.” The answer is intuitively yes; that strict implication is true in all cases. Therefore, the proof that the absolute Russell set exists is sound.

Paul E. Mokrzecki

Proof.

Lemma. If a statement is false, then it materially implies some contradiction exists.

Proof of Lemma. It is given that a statement is false. So, the hypothesis of the material implication “the statement materially implies some contradiction exists” is false. By the truth table for material implication, any material implication with a false hypothesis is true. So, the statement materially implies some contradiction exists. This concludes the proof of the lemma.

Assume the absolute Russell set exists. As I’ve proved in my post at http://www.ilovephilosophy.com/viewtopic.php?p=2699066#p2699066, the absolute Russell set both is and is not an element of itself. For that reason, some contradiction exists. It follows by ex contradictione quodlibet that no contradiction exists. Discharge the assumption. Thus, by implication introduction, it is materially true that if the absolute Russell set exists, no contradiction exists. So, as a step of the proof that may be questionable, it is not materially true that if the absolute Russell set exists, some contradiction exists. In other words, the absolute Russell set exists does not materially imply some contradiction exists. Using modus tollens with the lemma and previous statement, the statement “the absolute Russell set exists” is not false. Thus, the statement is true. Therefore, the absolute Russell set exists. This concludes the proof.

A possible question with the proof is whether “the absolute Russell set exists materially implies no contradiction exists” strictly implies “the absolute Russell set exists does not materially imply some contradiction exists.” It's fairly intuitive that if a statement implies no contradiction, then it does not imply some contradiction. “The absolute Russell set exists materially implies no contradiction exists” does not tautologically imply “the absolute Russell set exists does not materially imply some contradiction exists;" the four corner entries in the bottommost two rows and the rightmost two columns of the following truth table reveal that in the case where the absolute Russell set does not exist, the former statement is true, but the latter is false.

Truth Table.

p = “The absolute Russell set exists.”

q = “Some contradiction exists.”

p.....q..|..¬q.....p → q.....p → ¬q.....¬(p → q)

T.....T...|...F.........T............F...............F

T.....F...|...T.........F............T...............T

F.....T...|...F.........T............T...............F

F.....F...|...T.........T............T...............F

This concludes the truth table.

The question of interest, however, regards strict implication, not tautological implication. Both statements describe the epistemically possible case in which the absolute Russell set does exist. So, the question can be reconsidered as whether, in the epistemically possible case in which the absolute Russell set exists, “no contradiction exists” strictly implies “it is not true that some contradiction exists.” The answer is intuitively yes; that strict implication is true in all cases. Therefore, the proof that the absolute Russell set exists is sound.

Paul E. Mokrzecki

Paul E. Mokrzecki

- browser32
**Posts:**217**Joined:**Fri Jun 03, 2011 1:25 am**Location:**Hadley, Massachusetts, United States

I clicked on your link that the "absolute Russell set," whatever that is, exists. Sadly, there was no proof there ... just ANOTHER link to some claimed proof, which I did not click on. Fool me once dude.

So just to keep this discussion self-contained in this thread, please do me a favor.

1) Define what you mean by the "absolute Russell set," since I have no idea what that means nor does anyone else; and

2) Show your "proof" here.

If by "absolute Russell set" you mean the standard Russell set, namely the set of all sets that are not members of themselves, of course that set does NOT exist via an easy and standard proof.

So just to keep this discussion self-contained in this thread, please do me a favor.

1) Define what you mean by the "absolute Russell set," since I have no idea what that means nor does anyone else; and

2) Show your "proof" here.

If by "absolute Russell set" you mean the standard Russell set, namely the set of all sets that are not members of themselves, of course that set does NOT exist via an easy and standard proof.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

browser32 wrote:The absolute Russell set exists.

Proof.

Lemma. If a statement is false, then it materially implies some contradiction exists.

Proof of Lemma. It is given that a statement is false. So, the hypothesis of the material implication “the statement materially implies some contradiction exists” is false. By the truth table for material implication, any material implication with a false hypothesis is true. So, the statement materially implies some contradiction exists. This concludes the proof of the lemma.

Assume the absolute Russell set exists. As I’ve proved in my post at http://www.ilovephilosophy.com/viewtopic.php?p=2699066#p2699066, the absolute Russell set both is and is not an element of itself. For that reason, some contradiction exists. It follows by ex contradictione quodlibet that no contradiction exists. Discharge the assumption. Thus, by implication introduction, it is materially true that if the absolute Russell set exists, no contradiction exists. So, as a step of the proof that may be questionable, it is not materially true that if the absolute Russell set exists, some contradiction exists. In other words, the absolute Russell set exists does not materially imply some contradiction exists. Using modus tollens with the lemma and previous statement, the statement “the absolute Russell set exists” is not false. Thus, the statement is true. Therefore, the absolute Russell set exists. This concludes the proof.

A possible question with the proof is whether “the absolute Russell set exists materially implies no contradiction exists” strictly implies “the absolute Russell set exists does not materially imply some contradiction exists.” It's fairly intuitive that if a statement implies no contradiction, then it does not imply some contradiction. “The absolute Russell set exists materially implies no contradiction exists” does not tautologically imply “the absolute Russell set exists does not materially imply some contradiction exists;" the four corner entries in the bottommost two rows and the rightmost two columns of the following truth table reveal that in the case where the absolute Russell set does not exist, the former statement is true, but the latter is false.

Truth Table.

p = “The absolute Russell set exists.”

q = “Some contradiction exists.”

p.....q..|..¬q.....p → q.....p → ¬q.....¬(p → q)

T.....T...|...F.........T............F...............F

T.....F...|...T.........F............T...............T

F.....T...|...F.........T............T...............F

F.....F...|...T.........T............T...............F

This concludes the truth table.

The question of interest, however, regards strict implication, not tautological implication. Both statements describe the epistemically possible case in which the absolute Russell set does exist. So, the question can be reconsidered as whether, in the epistemically possible case in which the absolute Russell set exists, “no contradiction exists” strictly implies “it is not true that some contradiction exists.” The answer is intuitively yes; that strict implication is true in all cases. Therefore, the proof that the absolute Russell set exists is sound.

Paul E. Mokrzecki

If a statement is false, materially, regardless of the material, then that statement is neither true or false, for example that the Absolute Russell Set Exists/ ( or not); therefore there can not be a contradiction with whether it is contradictory or not.

If by absolute set is meant the idea or proposition , that an absolute set does or does not exist, that is a separate material, and that is a preoccupation of set theory, that is, whether the absolute set contains ALL sets, including the absolute set.

But the proposition that Russel's Absolute Set Exists, is not contradictory, therefore it does exist.

Topical and dynamic logic play a part here, and they are not necessarily connected, therefore there is no contradiction.

It seems so simple, but it was this 'discovery' , which made Frege exclaim, that he sees logic in a completely different way now.

- Meno_
- Philosopher
**Posts:**3119**Joined:**Tue Dec 08, 2015 2:39 am

Meno_ wrote:If a statement is false, materially, regardless of the material, then that statement is neither true or false

Sorry can you please clarify? If a statement is false then it's false.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

wtf wrote:Meno_ wrote:If a statement is false, materially, regardless of the material, then that statement is neither true or false

Sorry can you please clarify? If a statement is false then it's false.

A statement is neither true or false. It is merely a statement.

I can say that I am Einstein, and if You were to say that's obviously false, I could retort by saying, You are lacking certain facts, therefore my statement could be true or false, no contradiction there.

But if You were to press on further, I could retort just as well, and until that happens( You Pressing On) , no further clarification could be possible.

- Meno_
- Philosopher
**Posts:**3119**Joined:**Tue Dec 08, 2015 2:39 am

Meno_ wrote:wtf wrote:Meno_ wrote:If a statement is false, materially, regardless of the material, then that statement is neither true or false

Sorry can you please clarify? If a statement is false then it's false.

A statement is neither true or false. It is merely a statement.

I can say that I am Einstein, and if You were to say that's obviously false, I could retort by saying, You are lacking certain facts, therefore my statement could be true or false, no contradiction there.

But if You were to press on further, I could retort just as well, and until that happens( You Pressing On) , no further clarification could be possible.

We're in the realm of formal set theory. A statement along with an interpretation (model) has a truth value. If it's false it's false. Otherwise you're just confusing the issue and you're not helping browser to unconfuse himself.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

Meno_ wrote:wtf wrote:

Sorry can you please clarify? If a statement is false then it's false.

A statement is neither true or false. It is merely a statement.

I can say that I am Einstein, and if You were to say that's obviously false, I could retort by saying, You are lacking certain facts, therefore my statement could be true or false, no contradiction there.

But if You were to press on further, I could retort just as well, and until that happens( You Pressing On) , no further clarification could be possible.

We're in the realm of formal logic. A statement along with an interpretation (model) has a truth value. If it's false it's false. Otherwise you're just confusing the issue and you're not helping browser to unconfuse himself.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

Not at all. What is the issue? That it even befuddled Cantor himself to insanity?

Lets build this from the basement foundation to (through) the roof.

Lets build this from the basement foundation to (through) the roof.

- Meno_
- Philosopher
**Posts:**3119**Joined:**Tue Dec 08, 2015 2:39 am

Meno_ wrote:Not at all. What is the issue? That it even befuddled Cantor himself to insanity?

Utter nonsense. Cantor's pre-existing emotional issues were no doubt exacerbated by the hostility his theory of sets received, especially from his teacher Kronecker. But Cantor was certainly not confused by the Russell set. In fact your history is off. Cantor was not involved in that conversation at all. Frege published his book on logic that included unrestricted comprehension as an axiom. Russell showed that unrestricted comprehension leads to a contradiction. Russell's fix was type theory. Mainstream math instead adopted restricted comprehension.

Meno_ wrote:Lets build this from the basement foundation to (through) the roof.

Doubling down on browser's confusion won't get anyone anywhere.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

That Cantor was uninvolved , is a misnomer. He was involved in general set theory, as was Russell. That Leibnitz discovered the differential calculus independently, does not mean am absolute absence of total relations with Newton. The same with types of logic.

Whether set theory is simple or complex, is an ISSUE conjecture. That may ultimately be the issue of clarity.

Russell's contradiction was paradoxical, however.

Whether set theory is simple or complex, is an ISSUE conjecture. That may ultimately be the issue of clarity.

Russell's contradiction was paradoxical, however.

- Meno_
- Philosopher
**Posts:**3119**Joined:**Tue Dec 08, 2015 2:39 am

Meno_ wrote:That Cantor was uninvolved , is a misnomer. He was involved in general set theory, as was Russell. That Leibnitz discovered the differential calculus independently, does not mean am absolute absence of total relations with Newton. The same with types of logic.

Your claim that Cantor was confused and driven to insanity by Russell's paradox is pathetically wrong. That's a matter of historical fact.

Meno_ wrote:Whether set theory is simple or complex, is an ISSUE conjecture. That may ultimately be the issue of clarity.

English please. Whether set theory is simple or complex? What does that mean? What is an "ISSUE conjecture?" You are speaking nonsense.

Meno_ wrote:Russell's contradiction was paradoxical, however.

Not in the least. It simply showed that unrestricted comprehension leads to a contradiction; hence unrestricted comprehension may not be taken as an axiom of set theory if you wish set theory to be consistent. There's nothing paradoxical about that.

Euclid's proof of the infinitude of primes shows that if we have a finite list of the first n primes, there must be some prime not on the list. That's not paradoxical; it's a straightforward proof that there can't be only finitely many primes.

Likewise, Russell's proof shows that unrestricted set formation leads to a contradiction. Nothing paradoxical. It's just a straightforward proof of a mathematical fact.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

wtf wrote:Meno_ wrote:That Cantor was uninvolved , is a misnomer. He was involved in general set theory, as was Russell. That Leibnitz discovered the differential calculus independently, does not mean am absolute absence of total relations with Newton. The same with types of logic.

Your claim that Cantor was confused and driven to insanity by Russell's paradox is pathetically wrong. That's a matter of historical fact.Meno_ wrote:Whether set theory is simple or complex, is an ISSUE conjecture. That may ultimately be the issue of clarity.

English please. Whether set theory is simple or complex? What does that mean? What is an "ISSUE conjecture?" You are speaking nonsense.Meno_ wrote:Russell's contradiction was paradoxical, however.

Not in the least. It simply showed that unrestricted comprehension leads to a contradiction; hence unrestricted comprehension may not be taken as an axiom of set theory if you wish set theory to be consistent. There's nothing paradoxical about that.

Euclid's proof of the infinitude of primes shows that if we have a finite list of the first n primes, there must be some prime not on the list. That's not paradoxical; it's a straightforward proof that there can't be only finitely many primes.

Likewise, Russell's proof shows that unrestricted set formation leads to a contradiction. Nothing paradoxical. It's just a straightforward proof of a mathematical fact.

Since You asked, what is Absolute Russell Set, what is unrestricted set formation?

Is there a relation between the two concepts, or propositions?

- Meno_
- Philosopher
**Posts:**3119**Joined:**Tue Dec 08, 2015 2:39 am

Meno_ wrote:

Since You asked, what is Absolute Russell Set

I have no idea. That's a phrase browser used. I asked him to define it and I await his response.

The (standard, usual, historically accurate) Russell set is the set of all sets that are not elements of themselves. Symbolically:

$$R = \{x : x \notin x\}$$

Meno_ wrote:, what is unrestricted set formation?

Defining a set via a predicate. For example if P is a unary predicate such that P(x) = $$x \notin x$$ then we can rewrite the above definition as

$$R = \{x : P(x)\}$$

and as Russell noted, this leads to a contradiction. R both is and isn't an element of R.

Note that symbolically I wrote, "x such that P(x)". That's unrestricted, I'm allowing x to range over the entire universe. That leads to trouble.

The fix is to require restricted comprehension. ["Comprehension" is just a buzzword meaning "set formation."] When defining a set, we must write

$$\{x \in S : P(x)\}$$

In other words we have to start with some set S, and restrict x to range only over the elements of S.

Let's see how this fixes Russell's paradox. Say we let \(\mathbb Z\) be the set of integers. Then suppose we form a restricted Russell set:

$$R = \{x \in \mathbb Z : x \notin x\}$$

What are the elements of R? Well 3 is an integer. Is 3 an element of 3? No, so 3 is in R. Is 47 an element of 47? No, so 47 is in R. Continuing like this, we see that R is just the integers! So by requiring x to range over the elements of an already-existing set, we avoid the paradox.

So unresricted comprehension leads to a contradiction; and restricted comprehension doesn't.

See https://en.wikipedia.org/wiki/Axiom_sch ... cification for a longer discussion of the same point. It's simply that in order to legally form a set by taking all the objects that satisfy some predicate, we must restrict the objects to some set that's already known to exist.

Last edited by wtf on Sat Apr 21, 2018 1:32 am, edited 1 time in total.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

I understand this much in simpler terms , normal and abnormal sets are easier to understand without any contradiction between logic and language

Zemelo's treatment seems most informal for the purposes at hand.

Never the less, it is obvious that primal logic would lead to contradiction. But that such a nuance be so earth shaking to Frege shows , that it is not so easy to shift gears into sub typical nomenclature, from merely one of order as a way of differentiation.

I bring in calculus for the reason that reference to Euclid is as far removed ,do that lest we forget that 3000 years difference in thought can really up the ante.

Zemelo's treatment seems most informal for the purposes at hand.

Never the less, it is obvious that primal logic would lead to contradiction. But that such a nuance be so earth shaking to Frege shows , that it is not so easy to shift gears into sub typical nomenclature, from merely one of order as a way of differentiation.

I bring in calculus for the reason that reference to Euclid is as far removed ,do that lest we forget that 3000 years difference in thought can really up the ante.

- Meno_
- Philosopher
**Posts:**3119**Joined:**Tue Dec 08, 2015 2:39 am

Meno_ wrote:I understand this much in simpler terms

Are you being serious? Since I gave you a clear and simple explanation of restricted and unrestricted comprehension, a simple thank-you would suffice. Instead, you wrote a post of meaningless gobbledygook. Why? Can't you accept that I explained something to you that you asked about, and be happy that you learned something?

Meno_ wrote:, normal and abnormal sets are easier to understand

I have no idea what normal and abnormal sets are. There is no such terminology. There are of course many uses of "normal" in math, such as normal subgroups in group theory, and normal extensions in field theory. But none of these usages have any relation to set theory. There are simply no such things as normal and abnormal sets.

Meno_ wrote: without any contradiction between logic and language

What contradiction between logic and language do you think is involved in my explanation of restricted and unrestricted comprehension? Did you read the Wiki page I linked? Are you trying to learn anything? Or just trying to argue with the most basic of mathematical facts taught to undergraduates?

Meno_ wrote: Zemelo's treatment seems most informal for the pirposes at hand.

Really? I find that quite surprising, since it's Zermelo who did the most in the early days to formalize set theory. In fact Cantor gets the credit and Zermelo did most of the heavy lifting.

Tell me, which particular aspects of Zermelo's treatment of set theory do you find relevant here?

Meno_ wrote:Never the less, it is obvious that primal logic

I'm afraid that I don't know what primal logic is. Please define it.

Meno_ wrote: would lead to contradiction.

Until you define primal logic you're in no position to make such a claim.

Meno_ wrote:But that such a nuance be so earth shaking to Frege shows , that it is not so easy to shift gears into sub typical nomenclature, from merely one of order as a way of differentiation.

You appear to be speaking in bullshit-ese here. You're slinging words that have no referents. The above sentence says nothing and means nothing. Are you playing games?

Meno_ wrote:I bring in calculus

You did? When? Calculus has nothing at all to do with any of this.

Meno_ wrote: for the reason that reference to Euclid is as far removed

Euclid has nothing to do with any of this.

Meno_ wrote: , that we forget that 3000 years difference in thought can really up the ante.

I hope you won't mind if I'm direct. You're full of baloney and clearly not engaging in a good faith dialog. Have a nice day.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

Well, emotionalism doesen't enter here and I could refer you to normal and abnormal sets, but briefly, an abnormal set contains all sets including itself and a normal one does not. Primal logic is reducible logic which ends in contradiction, and all others do not.

If philosophy be a matter of naivete, it is as understandable as that, with which Cantor was concerned, yet surely one couldn't label him ignorant.

If Euclid has nothing to do with this, why did You bring it up?

I see he has everything to do with it& absolutely.

Of course I do not see anything in Your directness, but bravado, and that is perfectly forgivable, as far as I am concerned.

Its like I don't know who wrote it, Jane Austin, You can make sense without being sensible, or having sensibility. And remember my name , Meno, and You know how he learned.

I chose it somewhere along that line, for an obvious purpose.

And a very good day to you as well.

If philosophy be a matter of naivete, it is as understandable as that, with which Cantor was concerned, yet surely one couldn't label him ignorant.

If Euclid has nothing to do with this, why did You bring it up?

I see he has everything to do with it& absolutely.

Of course I do not see anything in Your directness, but bravado, and that is perfectly forgivable, as far as I am concerned.

Its like I don't know who wrote it, Jane Austin, You can make sense without being sensible, or having sensibility. And remember my name , Meno, and You know how he learned.

I chose it somewhere along that line, for an obvious purpose.

And a very good day to you as well.

- Meno_
- Philosopher
**Posts:**3119**Joined:**Tue Dec 08, 2015 2:39 am

wtf wrote:I clicked on your link that the "absolute Russell set," whatever that is, exists. Sadly, there was no proof there ... just ANOTHER link to some claimed proof, which I did not click on.

The link in my previous post should bring you to a proof. If you want to better understand the proof, then please follow the appropriate additional referrals. It's all there; you just have to click and scroll a bit.

wtf wrote:If by "absolute Russell set" you mean the standard Russell set, namely the set of all sets that are not members of themselves, of course that set does NOT exist via an easy and standard proof.

By the absolute Russell set, I do not mean the set of all sets that are not members of themselves; I mean the set of all things that are not members of themselves. As I have proved privately in first-order logic, as I explained in the past debate "The Absolute Russell Set Exists" I proposed at http://www.debate.org/debates/The-Absol ... -Exists/1/, which I have already indirectly referred to, and as may be common knowledge, the set does not "exist via an easy and standard proof." I, notwithstanding, present proof to the contrary.

wtf wrote:Russell showed that unrestricted comprehension leads to a contradiction.

And that contradiction, through ex contradictione quodlibet, leads to no contradiction at all. I brought that up in my original post.

wtf wrote:unrestricted comprehension may not be taken as an axiom of set theory if you wish set theory to be consistent.

That's not true. Set theory is consistent both with and without unrestricted comprehension. By the law of non-contradiction, set theory is consistent or inconsistent. If set theory is consistent, then by reiteration, it is consistent. If set theory is inconsistent, then by ex contradictione quodlibet, it is consistent. Therefore by disjunction elimination, set theory is consistent.

Paul E. Mokrzecki

- browser32
**Posts:**217**Joined:**Fri Jun 03, 2011 1:25 am**Location:**Hadley, Massachusetts, United States

Contradictione quidlobet is a shift to a functional approach to verifying a second order logic , grounding a utilitarian-positivism, meaning it tries to overcome the naturalistic fallacy, utilizing a neo-Kantism.

Marbourg school-Cassirer.

It is, as if in Ayer's behaviorist model, what signifies meaning. through the function of language where usage determines the logical structure , not some intrinsic property.

This structural hierarchy coincides with the epistemological significance of semantic usage, rather then the intractibly logical signifier, of a first order hierarchy , which does cause contradiction. The usage will will synthesize this by reasserting the logical unity between usage , function and meaning.(Typical logical progression. Is de-emphasized, in favor of logically ordered sequences).

This is the ontic, rather then the ontological take on it.

I hate to appear defensive, but a proof predicated on an assumption is only part of the story, and this is an effort to show the fallibility of that assumption.

I just learned this : in classical logic the principle of acceptance of contradictory statements is valid, but in relevance logic it is rejected. This backs up the idea that such a logical proof is not totally relevant therefore invalid.

The paradox comes in where two different types of logic are conflated.

It is an interesting mind game, however the claims upon which the assumptions are based have lost their significance.

But thanks for the opportunity !

Marbourg school-Cassirer.

It is, as if in Ayer's behaviorist model, what signifies meaning. through the function of language where usage determines the logical structure , not some intrinsic property.

This structural hierarchy coincides with the epistemological significance of semantic usage, rather then the intractibly logical signifier, of a first order hierarchy , which does cause contradiction. The usage will will synthesize this by reasserting the logical unity between usage , function and meaning.(Typical logical progression. Is de-emphasized, in favor of logically ordered sequences).

This is the ontic, rather then the ontological take on it.

I hate to appear defensive, but a proof predicated on an assumption is only part of the story, and this is an effort to show the fallibility of that assumption.

I just learned this : in classical logic the principle of acceptance of contradictory statements is valid, but in relevance logic it is rejected. This backs up the idea that such a logical proof is not totally relevant therefore invalid.

The paradox comes in where two different types of logic are conflated.

It is an interesting mind game, however the claims upon which the assumptions are based have lost their significance.

But thanks for the opportunity !

- Meno_
- Philosopher
**Posts:**3119**Joined:**Tue Dec 08, 2015 2:39 am

This post includes a second proof that the absolute Russell set exists. The second proof is a modified and arguably stronger version of the first proof. All conditional statements in the modified proof are strict; that differs from the first proof, in which some conditional statements were material.

Second Proof.

Lemma 2. If a statement is false, then if the statement is true, then some contradiction exists.

Proof of Lemma 2. It is given that a statement is false. Assume the statement is true. Then some contradiction exists. Discharge the assumption. So, by conditional introduction, if the statement is true, then some contradiction exists. This concludes the proof of the lemma.

Postulate 2. If one statement implies it is not true that a second statement, then the first statement does not imply the second statement.

Argument for Postulate 2. Postulate 2 seems intuitively true, despite the fact that Postulate 2 would traditionally be false due to the case in which the first statement is necessarily false, since, as suggested by Property 4 of Theorem 2.1 on page 11 of the March 23, 2014 Edition of Want Theory #6 by me, any conditional statement with a necessarily false hypothesis is true. However, a conditional statement with a necessarily false hypothesis is only vacuously true, and treats the false hypothesis as true, yielding a contradiction that by ex contradictione quodlibet implies and does not imply everything. So, such a conditional statement is true and false simultaneously. Thus, for the sake of Postulate 2, if the first statement is necessarily false, then the conditional statement "if the first statement, then the second statement" is considered false. This concludes the argument for Postulate 2.

Assume the statement "the absolute Russell set exists" is true. By simplification, the absolute Russell set exists. As I’ve proved in my post at http://www.ilovephilosophy.com/viewtopic.php?p=2699066#p2699066, the absolute Russell set both is and is not an element of itself. wtf was critical of that citation in my first proof, so I will explain the linked proof some more here. That proof invokes the fact that the absolute Russell set is an element of itself if and only if it is not an element of itself. That fact is readily derived from a property of the absolute Russell set: a thing is an element of the absolute Russell set if and only if it is not an element of itself. As I had done in the debate I previously proposed and referred to, for more context, I cite pages 432, 433, and 434 of Language, Proof and Logic (1999, 2000, 2002, 2003, 2007, 2008) by Jon Barwise and John Etchemendy, and https://en.wikipedia.org/wiki/Russell%27s_paradox#Formal_presentation.

Since the absolute Russell set is and is not an element of itself, some contradiction exists. It follows by ex contradictione quodlibet that no contradiction exists. Discharge the assumption. Thus, by conditional introduction, if the statement "the absolute Russell set exists" is true, then no contradiction exists. So, by Postulate 2, it is not true that if the statement "the absolute Russell set exists" is true, then some contradiction exists. Invoking modus tollens with Lemma 2 and the previous sentence, it is shown that the statement “the absolute Russell set exists” is not false. Thus, the statement is true. Therefore, the absolute Russell set exists. This concludes the second proof.

As to the first proof, although the truth table for material implication may be correct, it is incomplete. Material implication is not truth functional as has been believed. As suggested by my argument for Postulate 2, whenever the hypothesis of a material implication is false, the implication is not just true, but it is also false. That claim is supported by the fact that if a false statement is assumed true, then by ex contradictione quodlibet, all statements are true and false simultaneously. So, a false statement always materially implies and always does not materially imply every statement. For these reasons, the following Postulate 1, which was implicitly invoked in the first proof at the step I explicitly claimed may be questionable, is true.

Postulate 1. If one statement materially implies it is not true that a second statement, then the first statement does not materially imply the second statement.

Note that while the hypothesis and conclusion of Postulate 1 are a material implication and material nonimplication, respectively, Postulate 1 itself is a strict implication.

Second Proof.

Lemma 2. If a statement is false, then if the statement is true, then some contradiction exists.

Proof of Lemma 2. It is given that a statement is false. Assume the statement is true. Then some contradiction exists. Discharge the assumption. So, by conditional introduction, if the statement is true, then some contradiction exists. This concludes the proof of the lemma.

Postulate 2. If one statement implies it is not true that a second statement, then the first statement does not imply the second statement.

Argument for Postulate 2. Postulate 2 seems intuitively true, despite the fact that Postulate 2 would traditionally be false due to the case in which the first statement is necessarily false, since, as suggested by Property 4 of Theorem 2.1 on page 11 of the March 23, 2014 Edition of Want Theory #6 by me, any conditional statement with a necessarily false hypothesis is true. However, a conditional statement with a necessarily false hypothesis is only vacuously true, and treats the false hypothesis as true, yielding a contradiction that by ex contradictione quodlibet implies and does not imply everything. So, such a conditional statement is true and false simultaneously. Thus, for the sake of Postulate 2, if the first statement is necessarily false, then the conditional statement "if the first statement, then the second statement" is considered false. This concludes the argument for Postulate 2.

Assume the statement "the absolute Russell set exists" is true. By simplification, the absolute Russell set exists. As I’ve proved in my post at http://www.ilovephilosophy.com/viewtopic.php?p=2699066#p2699066, the absolute Russell set both is and is not an element of itself. wtf was critical of that citation in my first proof, so I will explain the linked proof some more here. That proof invokes the fact that the absolute Russell set is an element of itself if and only if it is not an element of itself. That fact is readily derived from a property of the absolute Russell set: a thing is an element of the absolute Russell set if and only if it is not an element of itself. As I had done in the debate I previously proposed and referred to, for more context, I cite pages 432, 433, and 434 of Language, Proof and Logic (1999, 2000, 2002, 2003, 2007, 2008) by Jon Barwise and John Etchemendy, and https://en.wikipedia.org/wiki/Russell%27s_paradox#Formal_presentation.

Since the absolute Russell set is and is not an element of itself, some contradiction exists. It follows by ex contradictione quodlibet that no contradiction exists. Discharge the assumption. Thus, by conditional introduction, if the statement "the absolute Russell set exists" is true, then no contradiction exists. So, by Postulate 2, it is not true that if the statement "the absolute Russell set exists" is true, then some contradiction exists. Invoking modus tollens with Lemma 2 and the previous sentence, it is shown that the statement “the absolute Russell set exists” is not false. Thus, the statement is true. Therefore, the absolute Russell set exists. This concludes the second proof.

As to the first proof, although the truth table for material implication may be correct, it is incomplete. Material implication is not truth functional as has been believed. As suggested by my argument for Postulate 2, whenever the hypothesis of a material implication is false, the implication is not just true, but it is also false. That claim is supported by the fact that if a false statement is assumed true, then by ex contradictione quodlibet, all statements are true and false simultaneously. So, a false statement always materially implies and always does not materially imply every statement. For these reasons, the following Postulate 1, which was implicitly invoked in the first proof at the step I explicitly claimed may be questionable, is true.

Postulate 1. If one statement materially implies it is not true that a second statement, then the first statement does not materially imply the second statement.

Note that while the hypothesis and conclusion of Postulate 1 are a material implication and material nonimplication, respectively, Postulate 1 itself is a strict implication.

Paul E. Mokrzecki

- browser32
**Posts:**217**Joined:**Fri Jun 03, 2011 1:25 am**Location:**Hadley, Massachusetts, United States

browser32 wrote:This post includes a second proof that the absolute Russell set exists.

Bizarre that you'd write all this word salad without bothering to define the "absolute Russell set," which has no meaning in standard math or logic.

But if you merely mean the class of all sets that are not members of themselves, I already showed you how to define it:

$$R = \{x : x \notin x\}$$

This is a perfectly well-defined proper class. It's just not a set.

Proper classes are formally defined in some versions of set theory. In ZFC where they're not formally defined, they're used informally, meaning a class that's "too big" to be a set. The class of all sets, the class of all Abelian groups, the class of all topological spaces, the class of all sets that are not members of themselves, etc. All those are proper classes either in the formal or informal sense.

With your interest in the subject, why don't you study some basic set theory and logic? You'd find it interesting and fun.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

wtf wrote:Bizarre that you'd write all this word salad without bothering to define the "absolute Russell set," which has no meaning in standard math or logic.

I do not have to define the absolute Russell set. A property I described it to have should have sufficed for the discussion so far. I now define the absolute Russell set as the set that has a property I have ascribed to it in my previous post; I define the absolute Russell set as the set such that

browser32 wrote:a thing is an element of the absolute Russell set if and only if it is not an element of itself.

wtf wrote:With your interest in the subject, why don't you study some basic set theory and logic?

I've already done that. And I am not convinced, despite formal proofs, that neither the absolute Russell set nor the universal set exist.

Just because a set is inconsistent, doesn't mean it doesn't exist.

Paul E. Mokrzecki

- browser32
**Posts:**217**Joined:**Fri Jun 03, 2011 1:25 am**Location:**Hadley, Massachusetts, United States

I've already showed you how to define the class of all sets (or things, if you prefer) that are not members of themselves:

$$R = \{x : x \notin x\}$$

As far as your saying you don't have to define something to show it exists, surely even you can see that's insane.

$$R = \{x : x \notin x\}$$

As far as your saying you don't have to define something to show it exists, surely even you can see that's insane.

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

I'm not interested in classes; I'm interested in sets. I'm skeptical of the difference between sets and classes. The difference seems rather artificial and unnecessary.

Just because a thing is not defined, doesn't mean it doesn't exist.

wtf wrote:As far as your saying you don't have to define something to show it exists, surely even you can see that's insane.

Just because a thing is not defined, doesn't mean it doesn't exist.

Paul E. Mokrzecki

- browser32
**Posts:**217**Joined:**Fri Jun 03, 2011 1:25 am**Location:**Hadley, Massachusetts, United States

browser32 wrote:I'm not interested in classes; I'm interested in sets. I'm skeptical of the difference between sets and classes. The difference seems rather artificial and unnecessary.wtf wrote:As far as your saying you don't have to define something to show it exists, surely even you can see that's insane.

Just because a thing is not defined, doesn't mean it doesn't exist.

Are there any sane people on this forum?

- wtf
**Posts:**229**Joined:**Sun Dec 06, 2015 5:47 am

wtf wrote:Are there any sane people on this forum?

Yes, I am a sane person on this forum.

There's a difference between a definition and a description. All definitions are descriptions, but not all descriptions are definitions. The description I had initially given of the absolute Russell set, while on another web page, exclusive, and sufficient for my purposes, was not regarded by me as a formal definition. I believe I may have regarded so intentionally, because I may have suspected the definition of the absolute Russell set was already given in a printed textbook, the one by Barwise and Etchemendy.

I should not have to define in this thread everything that has already been defined elsewhere. That decreases the value of this thread and increases its unattractiveness to people looking for original thought.

Paul E. Mokrzecki

- browser32
**Posts:**217**Joined:**Fri Jun 03, 2011 1:25 am**Location:**Hadley, Massachusetts, United States

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