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’s_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.