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