This post includes a third proof that the absolute Russell set exists. The third proof invokes an idea that was used in the second proof, but is simpler and more direct than the second proof. The idea is expressed in the following quotation from the argument for Postulate 2 in the second proof; the second proof is located at

viewtopic.php?p=2699904#p2699904.

browser32 wrote: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.

Third Proof.

Premises: s is a statement. Under the assumption that "

s and it is not true that

s,"

s is true. It is not true that: under the assumption that "

s and it is not true that

s,"

s is true.

Conclusion: The absolute Russell set exists.

Statements (Reasons)1.

s is a statement. (Premise)

2. Under the assumption that "

s and it is not true that

s,"

s is true. (Premise)

3. It is not true that: under the assumption that "

s and it is not true that

s,"

s is true. (Premise)

4. Some contradiction exists. ((3) is the negation of (2))

5. The absolute Russell set exists. (

Ex contradictione quodlibet)

This concludes the third proof.

Premise (1) is a prescribed description of

s. Each of premises (2) and (3) is evident by itself.