This is equivalent to the question "For which triangles the Pythagorean theorem applies?". If the hypothesis is false the conclusion could be true or false and the implication will still be true. So I would say the empty sets follows the axiom of regularity because the axiom is not interested in it even though it doesn't have a disjoint member with itself (cause it doesn't have any members).
Is there a typo in the statement of the regularity axiom? Every non-empty set A has a member A with m \cap A = \varnothing. Shall the second A be replaced by m? Thank you!
Does anyone have a reference for the formalisation of the proof of 2->3?
Very clear explanation! Thank you!
Why can't a set belong to itself in the first place? Will such set lead to any contradictions?
Russell paradox
Is empty set not follow the axiom of regularity?
This is equivalent to the question "For which triangles the Pythagorean theorem applies?". If the hypothesis is false the conclusion could be true or false and the implication will still be true. So I would say the empty sets follows the axiom of regularity because the axiom is not interested in it even though it doesn't have a disjoint member with itself (cause it doesn't have any members).
Is there a typo in the statement of the regularity axiom? Every non-empty set A has a member A with m \cap A = \varnothing. Shall the second A be replaced by m? Thank you!
Yes