Russell's Paradox (1901)

Where Φ denotes the axioms of Frege's naïve set theory, we can prove the following argument:

Φ :- {x|x∉x} ∈ {x|x∉x} & ~ {x|x∉x} ∈ {x|x∉x}

In other words, the axioms of Frege's naïve set theory are inconsistent.

A = {x | x ∉ x}, i.e. the set of sets which are not members of themselves.

If A ∈ A then A ∉ A.

If A ∉ A then A ∈ A.

Moral: Frege's naive set theory, based on the axioms of extensionality (two sets are equal iff they contain the same elements) and unlimited set abstraction, leads to a contradiction. Proposed solutions (1908): (a) Russell's type theory; (b) Zermelo's axiomatic set theory, which later evolved into the canonical Zermelo-Fraenkel (ZF) set theory.

Extensionality: ∀x∀y((∀z(z∈x ↔ z∈y)) ↔ x=y)

Reality consists solely of sets, along with a single primitive relation ∈ (any restrictions?). Equality is defined in terms of ∈ by means of extensionality.