**Proof Plan.** The argument is a direct self-referential diagonalisation. We assume $R = \{x : x \notin x\}$ exists as a [set](/page/Set) and ask whether $R \in R$. Both cases yield the opposite conclusion, giving a contradiction. There is only one logical step: instantiate the defining property of $R$ at $R$ itself.
**Step 1 (Derive the biconditional).** By the definition of $R$, membership in $R$ is governed by the property $P(x) :\Leftrightarrow x \notin x$. Applying this to $R$ itself gives
\begin{align*}
R \in R \iff R \notin R.
\end{align*}
**Step 2 (Extract the contradiction).** Suppose $R \in R$. Then by the biconditional, $R \notin R$ — contradicting the assumption. Suppose instead $R \notin R$. Then by the biconditional, $R \in R$ — again a contradiction. Since $R \in R \lor R \notin R$ is a tautology of classical logic, both cases are exhausted and both yield contradictions. Therefore no such set $R$ can exist. $\blacksquare$
