Russell's paradox ($R\in R⟺R\notin R$)

Unrestricted formation of a set $R=\{x:x\notin x\}$ contradicts its own membership condition: substituting $R$ for $x$ gives the displayed equivalence. Thus a predicate cannot always define a set of all objects satisfying it. Restricted set comprehension forms a subset of a previously given set, avoiding this argument. The contradiction does not assert that a set whose elements happen not to contain themselves is impossible; it rules out the universal collection with exactly the specified membership condition.

For a binary relation $S$, no object $b$ can satisfy $S(b,p)\leftrightarrow \neg S(p,p)$ for every $p$. Putting $p=b$ would give $s\leftrightarrow \neg s$ for $s=S(b,b)$. The forward direction derives $\neg s$ by a negation introduction subproof assuming $s$; the reverse direction then gives $s$, a contradiction. Thus this argument even avoids needing the law of excluded middle. The familiar interpretation reads $S(b,p)$ as “$b$ shaves $p$”; the logical obstruction is self-application rather than a physical peculiarity of shaving.