= Barber paradox
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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.
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