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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Suppose that there is a [[set|0SJ3RX9TXA]] $R$ such that, for each set $x$, $x \\in R$ if and only if $x \\notin x$.\n2. The object $R$ is itself a set. So we may take $x = R$ in step 1. This gives: $R \\in R$ if and only if $R \\notin R$.\n3. Suppose $R \\in R$. By step 2, $R \\notin R$. This contradicts the assumption $R \\in R$.\n4. So $R \\in R$ is false, that is, $R$ is not an [[element|0EZ758XE73]] of $R$. In symbols, $R \\notin R$.\n5. By step 2 and step 4, $R \\in R$.\n6. Steps 4 and 5 give both $R \\notin R$ and $R \\in R$, a contradiction.\n7. So the assumption of step 1 is false. There is no set $R$ such that, for each set $x$, $x \\in R$ if and only if $x \\notin x$.\n\n## Idea\n\nAsk whether $R$ contains itself. The defining property of $R$ says that $R \\in R$ exactly when $R \\notin R$. No statement can be equivalent to its own negation, so no such $R$ exists.\n",
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