[proofplan] Apply the Axiom of Foundation to the singleton set $\{x\}$. Foundation gives an element of $\{x\}$ that is disjoint from $\{x\}$; since the only element of $\{x\}$ is $x$, this says $x \cap \{x\} = \varnothing$. If $x \in x$, then $x$ would lie in both $x$ and $\{x\}$, contradicting that disjointness. [/proofplan] [step:Apply Foundation to the singleton containing $x$] Let $x$ be an arbitrary set. Define the singleton set \begin{align*} S := \{x\}. \end{align*} Then $S$ is nonempty because $x \in S$. By the Axiom of Foundation applied to the nonempty set $S$, there exists a set $a$ such that \begin{align*} a \in S \quad\text{and}\quad a \cap S = \varnothing. \end{align*} Since $S = \{x\}$, the relation $a \in S$ implies $a = x$. Substituting $a = x$ into the disjointness conclusion gives \begin{align*} x \cap S = \varnothing. \end{align*} [/step] [step:Derive the contradiction from self-membership] Assume, for contradiction, that $x \in x$. Since also $x \in S$ by the definition of $S$, the set $x$ belongs to both $x$ and $S$. Hence \begin{align*} x \in x \cap S. \end{align*} This contradicts the conclusion $x \cap S = \varnothing$ from the previous step. Therefore the assumption $x \in x$ is false, so $x \notin x$. Because $x$ was arbitrary, every set is not an element of itself. [/step]