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Metamath Proof Explorer

Theorem nssrex

Description: Negation of subclass relationship. (Contributed by Glauco Siliprandi, 3-Mar-2021)

		Ref	Expression
	Assertion	nssrex	$⊢ \neg A \subseteq B \leftrightarrow \exists x \in A \neg x \in B$

Step	Hyp	Ref	Expression
1		nss	$⊢ \neg A \subseteq B \leftrightarrow \exists x (x \in A \land \neg x \in B)$
2		df-rex	$⊢ \exists x \in A \neg x \in B \leftrightarrow \exists x (x \in A \land \neg x \in B)$
3	1 2	bitr4i	$⊢ \neg A \subseteq B \leftrightarrow \exists x \in A \neg x \in B$