Theorem nssrex

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

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

Proof

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