Theorem nssss

Description: Negation of subclass relationship. Compare nss . (Contributed by NM, 30-Jun-2004) (Proof shortened by Andrew Salmon, 25-Jul-2011)

	Ref	Expression
Assertion	nssss	$$⊢\neg A\subseteq B\leftrightarrow \exists x\left(x\subseteq A\wedge \neg x\subseteq B\right)$$

Proof

Step	Hyp	Ref	Expression
1		exanali	$$⊢\exists x\left(x\subseteq A\wedge \neg x\subseteq B\right)\leftrightarrow \neg \forall x\left(x\subseteq A\to x\subseteq B\right)$$
2		ssextss	$$⊢A\subseteq B\leftrightarrow \forall x\left(x\subseteq A\to x\subseteq B\right)$$
3	1 2	xchbinxr	$$⊢\exists x\left(x\subseteq A\wedge \neg x\subseteq B\right)\leftrightarrow \neg A\subseteq B$$
4	3	bicomi	$$⊢\neg A\subseteq B\leftrightarrow \exists x\left(x\subseteq A\wedge \neg x\subseteq B\right)$$