Theorem nss

Description: Negation of subclass relationship. Exercise 13 of TakeutiZaring p. 18. (Contributed by NM, 25-Feb-1996) (Proof shortened by Andrew Salmon, 21-Jun-2011)

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

Proof

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