Theorem dfss6

Description: Alternate definition of subclass relationship. (Contributed by RP, 16-Apr-2020)

	Ref	Expression
Assertion	dfss6	$$⊢A\subseteq B\leftrightarrow \neg \exists x\left(x\in A\wedge \neg x\in B\right)$$

Proof

Step	Hyp	Ref	Expression
1		df-ss	$$⊢A\subseteq B\leftrightarrow \forall x\left(x\in A\to x\in B\right)$$
2		notnotb	$$⊢\forall x\left(x\in A\to x\in B\right)\leftrightarrow \neg \neg \forall x\left(x\in A\to x\in B\right)$$
3	1 2	bitri	$$⊢A\subseteq B\leftrightarrow \neg \neg \forall x\left(x\in A\to x\in B\right)$$
4		exanali	$$⊢\exists x\left(x\in A\wedge \neg x\in B\right)\leftrightarrow \neg \forall x\left(x\in A\to x\in B\right)$$
5	3 4	xchbinxr	$$⊢A\subseteq B\leftrightarrow \neg \exists x\left(x\in A\wedge \neg x\in B\right)$$