Theorem abss

Description: Class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006)

	Ref	Expression
Assertion	abss	$$⊢\left\{x|\phi \right\}\subseteq A\leftrightarrow \forall x\left(\phi \to x\in A\right)$$

Proof

Step	Hyp	Ref	Expression
1		abid2	$$⊢\left\{x|x\in A\right\}=A$$
2	1	sseq2i	$$⊢\left\{x|\phi \right\}\subseteq \left\{x|x\in A\right\}\leftrightarrow \left\{x|\phi \right\}\subseteq A$$
3		ss2ab	$$⊢\left\{x|\phi \right\}\subseteq \left\{x|x\in A\right\}\leftrightarrow \forall x\left(\phi \to x\in A\right)$$
4	2 3	bitr3i	$$⊢\left\{x|\phi \right\}\subseteq A\leftrightarrow \forall x\left(\phi \to x\in A\right)$$