Theorem abssf

Description: Class abstraction in a subclass relationship. (Contributed by Glauco Siliprandi, 26-Jun-2021)

	Ref	Expression
Hypothesis	abssf.1	$$⊢\underset{\_}{Ⅎ}xA$$
Assertion	abssf	$$⊢\left\{x|\phi \right\}\subseteq A\leftrightarrow \forall x\left(\phi \to x\in A\right)$$

Proof

Step	Hyp	Ref	Expression
1		abssf.1	$$⊢\underset{\_}{Ⅎ}xA$$
2	1	abid2f	$$⊢\left\{x|x\in A\right\}=A$$
3	2	sseq2i	$$⊢\left\{x|\phi \right\}\subseteq \left\{x|x\in A\right\}\leftrightarrow \left\{x|\phi \right\}\subseteq A$$
4		ss2ab	$$⊢\left\{x|\phi \right\}\subseteq \left\{x|x\in A\right\}\leftrightarrow \forall x\left(\phi \to x\in A\right)$$
5	3 4	bitr3i	$$⊢\left\{x|\phi \right\}\subseteq A\leftrightarrow \forall x\left(\phi \to x\in A\right)$$