Theorem elrab

Description: Membership in a restricted class abstraction, using implicit substitution. (Contributed by NM, 21-May-1999) Remove dependency on ax-13 . (Revised by Steven Nguyen, 23-Nov-2022)

	Ref	Expression
Hypothesis	elrab.1	$$⊢x=A\to \left(\phi \leftrightarrow \psi \right)$$
Assertion	elrab	$$⊢A\in \left\{x\in B|\phi \right\}\leftrightarrow \left(A\in B\wedge \psi \right)$$

Proof

Step	Hyp	Ref	Expression
1		elrab.1	$$⊢x=A\to \left(\phi \leftrightarrow \psi \right)$$
2		elex	$$⊢A\in \left\{x\in B|\phi \right\}\to A\in \mathrm{V}$$
3		elex	$$⊢A\in B\to A\in \mathrm{V}$$
4	3	adantr	$$⊢\left(A\in B\wedge \psi \right)\to A\in \mathrm{V}$$
5		eleq1	$$⊢x=A\to \left(x\in B\leftrightarrow A\in B\right)$$
6	5 1	anbi12d	$$⊢x=A\to \left(\left(x\in B\wedge \phi \right)\leftrightarrow \left(A\in B\wedge \psi \right)\right)$$
7		df-rab	$$⊢\left\{x\in B|\phi \right\}=\left\{x|\left(x\in B\wedge \phi \right)\right\}$$
8	6 7	elab2g	$$⊢A\in \mathrm{V}\to \left(A\in \left\{x\in B|\phi \right\}\leftrightarrow \left(A\in B\wedge \psi \right)\right)$$
9	2 4 8	pm5.21nii	$$⊢A\in \left\{x\in B|\phi \right\}\leftrightarrow \left(A\in B\wedge \psi \right)$$