Theorem abrexexg

Description: Existence of a class abstraction of existentially restricted sets. The class B can be thought of as an expression in x (which is typically a free variable in the class expression substituted for B ) and the class abstraction appearing in the statement as the class of values B as x varies through A . If the "domain" A is a set, then the abstraction is also a set. Therefore, this statement is a kind of Replacement. This can be seen by tracing back through the path mptexg , funex , fnex , resfunexg , and funimaexg . See also abrexex2g . There are partial converses under additional conditions, see for instance abnexg . (Contributed by NM, 3-Nov-2003) (Proof shortened by Mario Carneiro, 31-Aug-2015)

	Ref	Expression
Assertion	abrexexg	$$⊢A\in V\to \left\{y|\exists x\in Ay=B\right\}\in \mathrm{V}$$

Proof

Step	Hyp	Ref	Expression
1		eqid	$$⊢\left(x\in A⟼B\right)=\left(x\in A⟼B\right)$$
2	1	rnmpt	$$⊢\mathrm{ran}\left(x\in A⟼B\right)=\left\{y|\exists x\in Ay=B\right\}$$
3		mptexg	$$⊢A\in V\to \left(x\in A⟼B\right)\in \mathrm{V}$$
4		rnexg	$$⊢\left(x\in A⟼B\right)\in \mathrm{V}\to \mathrm{ran}\left(x\in A⟼B\right)\in \mathrm{V}$$
5	3 4	syl	$$⊢A\in V\to \mathrm{ran}\left(x\in A⟼B\right)\in \mathrm{V}$$
6	2 5	eqeltrrid	$$⊢A\in V\to \left\{y|\exists x\in Ay=B\right\}\in \mathrm{V}$$