Metamath Proof Explorer


Theorem abeqin

Description: Intersection with class abstraction. (Contributed by Peter Mazsa, 21-Jul-2021)

Ref Expression
Hypotheses abeqin.1 A = B C
abeqin.2 B = x | φ
Assertion abeqin A = x C | φ

Proof

Step Hyp Ref Expression
1 abeqin.1 A = B C
2 abeqin.2 B = x | φ
3 2 ineq1i B C = x | φ C
4 dfrab2 x C | φ = x | φ C
5 3 1 4 3eqtr4i A = x C | φ