Metamath Proof Explorer


Theorem abeqin

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

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

Proof

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