Metamath Proof Explorer


Theorem elabgw

Description: Membership in a class abstraction, using two substitution hypotheses to avoid a disjoint variable condition on x and A . This is to elabg what sbievw2 is to sbievw . (Contributed by SN, 20-Apr-2024)

Ref Expression
Hypotheses elabgw.1 x = y φ ψ
elabgw.2 y = A ψ χ
Assertion elabgw A V A x | φ χ

Proof

Step Hyp Ref Expression
1 elabgw.1 x = y φ ψ
2 elabgw.2 y = A ψ χ
3 eleq1 y = A y x | φ A x | φ
4 df-clab y x | φ y x φ
5 1 sbievw y x φ ψ
6 4 5 bitri y x | φ ψ
7 3 2 6 vtoclbg A V A x | φ χ