Metamath Proof Explorer


Theorem elabd2

Description: Membership in a class abstraction, using implicit substitution. Deduction version of elab . (Contributed by Gino Giotto, 12-Oct-2024) (Revised by BJ, 16-Oct-2024)

Ref Expression
Hypotheses elabd2.ex φ A V
elabd2.eq φ B = x | ψ
elabd2.is φ x = A ψ χ
Assertion elabd2 φ A B χ

Proof

Step Hyp Ref Expression
1 elabd2.ex φ A V
2 elabd2.eq φ B = x | ψ
3 elabd2.is φ x = A ψ χ
4 2 eleq2d φ A B A x | ψ
5 elab6g A V A x | ψ x x = A ψ
6 4 5 sylan9bb φ A V A B x x = A ψ
7 elisset A V x x = A
8 3 pm5.74da φ x = A ψ x = A χ
9 8 albidv φ x x = A ψ x x = A χ
10 19.23v x x = A χ x x = A χ
11 9 10 bitrdi φ x x = A ψ x x = A χ
12 pm5.5 x x = A x x = A χ χ
13 11 12 sylan9bb φ x x = A x x = A ψ χ
14 7 13 sylan2 φ A V x x = A ψ χ
15 6 14 bitrd φ A V A B χ
16 1 15 mpdan φ A B χ