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 φAV
elabd2.eq φB=x|ψ
elabd2.is φx=Aψχ
Assertion elabd2 φABχ

Proof

Step Hyp Ref Expression
1 elabd2.ex φAV
2 elabd2.eq φB=x|ψ
3 elabd2.is φx=Aψχ
4 2 eleq2d φABAx|ψ
5 elab6g AVAx|ψxx=Aψ
6 4 5 sylan9bb φAVABxx=Aψ
7 elisset AVxx=A
8 3 pm5.74da φx=Aψx=Aχ
9 8 albidv φxx=Aψxx=Aχ
10 19.23v xx=Aχxx=Aχ
11 9 10 bitrdi φxx=Aψxx=Aχ
12 pm5.5 xx=Axx=Aχχ
13 11 12 sylan9bb φxx=Axx=Aψχ
14 7 13 sylan2 φAVxx=Aψχ
15 6 14 bitrd φAVABχ
16 1 15 mpdan φABχ