Metamath Proof Explorer


Theorem elabg

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of Quine p. 44. (Contributed by NM, 14-Apr-1995) Avoid ax-13 . (Revised by SN, 23-Nov-2022) Avoid ax-10 , ax-11 , ax-12 . (Revised by SN, 5-Oct-2024)

Ref Expression
Hypothesis elabg.1 x=Aφψ
Assertion elabg AVAx|φψ

Proof

Step Hyp Ref Expression
1 elabg.1 x=Aφψ
2 elab6g AVAx|φxx=Aφ
3 1 pm5.74i x=Aφx=Aψ
4 3 albii xx=Aφxx=Aψ
5 19.23v xx=Aψxx=Aψ
6 4 5 bitri xx=Aφxx=Aψ
7 elisset AVxx=A
8 pm5.5 xx=Axx=Aψψ
9 7 8 syl AVxx=Aψψ
10 6 9 bitrid AVxx=Aφψ
11 2 10 bitrd AVAx|φψ