Metamath Proof Explorer


Theorem elrab

Description: Membership in a restricted class abstraction, using implicit substitution. (Contributed by NM, 21-May-1999) Remove dependency on ax-13 . (Revised by Steven Nguyen, 23-Nov-2022)

Ref Expression
Hypothesis elrab.1 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
Assertion elrab ( 𝐴 ∈ { 𝑥𝐵𝜑 } ↔ ( 𝐴𝐵𝜓 ) )

Proof

Step Hyp Ref Expression
1 elrab.1 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
2 elex ( 𝐴 ∈ { 𝑥𝐵𝜑 } → 𝐴 ∈ V )
3 elex ( 𝐴𝐵𝐴 ∈ V )
4 3 adantr ( ( 𝐴𝐵𝜓 ) → 𝐴 ∈ V )
5 eleq1 ( 𝑥 = 𝐴 → ( 𝑥𝐵𝐴𝐵 ) )
6 5 1 anbi12d ( 𝑥 = 𝐴 → ( ( 𝑥𝐵𝜑 ) ↔ ( 𝐴𝐵𝜓 ) ) )
7 df-rab { 𝑥𝐵𝜑 } = { 𝑥 ∣ ( 𝑥𝐵𝜑 ) }
8 6 7 elab2g ( 𝐴 ∈ V → ( 𝐴 ∈ { 𝑥𝐵𝜑 } ↔ ( 𝐴𝐵𝜓 ) ) )
9 2 4 8 pm5.21nii ( 𝐴 ∈ { 𝑥𝐵𝜑 } ↔ ( 𝐴𝐵𝜓 ) )