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) Remove dependency on ax-13 . (Revised by Steven Nguyen, 23-Nov-2022)

	Ref	Expression
Hypothesis	elabg.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion	elabg	⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		elabg.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2		nfab1	⊢ Ⅎ 𝑥 { 𝑥 ∣ 𝜑 }
3	2	nfel2	⊢ Ⅎ 𝑥 𝐴 ∈ { 𝑥 ∣ 𝜑 }
4		nfv	⊢ Ⅎ 𝑥 𝜓
5	3 4	nfbi	⊢ Ⅎ 𝑥 ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 )
6		eleq1	⊢ ( 𝑥 = 𝐴 → ( 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝐴 ∈ { 𝑥 ∣ 𝜑 } ) )
7	6 1	bibi12d	⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜑 ) ↔ ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) ) )
8		abid	⊢ ( 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜑 )
9	5 7 8	vtoclg1f	⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )