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 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion	elabg	⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		elabg.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2		elab6g	⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ) )
3	1	pm5.74i	⊢ ( ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( 𝑥 = 𝐴 → 𝜓 ) )
4	3	albii	⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) )
5		19.23v	⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ↔ ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) )
6	4 5	bitri	⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) )
7		elisset	⊢ ( 𝐴 ∈ 𝑉 → ∃ 𝑥 𝑥 = 𝐴 )
8		pm5.5	⊢ ( ∃ 𝑥 𝑥 = 𝐴 → ( ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
9	7 8	syl	⊢ ( 𝐴 ∈ 𝑉 → ( ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
10	6 9	syl5bb	⊢ ( 𝐴 ∈ 𝑉 → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ 𝜓 ) )
11	2 10	bitrd	⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )