Theorem elabgt

Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg .) (Contributed by NM, 7-Nov-2005) (Proof shortened by Andrew Salmon, 8-Jun-2011) Reduce axiom usage. (Revised by GG, 12-Oct-2024) (Proof shortened by Wolf Lammen, 11-May-2025) (Proof shortened by SN, 1-Dec-2025)

	Ref	Expression
Assertion	elabgt	⊢ ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		elab6g	⊢ ( 𝐴 ∈ 𝐵 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ) )
2		pm5.74	⊢ ( ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ↔ ( ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( 𝑥 = 𝐴 → 𝜓 ) ) )
3	2	biimpi	⊢ ( ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) → ( ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( 𝑥 = 𝐴 → 𝜓 ) ) )
4	3	alimi	⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) → ∀ 𝑥 ( ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( 𝑥 = 𝐴 → 𝜓 ) ) )
5		albi	⊢ ( ∀ 𝑥 ( ( 𝑥 = 𝐴 → 𝜑 ) ↔ ( 𝑥 = 𝐴 → 𝜓 ) ) → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ) )
6	4 5	syl	⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ) )
7	1 6	sylan9bb	⊢ ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ) )
8		19.23v	⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ↔ ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) )
9		elisset	⊢ ( 𝐴 ∈ 𝐵 → ∃ 𝑥 𝑥 = 𝐴 )
10		pm5.5	⊢ ( ∃ 𝑥 𝑥 = 𝐴 → ( ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
11	9 10	syl	⊢ ( 𝐴 ∈ 𝐵 → ( ( ∃ 𝑥 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
12	8 11	bitrid	⊢ ( 𝐴 ∈ 𝐵 → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
13	12	adantr	⊢ ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜓 ) ↔ 𝜓 ) )
14	7 13	bitrd	⊢ ( ( 𝐴 ∈ 𝐵 ∧ ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) ) → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) )