Theorem elabgt 3316

Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 3320.) (Contributed by NM, 7-Nov-2005.) (Proof shortened by Andrew Salmon, 8-Jun-2011.)

Assertion

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

Distinct variable groups:   𝑥,𝐴   𝜓,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝐵(𝑥)

Proof of Theorem elabgt

Step	Hyp	Ref	Expression
1		nfcv 2751	. . 3 ⊢ Ⅎ𝑥𝐴
2		nfab1 2753	. . . . 5 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
3	2	nfel2 2767	. . . 4 ⊢ Ⅎ𝑥 𝐴 ∈ {𝑥 ∣ 𝜑}
4		nfv 1830	. . . 4 ⊢ Ⅎ𝑥𝜓
5	3, 4	nfbi 1821	. . 3 ⊢ Ⅎ𝑥(𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6		pm5.5 350	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)) ↔ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
7	1, 5, 6	spcgf 3261	. 2 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
8		abid 2598	. . . . . . 7 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
9		eleq1 2676	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
10	8, 9	syl5bbr 273	. . . . . 6 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
11	10	bibi1d 332	. . . . 5 ⊢ (𝑥 = 𝐴 → ((𝜑 ↔ 𝜓) ↔ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
12	11	biimpd 218	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝜑 ↔ 𝜓) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
13	12	a2i 14	. . 3 ⊢ ((𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
14	13	alimi 1730	. 2 ⊢ (∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
15	7, 14	impel 484	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓))) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383  ∀wal 1473   = wceq 1475   ∈ wcel 1977  {cab 2596

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175

This theorem is referenced by:  elrab3t  3330  dfrtrcl2  13650  abfmpeld  28834  abfmpel  28835  dftrcl3  37031  dfrtrcl3  37044