Theorem elabgt 3379

Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 3383.) (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 2793	. . 3 ⊢ Ⅎ𝑥𝐴
2		nfab1 2795	. . . . 5 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
3	2	nfel2 2810	. . . 4 ⊢ Ⅎ𝑥 𝐴 ∈ {𝑥 ∣ 𝜑}
4		nfv 1883	. . . 4 ⊢ Ⅎ𝑥𝜓
5	3, 4	nfbi 1873	. . 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 3319	. 2 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
8		abid 2639	. . . . . . 7 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
9		eleq1 2718	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
10	8, 9	syl5bbr 274	. . . . . 6 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
11	10	bibi1d 332	. . . . 5 ⊢ (𝑥 = 𝐴 → ((𝜑 ↔ 𝜓) ↔ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
12	11	biimpd 219	. . . 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 1779	. 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 196   ∧ wa 383  ∀wal 1521   = wceq 1523   ∈ wcel 2030  {cab 2637

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-nfc 2782  df-v 3233

This theorem is referenced by:  elrab3t  3395  dfrtrcl2  13846  abfmpeld  29582  abfmpel  29583  dftrcl3  38329  dfrtrcl3  38342