Theorem elabgt 3347

Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 3351.) (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 2764	. . 3 ⊢ Ⅎ𝑥𝐴
2		nfab1 2766	. . . . 5 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
3	2	nfel2 2781	. . . 4 ⊢ Ⅎ𝑥 𝐴 ∈ {𝑥 ∣ 𝜑}
4		nfv 1843	. . . 4 ⊢ Ⅎ𝑥𝜓
5	3, 4	nfbi 1833	. . 3 ⊢ Ⅎ𝑥(𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6		pm5.5 351	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)) ↔ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
7	1, 5, 6	spcgf 3288	. 2 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
8		abid 2610	. . . . . . 7 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
9		eleq1 2689	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
10	8, 9	syl5bbr 274	. . . . . 6 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
11	10	bibi1d 333	. . . . 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 1739	. 2 ⊢ (∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
15	7, 14	impel 485	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓))) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384  ∀wal 1481   = wceq 1483   ∈ wcel 1990  {cab 2608

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202

This theorem is referenced by:  elrab3t  3362  dfrtrcl2  13802  abfmpeld  29454  abfmpel  29455  dftrcl3  38012  dfrtrcl3  38025