Theorem elabgt 3596

Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 3599.) (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 2947	. . 3 ⊢ Ⅎ𝑥𝐴
2		nfab1 2949	. . . . 5 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
3	2	nfel2 2963	. . . 4 ⊢ Ⅎ𝑥 𝐴 ∈ {𝑥 ∣ 𝜑}
4		nfv 1890	. . . 4 ⊢ Ⅎ𝑥𝜓
5	3, 4	nfbi 1883	. . 3 ⊢ Ⅎ𝑥(𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6		pm5.5 363	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)) ↔ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
7	1, 5, 6	spcgf 3528	. 2 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
8		abid 2777	. . . . . . 7 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
9		eleq1 2868	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
10	8, 9	syl5bbr 286	. . . . . 6 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
11	10	bibi1d 345	. . . . 5 ⊢ (𝑥 = 𝐴 → ((𝜑 ↔ 𝜓) ↔ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
12	11	biimpd 230	. . . 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 1791	. 2 ⊢ (∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
15	7, 14	impel 506	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓))) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 207   ∧ wa 396  ∀wal 1518   = wceq 1520   ∈ wcel 2079  {cab 2773

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1775  ax-4 1789  ax-5 1886  ax-6 1945  ax-7 1990  ax-8 2081  ax-9 2089  ax-10 2110  ax-11 2124  ax-12 2139  ax-ext 2767

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-tru 1523  df-ex 1760  df-nf 1764  df-sb 2041  df-clab 2774  df-cleq 2786  df-clel 2861  df-nfc 2933  df-v 3434

This theorem is referenced by:  elrab3t  3612  dfrtrcl2  14243  abfmpeld  30062  abfmpel  30063  dftrcl3  39501  dfrtrcl3  39514