Theorem elabgt 3582

Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 3587.) (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 2919	. . 3 ⊢ Ⅎ𝑥𝐴
2		nfab1 2921	. . . . 5 ⊢ Ⅎ𝑥{𝑥 ∣ 𝜑}
3	2	nfel2 2937	. . . 4 ⊢ Ⅎ𝑥 𝐴 ∈ {𝑥 ∣ 𝜑}
4		nfv 1915	. . . 4 ⊢ Ⅎ𝑥𝜓
5	3, 4	nfbi 1904	. . 3 ⊢ Ⅎ𝑥(𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6		pm5.5 365	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)) ↔ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
7	1, 5, 6	spcgf 3508	. 2 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
8		abid 2739	. . . . . . 7 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)
9		eleq1 2839	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
10	8, 9	bitr3id 288	. . . . . 6 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
11	10	bibi1d 347	. . . . 5 ⊢ (𝑥 = 𝐴 → ((𝜑 ↔ 𝜓) ↔ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
12	11	biimpd 232	. . . 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 1813	. 2 ⊢ (∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → ∀𝑥(𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)))
15	7, 14	impel 509	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓))) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1536   = wceq 1538   ∈ wcel 2111  {cab 2735

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2729

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2736  df-cleq 2750  df-clel 2830  df-nfc 2901  df-v 3411

This theorem is referenced by:  elrab3t  3601  dfrtrcl2  14469  iinabrex  30430  abfmpeld  30515  abfmpel  30516  dftrcl3  40794  dfrtrcl3  40807