Theorem elabgt 2983

Description: Membership in a class abstraction, using implicit substitution. (Closed theorem version of elabg 2987.) (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		abid 2341	. . . . . . 7 ⊢ (x ∈ {x ∣ φ} ↔ φ)
2		eleq1 2413	. . . . . . 7 ⊢ (x = A → (x ∈ {x ∣ φ} ↔ A ∈ {x ∣ φ}))
3	1, 2	syl5bbr 250	. . . . . 6 ⊢ (x = A → (φ ↔ A ∈ {x ∣ φ}))
4	3	bibi1d 310	. . . . 5 ⊢ (x = A → ((φ ↔ ψ) ↔ (A ∈ {x ∣ φ} ↔ ψ)))
5	4	biimpd 198	. . . 4 ⊢ (x = A → ((φ ↔ ψ) → (A ∈ {x ∣ φ} ↔ ψ)))
6	5	a2i 12	. . 3 ⊢ ((x = A → (φ ↔ ψ)) → (x = A → (A ∈ {x ∣ φ} ↔ ψ)))
7	6	alimi 1559	. 2 ⊢ (∀x(x = A → (φ ↔ ψ)) → ∀x(x = A → (A ∈ {x ∣ φ} ↔ ψ)))
8		nfcv 2490	. . . 4 ⊢ ℲxA
9		nfab1 2492	. . . . . 6 ⊢ Ⅎx{x ∣ φ}
10	9	nfel2 2502	. . . . 5 ⊢ Ⅎx A ∈ {x ∣ φ}
11		nfv 1619	. . . . 5 ⊢ Ⅎxψ
12	10, 11	nfbi 1834	. . . 4 ⊢ Ⅎx(A ∈ {x ∣ φ} ↔ ψ)
13		pm5.5 326	. . . 4 ⊢ (x = A → ((x = A → (A ∈ {x ∣ φ} ↔ ψ)) ↔ (A ∈ {x ∣ φ} ↔ ψ)))
14	8, 12, 13	spcgf 2935	. . 3 ⊢ (A ∈ B → (∀x(x = A → (A ∈ {x ∣ φ} ↔ ψ)) → (A ∈ {x ∣ φ} ↔ ψ)))
15	14	imp 418	. 2 ⊢ ((A ∈ B ∧ ∀x(x = A → (A ∈ {x ∣ φ} ↔ ψ))) → (A ∈ {x ∣ φ} ↔ ψ))
16	7, 15	sylan2 460	1 ⊢ ((A ∈ B ∧ ∀x(x = A → (φ ↔ ψ))) → (A ∈ {x ∣ φ} ↔ ψ))

Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540   = wceq 1642   ∈ wcel 1710  {cab 2339

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862

This theorem is referenced by: (None)