Theorem cleqf 2514

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 7-Oct-2016.)

Hypotheses

Ref	Expression
cleqf.1	⊢ ℲxA
cleqf.2	⊢ ℲxB

Assertion

Ref	Expression
cleqf	⊢ (A = B ↔ ∀x(x ∈ A ↔ x ∈ B))

Proof of Theorem cleqf

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfcleq 2347	. 2 ⊢ (A = B ↔ ∀y(y ∈ A ↔ y ∈ B))
2		nfv 1619	. . 3 ⊢ Ⅎy(x ∈ A ↔ x ∈ B)
3		cleqf.1	. . . . 5 ⊢ ℲxA
4	3	nfcri 2484	. . . 4 ⊢ Ⅎx y ∈ A
5		cleqf.2	. . . . 5 ⊢ ℲxB
6	5	nfcri 2484	. . . 4 ⊢ Ⅎx y ∈ B
7	4, 6	nfbi 1834	. . 3 ⊢ Ⅎx(y ∈ A ↔ y ∈ B)
8		eleq1 2413	. . . 4 ⊢ (x = y → (x ∈ A ↔ y ∈ A))
9		eleq1 2413	. . . 4 ⊢ (x = y → (x ∈ B ↔ y ∈ B))
10	8, 9	bibi12d 312	. . 3 ⊢ (x = y → ((x ∈ A ↔ x ∈ B) ↔ (y ∈ A ↔ y ∈ B)))
11	2, 7, 10	cbval 1984	. 2 ⊢ (∀x(x ∈ A ↔ x ∈ B) ↔ ∀y(y ∈ A ↔ y ∈ B))
12	1, 11	bitr4i 243	1 ⊢ (A = B ↔ ∀x(x ∈ A ↔ x ∈ B))

This proof depends on syntax axioms:   ↔ wb 176  ∀wal 1540   = wceq 1642   ∈ wcel 1710  Ⅎwnfc 2477

This proof depends on 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 proof 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-cleq 2346  df-clel 2349  df-nfc 2479

This theorem is used by:  abid2f  2515  n0f  3559  iunab  4013  iinab  4028  sniota  4370