Theorem cleqf 2399

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions. See also cleqh 2331. (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 2225	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2		nfv 1576	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		cleqf.1	. . . . 5 ⊢ Ⅎ𝑥𝐴
4	3	nfcri 2368	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐴
5		cleqf.2	. . . . 5 ⊢ Ⅎ𝑥𝐵
6	5	nfcri 2368	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐵
7	4, 6	nfbi 1637	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)
8		eleq1 2294	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
9		eleq1 2294	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵))
10	8, 9	bibi12d 235	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)))
11	2, 7, 10	cbval 1802	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
12	1, 11	bitr4i 187	1 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))

Syntax hints:   ↔ wb 105  ∀wal 1395   = wceq 1397   ∈ wcel 2202  Ⅎwnfc 2361

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1811  df-cleq 2224  df-clel 2227  df-nfc 2363

This theorem is referenced by:  abid2f  2400  n0rf  3507  eq0  3513  iunab  4017  iinab  4032  sniota  5317