Theorem cleqf 2372

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions. See also cleqh 2304. (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 2198	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2		nfv 1550	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		cleqf.1	. . . . 5 ⊢ Ⅎ𝑥𝐴
4	3	nfcri 2341	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐴
5		cleqf.2	. . . . 5 ⊢ Ⅎ𝑥𝐵
6	5	nfcri 2341	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐵
7	4, 6	nfbi 1611	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)
8		eleq1 2267	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
9		eleq1 2267	. . . 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 1776	. 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 1370   = wceq 1372   ∈ wcel 2175  Ⅎwnfc 2334

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 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-tru 1375  df-nf 1483  df-sb 1785  df-cleq 2197  df-clel 2200  df-nfc 2336

This theorem is referenced by:  abid2f  2373  n0rf  3472  eq0  3478  iunab  3973  iinab  3988  sniota  5261