Theorem cleqf 3008

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions as in dfcleq 2813. See also cleqh 2934. (Contributed by NM, 26-May-1993.) (Revised by Mario Carneiro, 7-Oct-2016.) (Proof shortened by Wolf Lammen, 17-Nov-2019.) Avoid ax-13 2384. (Revised by Wolf Lammen, 10-May-2023.) Avoid ax-10 2139. (Revised by Gino Giotto, 20-Aug-2023.)

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 2813	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2		nfv 1909	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		cleqf.1	. . . . 5 ⊢ Ⅎ𝑥𝐴
4	3	nfcriv 2965	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐴
5		cleqf.2	. . . . 5 ⊢ Ⅎ𝑥𝐵
6	5	nfcriv 2965	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐵
7	4, 6	nfbi 1898	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)
8		eleq1w 2893	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
9		eleq1w 2893	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵))
10	8, 9	bibi12d 348	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)))
11	2, 7, 10	cbvalv1 2355	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
12	1, 11	bitr4i 280	1 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))

Syntax hints:   ↔ wb 208  ∀wal 1529   = wceq 1531   ∈ wcel 2108  Ⅎwnfc 2959

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1905  ax-6 1964  ax-7 2009  ax-8 2110  ax-9 2118  ax-11 2154  ax-12 2170  ax-ext 2791

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1534  df-ex 1775  df-nf 1779  df-cleq 2812  df-clel 2891  df-nfc 2961

This theorem is referenced by:  abid2f  3010  abeq2f  3011  eqvf  3502  eqrd  3984  eq0f  4303  iunab  4966  iinab  4981  mbfposr  24245  mbfinf  24258  itg1climres  24307  bnj1366  32094  bj-rabtrALT  34242  bj-rcleqf  34330  compab  40765  ssmapsn  41469  infnsuprnmpt  41512  pimrecltpos  42978  pimrecltneg  42992  smfaddlem1  43030  smflimsuplem7  43091  absnsb  43253