Theorem cleqf 2954

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions as in dfcleq 2756. See also cleqh 2893. (Contributed by NM, 26-May-1993.) (Revised by Mario Carneiro, 7-Oct-2016.) (Proof shortened by Wolf Lammen, 17-Nov-2019.) Avoid ax-13 2404. (Revised by Wolf Lammen, 10-May-2023.) Avoid ax-10 2180. (Revised by GG, 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 2756	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2		nfv 1949	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		cleqf.1	. . . . 5 ⊢ Ⅎ𝑥𝐴
4	3	nfcri 2918	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐴
5		cleqf.2	. . . . 5 ⊢ Ⅎ𝑥𝐵
6	5	nfcri 2918	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐵
7	4, 6	nfbi 1938	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)
8		eleq1w 2847	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
9		eleq1w 2847	. . . 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 2374	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
12	1, 11	bitr4i 281	1 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568   = wceq 1570   ∈ wcel 2147  Ⅎwnfc 2911

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-cleq 2755  df-clel 2839  df-nfc 2913

This theorem is used by:  eqabf  2955  abid2fOLD  2957  eqvf  3465  eqrd  3932  eq0f  4297  rspsn0  21577  mbfposr  26028  mbfinf  26041  itg1climres  26090  bnj1366  35570  bj-rabtrALT  38059  bj-rcleqf  38153  compab  45685  ssmapsn  46502  infnsuprnmpt  46535  pimrecltpos  47991  pimrecltneg  48007  smfaddlem1  48046  smflimsuplem7  48109  absnsb  48370