Theorem cleqf 2950

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions as in dfcleq 2753. See also cleqh 2889. (Contributed by NM, 26-May-1993.) (Revised by Mario Carneiro, 7-Oct-2016.) (Proof shortened by Wolf Lammen, 17-Nov-2019.) Avoid ax-13 2401. (Revised by Wolf Lammen, 10-May-2023.) Avoid ax-10 2178. (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 2753	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2		nfv 1947	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		cleqf.1	. . . . 5 ⊢ Ⅎ𝑥𝐴
4	3	nfcri 2914	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐴
5		cleqf.2	. . . . 5 ⊢ Ⅎ𝑥𝐵
6	5	nfcri 2914	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐵
7	4, 6	nfbi 1936	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)
8		eleq1w 2843	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
9		eleq1w 2843	. . . 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 2370	. 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 2145  Ⅎwnfc 2907

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-11 2194  ax-12 2213  ax-ext 2732

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 2752  df-clel 2835  df-nfc 2909

This theorem is used by:  eqabf  2951  abid2fOLD  2953  eqvf  3461  eqrd  3950  eq0f  4294  rspsn0  21436  mbfposr  25881  mbfinf  25894  itg1climres  25943  bnj1366  35339  bj-rabtrALT  37676  bj-rcleqf  37770  compab  45266  ssmapsn  46047  infnsuprnmpt  46080  pimrecltpos  47537  pimrecltneg  47553  smfaddlem1  47592  smflimsuplem7  47655  absnsb  47916