Theorem cleqh 2858

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions as in dfcleq 2720. See also cleqf 2929. (Contributed by NM, 26-May-1993.) (Proof shortened by Wolf Lammen, 14-Nov-2019.) Remove dependency on ax-13 2366. (Revised by BJ, 30-Nov-2020.)

Hypotheses

Ref	Expression
cleqh.1	⊢ (𝑦 ∈ 𝐴 → ∀𝑥 𝑦 ∈ 𝐴)
cleqh.2	⊢ (𝑦 ∈ 𝐵 → ∀𝑥 𝑦 ∈ 𝐵)

Assertion

Ref	Expression
cleqh	⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))

Distinct variable groups:   𝑦,𝐴   𝑦,𝐵   𝑥,𝑦

Allowed substitution hints:   𝐴(𝑥)   𝐵(𝑥)

Proof of Theorem cleqh

Step	Hyp	Ref	Expression
1		dfcleq 2720	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2		nfv 1910	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		cleqh.1	. . . . 5 ⊢ (𝑦 ∈ 𝐴 → ∀𝑥 𝑦 ∈ 𝐴)
4	3	nf5i 2135	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐴
5		cleqh.2	. . . . 5 ⊢ (𝑦 ∈ 𝐵 → ∀𝑥 𝑦 ∈ 𝐵)
6	5	nf5i 2135	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐵
7	4, 6	nfbi 1899	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)
8		eleq1w 2811	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
9		eleq1w 2811	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵))
10	8, 9	bibi12d 345	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)))
11	2, 7, 10	cbvalv1 2332	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
12	1, 11	bitr4i 278	1 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))

Syntax hints:   → wi 4   ↔ wb 205  ∀wal 1532   = wceq 1534   ∈ wcel 2099

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 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2164  ax-ext 2698

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-ex 1775  df-nf 1779  df-cleq 2719  df-clel 2805

This theorem is referenced by:  eqabbOLD  2869