Theorem cleqh 2724

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions. See also cleqf 2790. (Contributed by NM, 26-May-1993.) (Proof shortened by Wolf Lammen, 14-Nov-2019.) Remove dependency on ax-13 2246. (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 2616	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2		nfv 1843	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		cleqh.1	. . . . 5 ⊢ (𝑦 ∈ 𝐴 → ∀𝑥 𝑦 ∈ 𝐴)
4	3	nf5i 2024	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐴
5		cleqh.2	. . . . 5 ⊢ (𝑦 ∈ 𝐵 → ∀𝑥 𝑦 ∈ 𝐵)
6	5	nf5i 2024	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐵
7	4, 6	nfbi 1833	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)
8		eleq1 2689	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
9		eleq1 2689	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵))
10	8, 9	bibi12d 335	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)))
11	2, 7, 10	cbvalv1 2175	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
12	1, 11	bitr4i 267	1 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))

Syntax hints:   → wi 4   ↔ wb 196  ∀wal 1481   = wceq 1483   ∈ wcel 1990

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-cleq 2615  df-clel 2618

This theorem is referenced by:  abeq2  2732  abbi  2737  cleqf  2790  abeq2f  2792  bj-abeq2  32773