Theorem cleqh 2711

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions. See also cleqf 2776. (Contributed by NM, 26-May-1993.) (Proof shortened by Wolf Lammen, 14-Nov-2019.) Remove dependency on ax-13 2234. (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 2604	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2		nfv 1830	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		cleqh.1	. . . . 5 ⊢ (𝑦 ∈ 𝐴 → ∀𝑥 𝑦 ∈ 𝐴)
4	3	nf5i 2011	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐴
5		cleqh.2	. . . . 5 ⊢ (𝑦 ∈ 𝐵 → ∀𝑥 𝑦 ∈ 𝐵)
6	5	nf5i 2011	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐵
7	4, 6	nfbi 1821	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)
8		eleq1 2676	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
9		eleq1 2676	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵))
10	8, 9	bibi12d 334	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)))
11	2, 7, 10	cbvalv1 2163	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
12	1, 11	bitr4i 266	1 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))

Syntax hints:   → wi 4   ↔ wb 195  ∀wal 1473   = wceq 1475   ∈ wcel 1977

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-cleq 2603  df-clel 2606

This theorem is referenced by:  abeq2  2719  abbi  2724  cleqf  2776  abeq2f  2778  bj-abeq2  31961