Theorem cleqh 2895

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions as in dfcleq 2759. See also cleqf 2956. (Contributed by NM, 26-May-1993.) (Proof shortened by Wolf Lammen, 14-Nov-2019.) Remove dependency on ax-13 2407. (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 2759	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2		nfv 1947	. . 3 ⊢ Ⅎ𝑦(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
3		cleqh.1	. . . . 5 ⊢ (𝑦 ∈ 𝐴 → ∀𝑥 𝑦 ∈ 𝐴)
4	3	nf5i 2184	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐴
5		cleqh.2	. . . . 5 ⊢ (𝑦 ∈ 𝐵 → ∀𝑥 𝑦 ∈ 𝐵)
6	5	nf5i 2184	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ 𝐵
7	4, 6	nfbi 1936	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵)
8		eleq1w 2849	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
9		eleq1w 2849	. . . 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 2376	. 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:   → wi 4   ↔ wb 209  ∀wal 1568   = wceq 1570   ∈ wcel 2146

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 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2738

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 2758  df-clel 2841

This theorem is used by: (None)