Theorem cleqf 1167

Description: Establish equality between classes, requiring only that x not be 'free' in A and B (but not necessarily absent from them).

Hypotheses

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

Assertion

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

Distinct variable group(s):   y,A   y,B   x,y

Proof of Theorem cleqf

Step	Hyp	Ref	Expression
1		dfcleq 1098	. 2 ⊢ (A = B ↔ ∀y(y ∈ A ↔ y ∈ B))
2		ax-17 925	. . 3 ⊢ ((x ∈ A ↔ x ∈ B) → ∀y(x ∈ A ↔ x ∈ B))
3		cleqf.1	. . . 4 ⊢ (y ∈ A → ∀x y ∈ A)
4		cleqf.2	. . . 4 ⊢ (y ∈ B → ∀x y ∈ B)
5	3, 4	hbbi 705	. . 3 ⊢ ((y ∈ A ↔ y ∈ B) → ∀x(y ∈ A ↔ y ∈ B))
6		eleq1 1149	. . . 4 ⊢ (x = y → (x ∈ A ↔ y ∈ A))
7		eleq1 1149	. . . 4 ⊢ (x = y → (x ∈ B ↔ y ∈ B))
8	6, 7	bibi12d 477	. . 3 ⊢ (x = y → ((x ∈ A ↔ x ∈ B) ↔ (y ∈ A ↔ y ∈ B)))
9	2, 5, 8	cbval 848	. 2 ⊢ (∀x(x ∈ A ↔ x ∈ B) ↔ ∀y(y ∈ A ↔ y ∈ B))
10	1, 9	bitr4 154	1 ⊢ (A = B ↔ ∀x(x ∈ A ↔ x ∈ B))

Syntax hints:   → wi 2   ↔ wb 127  ∀wal 672   = weq 797   = wceq 1091   ∈ wcel 1092

This theorem is referenced by:  cleqab 1174  cleq2ab 1179  cbvab 1423  n0f 1713

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-12 802  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-cleq 1097  df-clel 1099

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