Theorem dfss2f 1499

Description: Equivalence for subclass relation requiring only that x not be free in A and B (but not necessarily absent from them).

Hypotheses

Ref	Expression
dfss2f.1	⊢ (y ∈ A → ∀x y ∈ A)
dfss2f.2	⊢ (y ∈ B → ∀x y ∈ B)

Assertion

Ref	Expression
dfss2f	⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))

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

Proof of Theorem dfss2f

Step	Hyp	Ref	Expression
1		dfss2 1497	. 2 ⊢ (A ⊆ B ↔ ∀y(y ∈ A → y ∈ B))
2		ax-17 925	. . 3 ⊢ ((x ∈ A → x ∈ B) → ∀y(x ∈ A → x ∈ B))
3		dfss2f.1	. . . 4 ⊢ (y ∈ A → ∀x y ∈ A)
4		dfss2f.2	. . . 4 ⊢ (y ∈ B → ∀x y ∈ B)
5	3, 4	hbim 702	. . 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	imbi12d 474	. . 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   ∈ wcel 1092   ⊆ wss 1487

This theorem is referenced by:  dfss3f 1500  hbss 1501  ss2ab 1551  fopab2 2891  iunon 2947  iinon 2948  ranklon 3540  scott0 3542

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-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-in 1491  df-ss 1492

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