Theorem sbcel2 1467

Description: Class substitution into a membership relation.

Assertion

Ref	Expression
sbcel2	⊢ (B ∈ C → ([B / x]A ∈ x ↔ A ∈ B))

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

Proof of Theorem sbcel2

Step	Hyp	Ref	Expression
1		elisset 1354	. 2 ⊢ (B ∈ C → B ∈ V)
2		elex 1356	. . 3 ⊢ (B ∈ V → ∃x x = B)
3		ax-17 925	. . . . . . 7 ⊢ (y ∈ B → ∀x y ∈ B)
4	3	hbsbc 1446	. . . . . 6 ⊢ ((B ∈ V → [B / x]A ∈ x) → ∀x(B ∈ V → [B / x]A ∈ x))
5		ax-17 925	. . . . . 6 ⊢ ((B ∈ V → A ∈ B) → ∀x(B ∈ V → A ∈ B))
6	4, 5	hbbi 705	. . . . 5 ⊢ (((B ∈ V → [B / x]A ∈ x) ↔ (B ∈ V → A ∈ B)) → ∀x((B ∈ V → [B / x]A ∈ x) ↔ (B ∈ V → A ∈ B)))
7		sbceq1 1443	. . . . . . 7 ⊢ (x = B → (A ∈ x ↔ [B / x]A ∈ x))
8		eleq2 1150	. . . . . . 7 ⊢ (x = B → (A ∈ x ↔ A ∈ B))
9	7, 8	bitr3d 408	. . . . . 6 ⊢ (x = B → ([B / x]A ∈ x ↔ A ∈ B))
10	9	imbi2d 464	. . . . 5 ⊢ (x = B → ((B ∈ V → [B / x]A ∈ x) ↔ (B ∈ V → A ∈ B)))
11	6, 10	19.23ai 746	. . . 4 ⊢ (∃x x = B → ((B ∈ V → [B / x]A ∈ x) ↔ (B ∈ V → A ∈ B)))
12	11	pm5.74rd 446	. . 3 ⊢ (∃x x = B → (B ∈ V → ([B / x]A ∈ x ↔ A ∈ B)))
13	2, 12	mpcom 49	. 2 ⊢ (B ∈ V → ([B / x]A ∈ x ↔ A ∈ B))
14	1, 13	syl 12	1 ⊢ (B ∈ C → ([B / x]A ∈ x ↔ A ∈ B))

Syntax hints:   → wi 2   ↔ wb 127  ∃wex 678   = wceq 1091   ∈ wcel 1092  Vcvv 1348  [wsbc 1440

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-v 1349  df-sbc 1441

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