Theorem sbc6g 1451

Description: An equivalence for class substitution.

Assertion

Ref	Expression
sbc6g	⊢ (A ∈ B → ([A / x]φ ↔ ∀x(x = A → φ)))

Distinct variable group(s):   x,A

Proof of Theorem sbc6g

Step	Hyp	Ref	Expression
1		ax-17 925	. . . . . 6 ⊢ (y ∈ A → ∀x y ∈ A)
2	1	hbsbc 1446	. . . . 5 ⊢ ((A ∈ V → [A / x]φ) → ∀x(A ∈ V → [A / x]φ))
3		sbceq1 1443	. . . . . 6 ⊢ (x = A → (φ ↔ [A / x]φ))
4	3	imbi2d 464	. . . . 5 ⊢ (x = A → ((A ∈ V → φ) ↔ (A ∈ V → [A / x]φ)))
5	2, 4	ceqsalg 1362	. . . 4 ⊢ (A ∈ V → (∀x(x = A → (A ∈ V → φ)) ↔ (A ∈ V → [A / x]φ)))
6		biimt 549	. . . . . . 7 ⊢ (A ∈ V → (φ ↔ (A ∈ V → φ)))
7	6	imbi2d 464	. . . . . 6 ⊢ (A ∈ V → ((x = A → φ) ↔ (x = A → (A ∈ V → φ))))
8	7	bialdv 935	. . . . 5 ⊢ (A ∈ V → (∀x(x = A → φ) ↔ ∀x(x = A → (A ∈ V → φ))))
9		biimt 549	. . . . 5 ⊢ (A ∈ V → (∀x(x = A → φ) ↔ (A ∈ V → ∀x(x = A → φ))))
10	8, 9	bitr3d 408	. . . 4 ⊢ (A ∈ V → (∀x(x = A → (A ∈ V → φ)) ↔ (A ∈ V → ∀x(x = A → φ))))
11	5, 10	bitr3d 408	. . 3 ⊢ (A ∈ V → ((A ∈ V → [A / x]φ) ↔ (A ∈ V → ∀x(x = A → φ))))
12	11	pm5.74rd 446	. 2 ⊢ (A ∈ V → (A ∈ V → ([A / x]φ ↔ ∀x(x = A → φ))))
13		elisset 1354	. 2 ⊢ (A ∈ B → A ∈ V)
14	12, 13, 13	sylc 62	1 ⊢ (A ∈ B → ([A / x]φ ↔ ∀x(x = A → φ)))

Syntax hints:   → wi 2   ↔ wb 127  ∀wal 672   = 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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