Theorem sbcel1gv 3106

Description: Class substitution into a membership relation. (Contributed by NM, 17-Nov-2006.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)

Assertion

Ref	Expression
sbcel1gv	⊢ (A ∈ V → ([̣A / x]̣x ∈ B ↔ A ∈ B))

Distinct variable group:   x,B

Allowed substitution hints:   A(x)   V(x)

Proof of Theorem sbcel1gv

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsbcq2 3050	. 2 ⊢ (y = A → ([y / x]x ∈ B ↔ [̣A / x]̣x ∈ B))
2		eleq1 2413	. 2 ⊢ (y = A → (y ∈ B ↔ A ∈ B))
3		clelsb1 2455	. 2 ⊢ ([y / x]x ∈ B ↔ y ∈ B)
4	1, 2, 3	vtoclbg 2916	1 ⊢ (A ∈ V → ([̣A / x]̣x ∈ B ↔ A ∈ B))

Syntax hints:   → wi 4   ↔ wb 176  [wsb 1648   ∈ wcel 1710  [̣wsbc 3047

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-sbc 3048

This theorem is referenced by: (None)