Theorem sbcn 1459

Description: Move negation in and out of class substitution.

Assertion

Ref	Expression
sbcn	|- (A e. B -> ([A / x] -. ph <-> -. [A / x]ph))

Proof of Theorem sbcn

Step	Hyp	Ref	Expression
1		dfsbcq 1442	. 2 |- (y = A -> ([y / x] -. ph <-> [A / x] -. ph))
2		dfsbcq 1442	. . 3 |- (y = A -> ([y / x]ph <-> [A / x]ph))
3	2	negbid 463	. 2 |- (y = A -> (-. [y / x]ph <-> -. [A / x]ph))
4		sbn 882	. 2 |- ([y / x] -. ph <-> -. [y / x]ph)
5	1, 3, 4	vtoclbg 1384	1 |- (A e. B -> ([A / x] -. ph <-> -. [A / x]ph))

Syntax hints:  -. wn 1   -> wi 2   <-> wb 127  [wsb 852   = wceq 1091   e. wcel 1092  [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-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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