Theorem sbcng 2495

Description: Move negation in and out of class substitution.

Assertion

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

Proof of Theorem sbcng

Step	Hyp	Ref	Expression
1		dfsbcq 2455	. 2 |- (y = A -> ([y / x] -. ph <-> [A / x] -. ph))
2		dfsbcq 2455	. . 3 |- (y = A -> ([y / x]ph <-> [A / x]ph))
3	2	notbid 673	. 2 |- (y = A -> (-. [y / x]ph <-> -. [A / x]ph))
4		sbn 1601	. 2 |- ([y / x] -. ph <-> -. [y / x]ph)
5	1, 3, 4	vtoclbg 2347	1 |- (A e. B -> ([A / x] -. ph <-> -. [A / x]ph))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 163   = wceq 1298   e. wcel 1300  [wsbc 1534

This theorem is referenced by:  sbcrext 2528  sbcrexgf 2530  ra4esbca 2538  rexpr 3082  sbcnel12g 16408  sbcne12g 16409

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-9 1307  ax-10 1308  ax-11 1309  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-v 2294  df-sbc 2454

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