Theorem sbcng 1965

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 1939	. 2 |- (y = A -> ([y / x] -. ph <-> [A / x] -. ph))
2		dfsbcq 1939	. . 3 |- (y = A -> ([y / x]ph <-> [A / x]ph))
3	2	negbid 610	. 2 |- (y = A -> (-. [y / x]ph <-> -. [A / x]ph))
4		sbn 1229	. 2 |- ([y / x] -. ph <-> -. [y / x]ph)
5	1, 3, 4	vtoclbg 1844	1 |- (A e. B -> ([A / x] -. ph <-> -. [A / x]ph))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   = wceq 954   e. wcel 956  [wsbc 1168

This theorem is referenced by:  sbcrext 1987  sbcrexgf 1989  ra4esbca 1995  rexpr 2425

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-11o 1216  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-sb 1170  df-clab 1462  df-cleq 1467  df-clel 1470  df-v 1808  df-sbc 1938

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