Theorem sbcng 1969

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 1943	. 2 |- (y = A -> ([y / x] -. ph <-> [A / x] -. ph))
2		dfsbcq 1943	. . 3 |- (y = A -> ([y / x]ph <-> [A / x]ph))
3	2	negbid 611	. 2 |- (y = A -> (-. [y / x]ph <-> -. [A / x]ph))
4		sbn 1231	. 2 |- ([y / x] -. ph <-> -. [y / x]ph)
5	1, 3, 4	vtoclbg 1848	1 |- (A e. B -> ([A / x] -. ph <-> -. [A / x]ph))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   = wceq 956   e. wcel 958  [wsbc 1170

This theorem is referenced by:  sbcrext 1991  sbcrexgf 1993  ra4esbca 1999  rexpr 2429

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-v 1812  df-sbc 1942

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