Theorem sbcng

Description: Move negation in and out of class substitution. (Contributed by NM, 16-Jan-2004)

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

Proof

Step	Hyp	Ref	Expression
1		dfsbcq2	|- ( y = A -> ( [ y / x ] -. ph <-> [. A / x ]. -. ph ) )
2		dfsbcq2	|- ( y = A -> ( [ y / x ] ph <-> [. A / x ]. ph ) )
3	2	notbid	|- ( y = A -> ( -. [ y / x ] ph <-> -. [. A / x ]. ph ) )
4		sbn	|- ( [ y / x ] -. ph <-> -. [ y / x ] ph )
5	1 3 4	vtoclbg	|- ( A e. V -> ( [. A / x ]. -. ph <-> -. [. A / x ]. ph ) )