Theorem spcv

Description: Rule of specialization, using implicit substitution. (Contributed by NM, 22-Jun-1994)

	Ref	Expression
Hypotheses	spcv.1	|- A e. _V
	spcv.2	|- ( x = A -> ( ph <-> ps ) )
Assertion	spcv	|- ( A. x ph -> ps )

Proof

Step	Hyp	Ref	Expression
1		spcv.1	|- A e. _V
2		spcv.2	|- ( x = A -> ( ph <-> ps ) )
3	2	spcgv	|- ( A e. _V -> ( A. x ph -> ps ) )
4	1 3	ax-mp	|- ( A. x ph -> ps )