Theorem spesbcd

Description: form of spsbc . (Contributed by Mario Carneiro, 9-Feb-2017)

	Ref	Expression
Hypothesis	spesbcd.1	|- ( ph -> [. A / x ]. ps )
Assertion	spesbcd	|- ( ph -> E. x ps )

Proof

Step	Hyp	Ref	Expression
1		spesbcd.1	|- ( ph -> [. A / x ]. ps )
2		spesbc	|- ( [. A / x ]. ps -> E. x ps )
3	1 2	syl	|- ( ph -> E. x ps )