Theorem rals1d

Description: Deduction rule: Given "all some" applied to a class, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018) (Revised by David A. Wheeler, 12-Jul-2026)

	Ref	Expression
Hypothesis	rals1d.1	|- ( ph -> AE x e. A ( ps -> ch ) )
Assertion	rals1d	|- ( ph -> A. x e. A ( ps -> ch ) )

Proof

Step	Hyp	Ref	Expression
1		rals1d.1	|- ( ph -> AE x e. A ( ps -> ch ) )
2		df-rals	|- ( AE x e. A ( ps -> ch ) <-> ( A. x e. A ( ps -> ch ) /\ E. x e. A ps ) )
3	1 2	sylib	|- ( ph -> ( A. x e. A ( ps -> ch ) /\ E. x e. A ps ) )
4	3	simpld	|- ( ph -> A. x e. A ( ps -> ch ) )