Theorem alsc1d 13959

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.)

Hypothesis

Ref	Expression
alsc1d.1	|- ( ph -> A.! x e. A ps )

Assertion

Ref	Expression
alsc1d	|- ( ph -> A. x e. A ps )

Proof of Theorem alsc1d

Step	Hyp	Ref	Expression
1		alsc1d.1	. . 3 |- ( ph -> A.! x e. A ps )
2		df-alsc 13955	. . 3 |- ( A.! x e. A ps <-> ( A. x e. A ps /\ E. x x e. A ) )
3	1, 2	sylib 121	. 2 |- ( ph -> ( A. x e. A ps /\ E. x x e. A ) )
4	3	simpld 111	1 |- ( ph -> A. x e. A ps )

Syntax hints:   -> wi 4   /\ wa 103   E.wex 1480   e. wcel 2136   A.wral 2444   A.!walsc 13953

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105

This theorem depends on definitions:  df-bi 116  df-alsc 13955

This theorem is referenced by: (None)