Theorem alsconv 10119

Description: There is an equivalence between the two "all some" forms. (Contributed by David A. Wheeler, 22-Oct-2018.)

Assertion

Ref	Expression
alsconv	|- ( A.! x ( x e. A -> ph ) <-> A.! x e. A ph )

Proof of Theorem alsconv

Step	Hyp	Ref	Expression
1		df-ral 2311	. . 3 |- ( A. x e. A ph <-> A. x ( x e. A -> ph ) )
2	1	anbi1i 431	. 2 |- ( ( A. x e. A ph /\ E. x x e. A ) <-> ( A. x ( x e. A -> ph ) /\ E. x x e. A ) )
3		df-alsc 10118	. 2 |- ( A.! x e. A ph <-> ( A. x e. A ph /\ E. x x e. A ) )
4		df-alsi 10117	. 2 |- ( A.! x ( x e. A -> ph ) <-> ( A. x ( x e. A -> ph ) /\ E. x x e. A ) )
5	2, 3, 4	3bitr4ri 202	1 |- ( A.! x ( x e. A -> ph ) <-> A.! x e. A ph )

Syntax hints:   -> wi 4   /\ wa 97   <-> wb 98   A.wal 1241   E.wex 1381   e. wcel 1393   A.wral 2306   A.!walsi 10115   A.!walsc 10116

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101

This theorem depends on definitions:  df-bi 110  df-ral 2311  df-alsi 10117  df-alsc 10118

This theorem is referenced by: (None)