Theorem alsconv 14109

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 2453	. . 3 |- ( A. x e. A ph <-> A. x ( x e. A -> ph ) )
2	1	anbi1i 455	. 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 14108	. 2 |- ( A.! x e. A ph <-> ( A. x e. A ph /\ E. x x e. A ) )
4		df-alsi 14107	. 2 |- ( A.! x ( x e. A -> ph ) <-> ( A. x ( x e. A -> ph ) /\ E. x x e. A ) )
5	2, 3, 4	3bitr4ri 212	1 |- ( A.! x ( x e. A -> ph ) <-> A.! x e. A ph )

Syntax hints:   -> wi 4   /\ wa 103   <-> wb 104   A.wal 1346   E.wex 1485   e. wcel 2141   A.wral 2448   A.!walsi 14105   A.!walsc 14106

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116  df-ral 2453  df-alsi 14107  df-alsc 14108

This theorem is referenced by: (None)