Theorem alsconv 15125

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 2470	. . 3 |- ( A. x e. A ph <-> A. x ( x e. A -> ph ) )
2	1	anbi1i 458	. 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 15124	. 2 |- ( A.! x e. A ph <-> ( A. x e. A ph /\ E. x x e. A ) )
4		df-alsi 15123	. 2 |- ( A.! x ( x e. A -> ph ) <-> ( A. x ( x e. A -> ph ) /\ E. x x e. A ) )
5	2, 3, 4	3bitr4ri 213	1 |- ( A.! x ( x e. A -> ph ) <-> A.! x e. A ph )

Syntax hints:   -> wi 4   /\ wa 104   <-> wb 105   A.wal 1361   E.wex 1502   e. wcel 2158   A.wral 2465   A.!walsi 15121   A.!walsc 15122

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117  df-ral 2470  df-alsi 15123  df-alsc 15124

This theorem is referenced by: (None)