Theorem alsconv 10791

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 2353	. . 3 |- ( A. x e. A ph <-> A. x ( x e. A -> ph ) )
2	1	anbi1i 445	. 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 10790	. 2 |- ( A.! x e. A ph <-> ( A. x e. A ph /\ E. x x e. A ) )
4		df-alsi 10789	. 2 |- ( A.! x ( x e. A -> ph ) <-> ( A. x ( x e. A -> ph ) /\ E. x x e. A ) )
5	2, 3, 4	3bitr4ri 211	1 |- ( A.! x ( x e. A -> ph ) <-> A.! x e. A ph )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103   A.wal 1282   E.wex 1421   e. wcel 1433   A.wral 2348   A.!walsi 10787   A.!walsc 10788

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106

This theorem depends on definitions:  df-bi 115  df-ral 2353  df-alsi 10789  df-alsc 10790

This theorem is referenced by: (None)