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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
StepHypRef Expression
1 df-ral 2470 . . 3  |-  ( A. x  e.  A  ph  <->  A. x
( x  e.  A  ->  ph ) )
21anbi1i 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
) )
52, 3, 43bitr4ri 213 1  |-  ( A.! x ( x  e.  A  ->  ph )  <->  A.! x  e.  A ph )
Colors of variables: wff set class
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)
  Copyright terms: Public domain W3C validator