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Theorem alsconv 11058
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 2358 . . 3  |-  ( A. x  e.  A  ph  <->  A. x
( x  e.  A  ->  ph ) )
21anbi1i 446 . 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 11057 . 2  |-  ( A.! x  e.  A ph  <->  ( A. x  e.  A  ph 
/\  E. x  x  e.  A ) )
4 df-alsi 11056 . 2  |-  ( A.! x ( x  e.  A  ->  ph )  <->  ( A. x ( x  e.  A  ->  ph )  /\  E. x  x  e.  A
) )
52, 3, 43bitr4ri 211 1  |-  ( A.! x ( x  e.  A  ->  ph )  <->  A.! x  e.  A ph )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102    <-> wb 103   A.wal 1283   E.wex 1422    e. wcel 1434   A.wral 2353   A.!walsi 11054   A.!walsc 11055
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 2358  df-alsi 11056  df-alsc 11057
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator