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Theorem 2albiim 1599
Description: Split a biconditional and distribute 2 quantifiers. (Contributed by NM, 3-Feb-2005.)
Assertion
Ref Expression
2albiim  |-  ( A. x A. y ( ph  <->  ps )  <->  ( A. x A. y ( ph  ->  ps )  /\  A. x A. y ( ps  ->  ph ) ) )

Proof of Theorem 2albiim
StepHypRef Expression
1 albiim 1598 . . 3  |-  ( A. y ( ph  <->  ps )  <->  ( A. y ( ph  ->  ps )  /\  A. y ( ps  ->  ph ) ) )
21albii 1553 . 2  |-  ( A. x A. y ( ph  <->  ps )  <->  A. x ( A. y ( ph  ->  ps )  /\  A. y
( ps  ->  ph )
) )
3 19.26 1580 . 2  |-  ( A. x ( A. y
( ph  ->  ps )  /\  A. y ( ps 
->  ph ) )  <->  ( A. x A. y ( ph  ->  ps )  /\  A. x A. y ( ps 
->  ph ) ) )
42, 3bitri 240 1  |-  ( A. x A. y ( ph  <->  ps )  <->  ( A. x A. y ( ph  ->  ps )  /\  A. x A. y ( ps  ->  ph ) ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 176    /\ wa 358   A.wal 1527
This theorem is referenced by:  sbnf2  2047  2eu6  2228  eqopab2b  4294  eqrel  4777  eqrelrel  4788  eqoprab2b  5906  pm14.123a  27625
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544
This theorem depends on definitions:  df-bi 177  df-an 360
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