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Theorem 2albiim 1667
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 1666 . . 3  |-  ( A. y ( ph  <->  ps )  <->  ( A. y ( ph  ->  ps )  /\  A. y ( ps  ->  ph ) ) )
21albii 1611 . 2  |-  ( A. x A. y ( ph  <->  ps )  <->  A. x ( A. y ( ph  ->  ps )  /\  A. y
( ps  ->  ph )
) )
3 19.26 1648 . 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 249 1  |-  ( A. x A. y ( ph  <->  ps )  <->  ( A. x A. y ( ph  ->  ps )  /\  A. x A. y ( ps  ->  ph ) ) )
Colors of variables: wff setvar class
Syntax hints:    -> wi 4    <-> wb 184    /\ wa 369   A.wal 1368
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1592  ax-4 1603
This theorem depends on definitions:  df-bi 185  df-an 371
This theorem is referenced by:  sbnf2  2153  sbnf2OLD  2154  2eu6  2380  2eu6OLD  2381  eqopab2b  4729  eqrel  5040  eqrelrel  5052  eqoprab2b  6256  eqrelrd2  26117  pm14.123a  29847
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