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Theorem 2albiim 1447
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 1446 . . 3  |-  ( A. y ( ph  <->  ps )  <->  ( A. y ( ph  ->  ps )  /\  A. y ( ps  ->  ph ) ) )
21albii 1429 . 2  |-  ( A. x A. y ( ph  <->  ps )  <->  A. x ( A. y ( ph  ->  ps )  /\  A. y
( ps  ->  ph )
) )
3 19.26 1440 . 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 183 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    /\ wa 103    <-> wb 104   A.wal 1312
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1406  ax-gen 1408
This theorem depends on definitions:  df-bi 116
This theorem is referenced by:  sbnf2  1932  eqopab2b  4169  eqrel  4596  eqrelrel  4608  eqoprab2b  5795
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