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Theorem 2albiim 1602
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 1601 . . 3  |-  ( A. y ( ph  <->  ps )  <->  ( A. y ( ph  ->  ps )  /\  A. y ( ps  ->  ph ) ) )
21albii 1556 . 2  |-  ( A. x A. y ( ph  <->  ps )  <->  A. x ( A. y ( ph  ->  ps )  /\  A. y
( ps  ->  ph )
) )
3 19.26 1583 . 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 1530
This theorem is referenced by:  sbnf2  2060  2eu6  2241  eqopab2b  4310  eqrel  4793  eqrelrel  4804  eqoprab2b  5922  pm14.123a  27728  sbnf2NEW7  29580
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1536  ax-5 1547
This theorem depends on definitions:  df-bi 177  df-an 360
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