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Theorem spsbbi 1769
Description: Specialization of biconditional. (Contributed by NM, 5-Aug-1993.) (Proof rewritten by Jim Kingdon, 21-Jan-2018.)
Assertion
Ref Expression
spsbbi  |-  ( A. x ( ph  <->  ps )  ->  ( [ y  /  x ] ph  <->  [ y  /  x ] ps )
)

Proof of Theorem spsbbi
StepHypRef Expression
1 spsbim 1768 . . 3  |-  ( A. x ( ph  ->  ps )  ->  ( [
y  /  x ] ph  ->  [ y  /  x ] ps ) )
2 spsbim 1768 . . 3  |-  ( A. x ( ps  ->  ph )  ->  ( [
y  /  x ] ps  ->  [ y  /  x ] ph ) )
31, 2anim12i 331 . 2  |-  ( ( A. x ( ph  ->  ps )  /\  A. x ( ps  ->  ph ) )  ->  (
( [ y  /  x ] ph  ->  [ y  /  x ] ps )  /\  ( [ y  /  x ] ps  ->  [ y  /  x ] ph ) ) )
4 albiim 1419 . 2  |-  ( A. x ( ph  <->  ps )  <->  ( A. x ( ph  ->  ps )  /\  A. x ( ps  ->  ph ) ) )
5 dfbi2 380 . 2  |-  ( ( [ y  /  x ] ph  <->  [ y  /  x ] ps )  <->  ( ( [ y  /  x ] ph  ->  [ y  /  x ] ps )  /\  ( [ y  /  x ] ps  ->  [ y  /  x ] ph ) ) )
63, 4, 53imtr4i 199 1  |-  ( A. x ( ph  <->  ps )  ->  ( [ y  /  x ] ph  <->  [ y  /  x ] ps )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102    <-> wb 103   A.wal 1285   [wsb 1689
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1379  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-4 1443  ax-ial 1470
This theorem depends on definitions:  df-bi 115  df-sb 1690
This theorem is referenced by:  sbbidh  1770  sbbid  1771  relelfvdm  5299
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