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Theorem sbbid 2111
Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 5-Aug-1993.)
Hypotheses
Ref Expression
sbbid.1  |-  F/ x ph
sbbid.2  |-  ( ph  ->  ( ps  <->  ch )
)
Assertion
Ref Expression
sbbid  |-  ( ph  ->  ( [ y  /  x ] ps  <->  [ y  /  x ] ch )
)

Proof of Theorem sbbid
StepHypRef Expression
1 sbbid.1 . . 3  |-  F/ x ph
2 sbbid.2 . . 3  |-  ( ph  ->  ( ps  <->  ch )
)
31, 2alrimi 1773 . 2  |-  ( ph  ->  A. x ( ps  <->  ch ) )
4 spsbbi 2110 . 2  |-  ( A. x ( ps  <->  ch )  ->  ( [ y  /  x ] ps  <->  [ y  /  x ] ch )
)
53, 4syl 16 1  |-  ( ph  ->  ( [ y  /  x ] ps  <->  [ y  /  x ] ch )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 177   A.wal 1546   F/wnf 1550   [wsb 1655
This theorem is referenced by:  sbcom  2122  sbcom2  2147
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1552  ax-5 1563  ax-17 1623  ax-9 1661  ax-8 1682  ax-6 1736  ax-7 1741  ax-11 1753  ax-12 1939
This theorem depends on definitions:  df-bi 178  df-an 361  df-tru 1325  df-ex 1548  df-nf 1551  df-sb 1656
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