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Theorem sbbid 1802
Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-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 1487 . 2  |-  ( ph  ->  A. x ( ps  <->  ch ) )
4 spsbbi 1800 . 2  |-  ( A. x ( ps  <->  ch )  ->  ( [ y  /  x ] ps  <->  [ y  /  x ] ch )
)
53, 4syl 14 1  |-  ( ph  ->  ( [ y  /  x ] ps  <->  [ y  /  x ] ch )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 104   A.wal 1314   F/wnf 1421   [wsb 1720
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 1408  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-4 1472  ax-ial 1499
This theorem depends on definitions:  df-bi 116  df-nf 1422  df-sb 1721
This theorem is referenced by:  bezoutlemmain  11613
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