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Theorem sbbid 2105
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 1816 . 2  |-  ( ph  ->  A. x ( ps  <->  ch ) )
4 spsbbi 2104 . 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 setvar class
Syntax hints:    -> wi 4    <-> wb 184   A.wal 1368   F/wnf 1590   [wsb 1702
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1592  ax-4 1603  ax-5 1671  ax-6 1710  ax-7 1730  ax-10 1777  ax-12 1794  ax-13 1955
This theorem depends on definitions:  df-bi 185  df-or 370  df-an 371  df-ex 1588  df-nf 1591  df-sb 1703
This theorem is referenced by:  sbcom3  2115  sbco3  2124  sbcom2  2160  sbcom2OLD  2161  sbal  2183  wl-equsb3  28548  wl-sbcom2d-lem1  28553  wl-sbcom3  28579
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