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Theorem sbbid 2018
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 1745 . 2  |-  ( ph  ->  A. x ( ps  <->  ch ) )
4 spsbbi 2017 . 2  |-  ( A. x ( ps  <->  ch )  ->  ( [ y  /  x ] ps  <->  [ y  /  x ] ch )
)
53, 4syl 15 1  |-  ( ph  ->  ( [ y  /  x ] ps  <->  [ y  /  x ] ch )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 176   A.wal 1527   F/wnf 1531   [wsb 1629
This theorem is referenced by:  sbcom  2029  sbcom2  2053
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866
This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1310  df-ex 1529  df-nf 1532  df-sb 1630
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