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Theorem sbbidh 1773
Description: Deduction substituting both sides of a biconditional. New proofs should use sbbid 1774 instead. (Contributed by NM, 5-Aug-1993.) (New usage is discouraged.)
Hypotheses
Ref Expression
sbbidh.1  |-  ( ph  ->  A. x ph )
sbbidh.2  |-  ( ph  ->  ( ps  <->  ch )
)
Assertion
Ref Expression
sbbidh  |-  ( ph  ->  ( [ y  /  x ] ps  <->  [ y  /  x ] ch )
)

Proof of Theorem sbbidh
StepHypRef Expression
1 sbbidh.1 . . 3  |-  ( ph  ->  A. x ph )
2 sbbidh.2 . . 3  |-  ( ph  ->  ( ps  <->  ch )
)
31, 2alrimih 1403 . 2  |-  ( ph  ->  A. x ( ps  <->  ch ) )
4 spsbbi 1772 . 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 103   A.wal 1287   [wsb 1692
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 1381  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-4 1445  ax-ial 1472
This theorem depends on definitions:  df-bi 115  df-sb 1693
This theorem is referenced by:  sbcomxyyz  1894
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