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Theorem sbbidv 1896
Description: Deduction substituting both sides of a biconditional, with  ph and  x disjoint. See also sbbid 1857. (Contributed by Wolf Lammen, 6-May-2023.) (Proof shortened by Steven Nguyen, 6-Jul-2023.)
Hypothesis
Ref Expression
sbbidv.1  |-  ( ph  ->  ( ps  <->  ch )
)
Assertion
Ref Expression
sbbidv  |-  ( ph  ->  ( [ t  /  x ] ps  <->  [ t  /  x ] ch )
)
Distinct variable group:    ph, x
Allowed substitution hints:    ph( t)    ps( x, t)    ch( x, t)

Proof of Theorem sbbidv
StepHypRef Expression
1 sbbidv.1 . . 3  |-  ( ph  ->  ( ps  <->  ch )
)
21alrimiv 1885 . 2  |-  ( ph  ->  A. x ( ps  <->  ch ) )
3 spsbbi 1855 . 2  |-  ( A. x ( ps  <->  ch )  ->  ( [ t  /  x ] ps  <->  [ t  /  x ] ch )
)
42, 3syl 14 1  |-  ( ph  ->  ( [ t  /  x ] ps  <->  [ t  /  x ] ch )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 105   A.wal 1362   [wsb 1773
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-4 1521  ax-17 1537  ax-ial 1545
This theorem depends on definitions:  df-bi 117  df-sb 1774
This theorem is referenced by:  eqabdv  2322
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