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Theorem sbrbif 1911
Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)
Hypotheses
Ref Expression
sbrbif.1  |-  ( ch 
->  A. x ch )
sbrbif.2  |-  ( [ y  /  x ] ph 
<->  ps )
Assertion
Ref Expression
sbrbif  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( ps  <->  ch ) )

Proof of Theorem sbrbif
StepHypRef Expression
1 sbrbif.2 . . 3  |-  ( [ y  /  x ] ph 
<->  ps )
21sbrbis 1910 . 2  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( ps  <->  [ y  /  x ] ch ) )
3 sbrbif.1 . . . 4  |-  ( ch 
->  A. x ch )
43sbh 1732 . . 3  |-  ( [ y  /  x ] ch 
<->  ch )
54bibi2i 226 . 2  |-  ( ( ps  <->  [ y  /  x ] ch )  <->  ( ps  <->  ch ) )
62, 5bitri 183 1  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( ps  <->  ch ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 104   A.wal 1312   [wsb 1718
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-io 681  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-10 1466  ax-11 1467  ax-i12 1468  ax-4 1470  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-i5r 1498
This theorem depends on definitions:  df-bi 116  df-nf 1420  df-sb 1719
This theorem is referenced by: (None)
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