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Theorem sbrbif 2027
Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)
Hypotheses
Ref Expression
sbrbif.1  |-  F/ 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 2026 . 2  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( ps  <->  [ y  /  x ] ch ) )
3 sbrbif.1 . . . 4  |-  F/ x ch
43sbf 1979 . . 3  |-  ( [ y  /  x ] ch 
<->  ch )
54bibi2i 304 . 2  |-  ( ( ps  <->  [ y  /  x ] ch )  <->  ( ps  <->  ch ) )
62, 5bitri 240 1  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( ps  <->  ch ) )
Colors of variables: wff set class
Syntax hints:    <-> wb 176   F/wnf 1534   [wsb 1638
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1536  ax-5 1547  ax-17 1606  ax-9 1644  ax-8 1661  ax-6 1715  ax-7 1720  ax-11 1727  ax-12 1878
This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1310  df-ex 1532  df-nf 1535  df-sb 1639
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