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Theorem sbrbis 1878
Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)
Hypothesis
Ref Expression
sbrbis.1  |-  ( [ y  /  x ] ph 
<->  ps )
Assertion
Ref Expression
sbrbis  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( ps  <->  [ y  /  x ] ch ) )

Proof of Theorem sbrbis
StepHypRef Expression
1 sbbi 1876 . 2  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( [
y  /  x ] ph 
<->  [ y  /  x ] ch ) )
2 sbrbis.1 . . 3  |-  ( [ y  /  x ] ph 
<->  ps )
32bibi1i 226 . 2  |-  ( ( [ y  /  x ] ph  <->  [ y  /  x ] ch )  <->  ( ps  <->  [ y  /  x ] ch ) )
41, 3bitri 182 1  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( ps  <->  [ y  /  x ] ch ) )
Colors of variables: wff set class
Syntax hints:    <-> wb 103   [wsb 1687
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-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469
This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1688
This theorem is referenced by:  sbrbif  1879  sbabel  2248
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