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Theorem sbrbis 2013
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 2011 . 2  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( [
y  /  x ] ph 
<->  [ y  /  x ] ch ) )
2 sbrbis.1 . . 3  |-  ( [ y  /  x ] ph 
<->  ps )
32bibi1i 305 . 2  |-  ( ( [ y  /  x ] ph  <->  [ y  /  x ] ch )  <->  ( ps  <->  [ y  /  x ] ch ) )
41, 3bitri 240 1  |-  ( [ y  /  x ]
( ph  <->  ch )  <->  ( ps  <->  [ y  /  x ] ch ) )
Colors of variables: wff set class
Syntax hints:    <-> wb 176   [wsb 1629
This theorem is referenced by:  sbrbif  2014  sbabel  2445
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866
This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1310  df-ex 1529  df-nf 1532  df-sb 1630
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