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Theorem sb3an 2010
Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006.)
Assertion
Ref Expression
sb3an  |-  ( [ y  /  x ]
( ph  /\  ps  /\  ch )  <->  ( [ y  /  x ] ph  /\ 
[ y  /  x ] ps  /\  [ y  /  x ] ch ) )

Proof of Theorem sb3an
StepHypRef Expression
1 df-3an 936 . . 3  |-  ( (
ph  /\  ps  /\  ch ) 
<->  ( ( ph  /\  ps )  /\  ch )
)
21sbbii 1634 . 2  |-  ( [ y  /  x ]
( ph  /\  ps  /\  ch )  <->  [ y  /  x ] ( ( ph  /\ 
ps )  /\  ch ) )
3 sban 2009 . 2  |-  ( [ y  /  x ]
( ( ph  /\  ps )  /\  ch )  <->  ( [ y  /  x ] ( ph  /\  ps )  /\  [ y  /  x ] ch ) )
4 sban 2009 . . . 4  |-  ( [ y  /  x ]
( ph  /\  ps )  <->  ( [ y  /  x ] ph  /\  [ y  /  x ] ps ) )
54anbi1i 676 . . 3  |-  ( ( [ y  /  x ] ( ph  /\  ps )  /\  [ y  /  x ] ch ) 
<->  ( ( [ y  /  x ] ph  /\ 
[ y  /  x ] ps )  /\  [
y  /  x ] ch ) )
6 df-3an 936 . . 3  |-  ( ( [ y  /  x ] ph  /\  [ y  /  x ] ps  /\ 
[ y  /  x ] ch )  <->  ( ( [ y  /  x ] ph  /\  [ y  /  x ] ps )  /\  [ y  /  x ] ch ) )
75, 6bitr4i 243 . 2  |-  ( ( [ y  /  x ] ( ph  /\  ps )  /\  [ y  /  x ] ch ) 
<->  ( [ y  /  x ] ph  /\  [
y  /  x ] ps  /\  [ y  /  x ] ch ) )
82, 3, 73bitri 262 1  |-  ( [ y  /  x ]
( ph  /\  ps  /\  ch )  <->  ( [ y  /  x ] ph  /\ 
[ y  /  x ] ps  /\  [ y  /  x ] ch ) )
Colors of variables: wff set class
Syntax hints:    <-> wb 176    /\ wa 358    /\ w3a 934   [wsb 1629
This theorem is referenced by:  sbc3ang  3049
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-3an 936  df-tru 1310  df-ex 1529  df-nf 1532  df-sb 1630
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