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Theorem sb3an 1907
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 947 . . 3  |-  ( (
ph  /\  ps  /\  ch ) 
<->  ( ( ph  /\  ps )  /\  ch )
)
21sbbii 1721 . 2  |-  ( [ y  /  x ]
( ph  /\  ps  /\  ch )  <->  [ y  /  x ] ( ( ph  /\ 
ps )  /\  ch ) )
3 sban 1904 . 2  |-  ( [ y  /  x ]
( ( ph  /\  ps )  /\  ch )  <->  ( [ y  /  x ] ( ph  /\  ps )  /\  [ y  /  x ] ch ) )
4 sban 1904 . . . 4  |-  ( [ y  /  x ]
( ph  /\  ps )  <->  ( [ y  /  x ] ph  /\  [ y  /  x ] ps ) )
54anbi1i 451 . . 3  |-  ( ( [ y  /  x ] ( ph  /\  ps )  /\  [ y  /  x ] ch ) 
<->  ( ( [ y  /  x ] ph  /\ 
[ y  /  x ] ps )  /\  [
y  /  x ] ch ) )
6 df-3an 947 . . 3  |-  ( ( [ y  /  x ] ph  /\  [ y  /  x ] ps  /\ 
[ y  /  x ] ch )  <->  ( ( [ y  /  x ] ph  /\  [ y  /  x ] ps )  /\  [ y  /  x ] ch ) )
75, 6bitr4i 186 . 2  |-  ( ( [ y  /  x ] ( ph  /\  ps )  /\  [ y  /  x ] ch ) 
<->  ( [ y  /  x ] ph  /\  [
y  /  x ] ps  /\  [ y  /  x ] ch ) )
82, 3, 73bitri 205 1  |-  ( [ y  /  x ]
( ph  /\  ps  /\  ch )  <->  ( [ y  /  x ] ph  /\ 
[ y  /  x ] ps  /\  [ y  /  x ] ch ) )
Colors of variables: wff set class
Syntax hints:    /\ wa 103    <-> wb 104    /\ w3a 945   [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-3an 947  df-nf 1420  df-sb 1719
This theorem is referenced by: (None)
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