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Theorem sb3an 1880
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 926 . . 3  |-  ( (
ph  /\  ps  /\  ch ) 
<->  ( ( ph  /\  ps )  /\  ch )
)
21sbbii 1695 . 2  |-  ( [ y  /  x ]
( ph  /\  ps  /\  ch )  <->  [ y  /  x ] ( ( ph  /\ 
ps )  /\  ch ) )
3 sban 1877 . 2  |-  ( [ y  /  x ]
( ( ph  /\  ps )  /\  ch )  <->  ( [ y  /  x ] ( ph  /\  ps )  /\  [ y  /  x ] ch ) )
4 sban 1877 . . . 4  |-  ( [ y  /  x ]
( ph  /\  ps )  <->  ( [ y  /  x ] ph  /\  [ y  /  x ] ps ) )
54anbi1i 446 . . 3  |-  ( ( [ y  /  x ] ( ph  /\  ps )  /\  [ y  /  x ] ch ) 
<->  ( ( [ y  /  x ] ph  /\ 
[ y  /  x ] ps )  /\  [
y  /  x ] ch ) )
6 df-3an 926 . . 3  |-  ( ( [ y  /  x ] ph  /\  [ y  /  x ] ps  /\ 
[ y  /  x ] ch )  <->  ( ( [ y  /  x ] ph  /\  [ y  /  x ] ps )  /\  [ y  /  x ] ch ) )
75, 6bitr4i 185 . 2  |-  ( ( [ y  /  x ] ( ph  /\  ps )  /\  [ y  /  x ] ch ) 
<->  ( [ y  /  x ] ph  /\  [
y  /  x ] ps  /\  [ y  /  x ] ch ) )
82, 3, 73bitri 204 1  |-  ( [ y  /  x ]
( ph  /\  ps  /\  ch )  <->  ( [ y  /  x ] ph  /\ 
[ y  /  x ] ps  /\  [ y  /  x ] ch ) )
Colors of variables: wff set class
Syntax hints:    /\ wa 102    <-> wb 103    /\ w3a 924   [wsb 1692
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 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473
This theorem depends on definitions:  df-bi 115  df-3an 926  df-nf 1395  df-sb 1693
This theorem is referenced by: (None)
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