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Theorem nfbid 1525
Description: If in a context  x is not free in  ps and  ch, then it is not free in  ( ps  <->  ch ). (Contributed by Mario Carneiro, 24-Sep-2016.) (Proof shortened by Wolf Lammen, 29-Dec-2017.)
Hypotheses
Ref Expression
nfbid.1  |-  ( ph  ->  F/ x ps )
nfbid.2  |-  ( ph  ->  F/ x ch )
Assertion
Ref Expression
nfbid  |-  ( ph  ->  F/ x ( ps  <->  ch ) )

Proof of Theorem nfbid
StepHypRef Expression
1 dfbi2 380 . 2  |-  ( ( ps  <->  ch )  <->  ( ( ps  ->  ch )  /\  ( ch  ->  ps )
) )
2 nfbid.1 . . . 4  |-  ( ph  ->  F/ x ps )
3 nfbid.2 . . . 4  |-  ( ph  ->  F/ x ch )
42, 3nfimd 1522 . . 3  |-  ( ph  ->  F/ x ( ps 
->  ch ) )
53, 2nfimd 1522 . . 3  |-  ( ph  ->  F/ x ( ch 
->  ps ) )
64, 5nfand 1505 . 2  |-  ( ph  ->  F/ x ( ( ps  ->  ch )  /\  ( ch  ->  ps ) ) )
71, 6nfxfrd 1409 1  |-  ( ph  ->  F/ x ( ps  <->  ch ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102    <-> wb 103   F/wnf 1394
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-5 1381  ax-gen 1383  ax-4 1445  ax-ial 1472  ax-i5r 1473
This theorem depends on definitions:  df-bi 115  df-nf 1395
This theorem is referenced by:  nfbi  1526  nfeudv  1963  nfeqd  2243  nfiotadxy  4983  iota2df  5004  bdsepnft  11778  strcollnft  11879
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