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Theorem nfbid 1581
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 386 . 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 1578 . . 3  |-  ( ph  ->  F/ x ( ps 
->  ch ) )
53, 2nfimd 1578 . . 3  |-  ( ph  ->  F/ x ( ch 
->  ps ) )
64, 5nfand 1561 . 2  |-  ( ph  ->  F/ x ( ( ps  ->  ch )  /\  ( ch  ->  ps ) ) )
71, 6nfxfrd 1468 1  |-  ( ph  ->  F/ x ( ps  <->  ch ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 103    <-> wb 104   F/wnf 1453
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-5 1440  ax-gen 1442  ax-4 1503  ax-ial 1527  ax-i5r 1528
This theorem depends on definitions:  df-bi 116  df-nf 1454
This theorem is referenced by:  nfbi  1582  nfeudv  2034  nfeqd  2327  nfiotadw  5161  iota2df  5182  bdsepnft  13882
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