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Theorem nfd 1459
Description: Deduce that  x is not free in  ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)
Hypotheses
Ref Expression
nfd.1  |-  F/ x ph
nfd.2  |-  ( ph  ->  ( ps  ->  A. x ps ) )
Assertion
Ref Expression
nfd  |-  ( ph  ->  F/ x ps )

Proof of Theorem nfd
StepHypRef Expression
1 nfd.1 . . . 4  |-  F/ x ph
21nfri 1455 . . 3  |-  ( ph  ->  A. x ph )
3 nfd.2 . . 3  |-  ( ph  ->  ( ps  ->  A. x ps ) )
42, 3alrimih 1401 . 2  |-  ( ph  ->  A. x ( ps 
->  A. x ps )
)
5 df-nf 1393 . 2  |-  ( F/ x ps  <->  A. x
( ps  ->  A. x ps ) )
64, 5sylibr 132 1  |-  ( ph  ->  F/ x ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1285   F/wnf 1392
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 1379  ax-gen 1381  ax-4 1443
This theorem depends on definitions:  df-bi 115  df-nf 1393
This theorem is referenced by:  nfdh  1460  nfrimi  1461  nfnt  1589  cbv1h  1677  nfald  1687  a16nf  1791  dvelimALT  1931  dvelimfv  1932  nfsb4t  1935  hbeud  1967
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