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Theorem nfor 1507
Description: If  x is not free in  ph and  ps, it is not free in  ( ph  \/  ps ). (Contributed by Jim Kingdon, 11-Mar-2018.)
Hypotheses
Ref Expression
nfor.1  |-  F/ x ph
nfor.2  |-  F/ x ps
Assertion
Ref Expression
nfor  |-  F/ x
( ph  \/  ps )

Proof of Theorem nfor
StepHypRef Expression
1 nfor.1 . . . 4  |-  F/ x ph
21nfri 1453 . . 3  |-  ( ph  ->  A. x ph )
3 nfor.2 . . . 4  |-  F/ x ps
43nfri 1453 . . 3  |-  ( ps 
->  A. x ps )
52, 4hbor 1479 . 2  |-  ( (
ph  \/  ps )  ->  A. x ( ph  \/  ps ) )
65nfi 1392 1  |-  F/ x
( ph  \/  ps )
Colors of variables: wff set class
Syntax hints:    \/ wo 662   F/wnf 1390
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 663  ax-5 1377  ax-gen 1379  ax-4 1441
This theorem depends on definitions:  df-bi 115  df-nf 1391
This theorem is referenced by:  nfdc  1590  nfun  3129  nfpr  3444  nfso  4059  nffrec  6039  indpi  6583  nfsum1  10320  nfsum  10321  bj-findis  10917
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