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Theorem nfor 1536
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 1482 . . 3  |-  ( ph  ->  A. x ph )
3 nfor.2 . . . 4  |-  F/ x ps
43nfri 1482 . . 3  |-  ( ps 
->  A. x ps )
52, 4hbor 1508 . 2  |-  ( (
ph  \/  ps )  ->  A. x ( ph  \/  ps ) )
65nfi 1421 1  |-  F/ x
( ph  \/  ps )
Colors of variables: wff set class
Syntax hints:    \/ wo 680   F/wnf 1419
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-io 681  ax-5 1406  ax-gen 1408  ax-4 1470
This theorem depends on definitions:  df-bi 116  df-nf 1420
This theorem is referenced by:  nfdc  1620  nfun  3200  nfpr  3541  nfso  4192  nffrec  6259  indpi  7114  nfsum1  11065  nfsum  11066  bj-findis  12979  isomninnlem  13027
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