Theorem nfor 1562

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

Step	Hyp	Ref	Expression
1		nfor.1	. . . 4 |- F/ x ph
2	1	nfri 1507	. . 3 |- ( ph -> A. x ph )
3		nfor.2	. . . 4 |- F/ x ps
4	3	nfri 1507	. . 3 |- ( ps -> A. x ps )
5	2, 4	hbor 1534	. 2 |- ( ( ph \/ ps ) -> A. x ( ph \/ ps ) )
6	5	nfi 1450	1 |- F/ x ( ph \/ ps )

Syntax hints:   \/ wo 698   F/wnf 1448

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 699  ax-5 1435  ax-gen 1437  ax-4 1498

This theorem depends on definitions:  df-bi 116  df-nf 1449

This theorem is referenced by:  nfdc  1647  nfun  3278  nfpr  3626  nfso  4280  nffrec  6364  indpi  7283  nfsum1  11297  nfsum  11298  nfcprod1  11495  nfcprod  11496  bj-findis  13861  isomninnlem  13909  trirec0  13923