Theorem nfor 1597

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 1542	. . 3 |- ( ph -> A. x ph )
3		nfor.2	. . . 4 |- F/ x ps
4	3	nfri 1542	. . 3 |- ( ps -> A. x ps )
5	2, 4	hbor 1569	. 2 |- ( ( ph \/ ps ) -> A. x ( ph \/ ps ) )
6	5	nfi 1485	1 |- F/ x ( ph \/ ps )

Syntax hints:   \/ wo 710   F/wnf 1483

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1470  ax-gen 1472  ax-4 1533

This theorem depends on definitions:  df-bi 117  df-nf 1484

This theorem is referenced by:  nfdc  1682  nfun  3329  nfpr  3683  nfso  4349  nffrec  6482  indpi  7455  nfsum1  11667  nfsum  11668  nfcprod1  11865  nfcprod  11866  bj-findis  15915  isomninnlem  15969  trirec0  15983