Theorem nfor 1506

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 1452	. . 3 |- ( ph -> A. x ph )
3		nfor.2	. . . 4 |- F/ x ps
4	3	nfri 1452	. . 3 |- ( ps -> A. x ps )
5	2, 4	hbor 1478	. 2 |- ( ( ph \/ ps ) -> A. x ( ph \/ ps ) )
6	5	nfi 1391	1 |- F/ x ( ph \/ ps )

Syntax hints:   \/ wo 661   F/wnf 1389

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 662  ax-5 1376  ax-gen 1378  ax-4 1440

This theorem depends on definitions:  df-bi 115  df-nf 1390

This theorem is referenced by:  nfdc  1589  nfun  3128  nfpr  3442  nfso  4057  nffrec  6005  indpi  6532  nfsum1  10193  nfsum  10194  bj-findis  10774