Theorem hb3or 1563

Description: If x is not free in ph, ps, and ch, it is not free in ( ph \/ ps \/ ch ). (Contributed by NM, 14-Sep-2003.)

Hypotheses

Ref	Expression
hb.1	|- ( ph -> A. x ph )
hb.2	|- ( ps -> A. x ps )
hb.3	|- ( ch -> A. x ch )

Assertion

Ref	Expression
hb3or	|- ( ( ph \/ ps \/ ch ) -> A. x ( ph \/ ps \/ ch ) )

Proof of Theorem hb3or

Step	Hyp	Ref	Expression
1		df-3or 981	. 2 |- ( ( ph \/ ps \/ ch ) <-> ( ( ph \/ ps ) \/ ch ) )
2		hb.1	. . . 4 |- ( ph -> A. x ph )
3		hb.2	. . . 4 |- ( ps -> A. x ps )
4	2, 3	hbor 1560	. . 3 |- ( ( ph \/ ps ) -> A. x ( ph \/ ps ) )
5		hb.3	. . 3 |- ( ch -> A. x ch )
6	4, 5	hbor 1560	. 2 |- ( ( ( ph \/ ps ) \/ ch ) -> A. x ( ( ph \/ ps ) \/ ch ) )
7	1, 6	hbxfrbi 1486	1 |- ( ( ph \/ ps \/ ch ) -> A. x ( ph \/ ps \/ ch ) )

Syntax hints:   -> wi 4   \/ wo 709   \/ w3o 979   A.wal 1362

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 710  ax-5 1461  ax-gen 1463

This theorem depends on definitions:  df-bi 117  df-3or 981

This theorem is referenced by: (None)