Theorem hbor 1560

Description: If x is not free in ph and ps, it is not free in ( ph \/ ps ). (Contributed by NM, 5-Aug-1993.) (Revised by NM, 2-Feb-2015.)

Hypotheses

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

Assertion

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

Proof of Theorem hbor

Step	Hyp	Ref	Expression
1		hb.1	. . 3 |- ( ph -> A. x ph )
2		orc 713	. . . 4 |- ( ph -> ( ph \/ ps ) )
3	2	alimi 1469	. . 3 |- ( A. x ph -> A. x ( ph \/ ps ) )
4	1, 3	syl 14	. 2 |- ( ph -> A. x ( ph \/ ps ) )
5		hb.2	. . 3 |- ( ps -> A. x ps )
6		olc 712	. . . 4 |- ( ps -> ( ph \/ ps ) )
7	6	alimi 1469	. . 3 |- ( A. x ps -> A. x ( ph \/ ps ) )
8	5, 7	syl 14	. 2 |- ( ps -> A. x ( ph \/ ps ) )
9	4, 8	jaoi 717	1 |- ( ( ph \/ ps ) -> A. x ( ph \/ ps ) )

Syntax hints:   -> wi 4   \/ wo 709   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

This theorem is referenced by:  hb3or  1563  nfor  1588  19.43  1642