Theorem hbor 1006

Description: If x is not free in ph and ps, it is not free in (ph \/ ps).

Hypotheses

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

Assertion

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

Proof of Theorem hbor

Step	Hyp	Ref	Expression
1		hb.1	. . . 4 |- (ph -> A.xph)
2	1	hbn 1002	. . 3 |- (-. ph -> A.x -. ph)
3		hb.2	. . 3 |- (ps -> A.xps)
4	2, 3	hbim 1005	. 2 |- ((-. ph -> ps) -> A.x(-. ph -> ps))
5		df-or 224	. 2 |- ((ph \/ ps) <-> (-. ph -> ps))
6	5	albii 997	. 2 |- (A.x(ph \/ ps) <-> A.x(-. ph -> ps))
7	4, 5, 6	3imtr4 219	1 |- ((ph \/ ps) -> A.x(ph \/ ps))

Syntax hints:  -. wn 2   -> wi 3   \/ wo 222  A.wal 952

This theorem is referenced by:  hb3or 1009  hbun 2182  hbif 2369  hbpr 2422  hbsuc 3035

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 961  ax-4 971  ax-5o 973  ax-6o 976

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225

Copyright terms: Public domain