Correction to fourth row of transitivity table

[GoogleCodeExporter]

Your correction to fourth row:
IF
A di B
B d C
THEN
A ni C (using your notation of "ni" for all relationships)

Pictorial counterexample:
AAAAAAAA_________
__BBBB___________
_________CCCCCCCC

Or in other words; if A < C, then there is no possibility that premise B d C 
could be true.  Therefore, some scenarios simply can not exist and thus A ni C 
(which contains relationships A {<, >, m, mi} C) is a false conclusion.  

However, I do agree that Allen's notation is fuzzy when he says "o oi dur con 
=".  If he meant "o oi d di =", then why change notation to be different from 
the rest of the table?  I'm just guessing widly, but maybe he didn't have 
enough space to write "o oi d di s si f fi =" in one table cell.  I believe it 
should be {o, oi, d, di, s, si, f, fi, =} which you already picked up on in 
later rows.

Original issue reported on code.google.com by justyn.b...@gmail.com on 22 Aug 2011 at 12:09

[GoogleCodeExporter]

Missed a few important words: "if conclusion A < C is true (a subset of A ni C) 
and premise A di B is true, then there is no possibility that premise B d C 
could be true".

Original comment by justyn.b...@gmail.com on 22 Aug 2011 at 12:23

[GoogleCodeExporter]

Dear Justyn,

I think you are right. Probably I was confused by the notation change. I 
changed the source code and the jar.

I am very sorry for the late answer. Apperantly, I was not notified about this.

Thank you very much for your detailed comment. Please do not hesitate to 
contact me, if you have questions.

Best regards

Original comment by jornfra...@gmail.com on 23 Dec 2011 at 10:55

[GoogleCodeExporter]

Original comment by jornfra...@gmail.com on 23 Dec 2011 at 10:56

• Changed state: Fixed