Re: OR NOT Logic
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Peter P. Benac wrote: [snip] > > And your domain is my Mother's Maiden Name :) > > Regards, > Pete Remind me who you bank with? ;-) C. - -- Craig McLeanhttp://fukka.co.uk [EMAIL PROTECTED] Where the fun never starts Powered by FreeBSD, and GIN! -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.3 (GNU/Linux) iD8DBQFEWfgdMDDagS2VwJ4RAvgsAKCcYO497ItTZVLoruJ77tY1L4vyHACePYe3 QHIUcm3EhUMpycvS9Ek5NWY= =D6jx -END PGP SIGNATURE-
Re: OR NOT Logic
Thank you Mouss and to everyone for answering my logic question. I now understand this better than I ever thought possible. Dan On May 3, 2006, at 15:14, mouss wrote: Dan wrote: Is: A && (B || C || D || E || F) equivalent to?: A && (!B && !C && !D && !E && !F) No. The DeMorgan laws are: ! ( A && B)= !A || !B !(A || B) == !A && !B so you _dispatch_ the '!' and replace || with && and conversely. example: if it is _not_ spam _and_ it is not virus, deliver it; else quarantine it is the same as if it is _either_ spam _or_ virus, quarantine it. else deliver it
Re: OR NOT Logic
Dan wrote: Is: A && (B || C || D || E || F) equivalent to?: A && (!B && !C && !D && !E && !F) No. The DeMorgan laws are: ! ( A && B)= !A || !B !(A || B) == !A && !B so you _dispatch_ the '!' and replace || with && and conversely. example: if it is _not_ spam _and_ it is not virus, deliver it; else quarantine it is the same as if it is _either_ spam _or_ virus, quarantine it. else deliver it
Re: OR NOT Logic
> Is: > > A && (B || C || D || E || F) if TRUE OR TRUE true if either is true > equivalent to?: > > A && (!B && !C && !D && !E && !F) if FALSE AND FALSE true if both are false So no, they aren't equivalent. Loren
Re: OR NOT Logic
>... >I believe that's a fundamental logic rule, so yes. > >A && B == ~A || ~B > >--Russell Almost: -- Not to confuse things with C's short ciruit operations | v ( A and B ) equals ( not ( ( not A ) or ( not B ) ) ) ^ | Also known as one case of the "contrapositive". Paul Shupak [EMAIL PROTECTED]
Re: OR NOT Logic
any(@criteria) = not all(not @criteria) Consider Lifeboat1-has-a-seat OR Lifeboat2-has-a-seat OR ... LifeboatN-has-a-seat vs. Lifeboat1-is-full AND Lifeboat2-is-full AND ... LifeboatN-is-full Thanks for the examples Matt, time for some testing Dan
RE: OR NOT Logic
Dan wrote: > Wow, > > I must be confusing this with any/all is/isn't. In various software > (mail scripts, iTunes smart playlists, etc): > > any IS IS IS IS > > equals > > all NOT NOT NOT NOT Exactly backwards. any(@criteria) = not all(not @criteria) Consider Lifeboat1-has-a-seat OR Lifeboat2-has-a-seat OR ... LifeboatN-has-a-seat vs. Lifeboat1-is-full AND Lifeboat2-is-full AND ... LifeboatN-is-full
Re: OR NOT Logic
Wow, I must be confusing this with any/all is/isn't. In various software (mail scripts, iTunes smart playlists, etc): any IS IS IS IS equals all NOT NOT NOT NOT Dan On May 2, 2006, at 15:11, Matt Kettler wrote: Russell Miller wrote: On Tuesday 02 May 2006 14:59, Dan wrote: Is: A && (B || C || D || E || F) equivalent to?: A && (!B && !C && !D && !E && !F) No the two are NOT equivalent. The first statement will be true if A and any one of B-F is true. The second statement will be true if A and all of B-F are false. It would be true if you modified the second rule to be: A && !(!B && !C && !D && !E && !F) I believe that's a fundamental logic rule, so yes. A && B == ~A || ~B No, that is not a fundamental logic rule. It is not even true, they are in fact exact opposites of each other. I'm going to change the use of ~ to !, just to match the OP's syntax of the NOT operation. ( A && B == !A || !B ) Truth tables: A B (A && B) ( !A || !B) 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 The correct rule is: !( A && B) == !A || !B It's called DeMorgan's theorem. Note the difference being that there's negation on both sides. A B !(A && B) ( !A || !B) 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 0
Re: OR NOT Logic
Russell Miller wrote: > On Tuesday 02 May 2006 14:59, Dan wrote: >> Is: >> >> A && (B || C || D || E || F) >> >> equivalent to?: >> >> A && (!B && !C && !D && !E && !F) No the two are NOT equivalent. The first statement will be true if A and any one of B-F is true. The second statement will be true if A and all of B-F are false. It would be true if you modified the second rule to be: A && !(!B && !C && !D && !E && !F) >> > I believe that's a fundamental logic rule, so yes. > > A && B == ~A || ~B No, that is not a fundamental logic rule. It is not even true, they are in fact exact opposites of each other. I'm going to change the use of ~ to !, just to match the OP's syntax of the NOT operation. ( A && B == !A || !B ) Truth tables: A B (A && B) ( !A || !B) 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 The correct rule is: !( A && B) == !A || !B It's called DeMorgan's theorem. Note the difference being that there's negation on both sides. A B !(A && B) ( !A || !B) 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 0
Re: OR NOT Logic
No!1 In the first example if A is postive and any one of the () variable are positive it will pass. In the second example A would have to be positive and all the () variable would have to be negative to pass. And your domain is my Mother's Maiden Name :) Regards, Pete > > Is: > > A && (B || C || D || E || F) > > equivalent to?: > > A && (!B && !C && !D && !E && !F) > > as in: > > meta __FORGED_OUTLOOK_DOLLARS (__OUTLOOK_DOLLARS_MUA && ! > __OUTLOOK_DOLLARS_MSGID && !__OUTLOOK_DOLLARS_OTHER && !__IMS_MSGID > && !__UNUSABLE_MSGID) > > > Thanks, > Dan > Peter P. Benac, CCNA Emacolet Networking Services, Inc Providing Network and Systems Project Management and Installation and Web Hosting. Phone: 919-618-2557 Web: http://www.emacolet.com Need quick reliable Systems or Network Management advice visit http://www.nmsusers.org To have principles... First have courage.. With principles comes integrity!!!
Re: OR NOT Logic
On Tuesday 02 May 2006 14:59, Dan wrote: > Is: > > A && (B || C || D || E || F) > > equivalent to?: > > A && (!B && !C && !D && !E && !F) > I believe that's a fundamental logic rule, so yes. A && B == ~A || ~B --Russell > as in: > > meta __FORGED_OUTLOOK_DOLLARS (__OUTLOOK_DOLLARS_MUA && ! > __OUTLOOK_DOLLARS_MSGID && !__OUTLOOK_DOLLARS_OTHER && !__IMS_MSGID > && !__UNUSABLE_MSGID) > > > Thanks, > Dan
OR NOT Logic
Is: A && (B || C || D || E || F) equivalent to?: A && (!B && !C && !D && !E && !F) as in: meta __FORGED_OUTLOOK_DOLLARS (__OUTLOOK_DOLLARS_MUA && ! __OUTLOOK_DOLLARS_MSGID && !__OUTLOOK_DOLLARS_OTHER && !__IMS_MSGID && !__UNUSABLE_MSGID) Thanks, Dan
