Date: Wed, 19 Dec 2001 21:55:48 -0600 From: Jonathan Lemon <jlemon@flugsvamp.com> To: "Gary W. Swearingen" <swear@blarg.net> Cc: Jonathan Lemon <jlemon@flugsvamp.com>, Brett Glass <brett@lariat.org>, chat@FreeBSD.ORG Subject: Re: GPL nonsense: time to stop Message-ID: <20011219215548.D76354@prism.flugsvamp.com> In-Reply-To: <0en10ey5jo.10e@localhost.localdomain> References: <local.mail.freebsd-chat/Pine.LNX.4.43.0112181134500.21473-100000@pilchuck.reedmedia.net> <local.mail.freebsd-chat/20011218110645.A2061@tisys.org> <200112182010.fBIKA9739621@prism.flugsvamp.com> <4.3.2.7.2.20011218180720.00d6e520@localhost> <20011219091631.Q377@prism.flugsvamp.com> <0en10ey5jo.10e@localhost.localdomain>
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On Wed, Dec 19, 2001 at 06:43:55PM -0800, Gary W. Swearingen wrote:
> Jonathan Lemon <jlemon@flugsvamp.com> writes:
>
> > E.g.: kernel + (N)"options XXX" = non-GPL'd kernel.
> > kernel + (N)"options XXX" + "options EXT2FS" = GPL'd kernel.
> >
> >
> > We are in agreement here, right?
>
> (Can you get your mail reader to not use tabs? I can't get my mail
> reader to quote them correctly and I doubt if many can. Thanks.)
I guess I'll tell 'vi' to use spaces instead of tabs.
> We are not in agreement. When you have the second equation, you must
> also have:
> kernel = GPL'd kernel
> and
> (N)"options XXX" = GPL'd (N)"options XXX"
> and therefor you will have
> kernel + (N)"options XXX" = GPL'd kernel.
>
> You can't distribute a whole (a GPL term) under the GPL without
> distributing its parts under the GPL. It seems like basic logic to
> me. Nothing to do with the GPL. Please explain carefully if you
> disagree.
Well, I agree with the above 4 sentences, but not the prior argument.
I imagine that this point is where you (and Brett, probably) lose most
of your readers.
The concept (to me anyway) is simple:
1. There exists a GPL encumbered source. Call this A.
2. I have some pure BSD kernel sources. Call this B.
3. Make a copy of the BSD code.
cp -R /usr/src /usr/src2. Call this C.
3'. (optional) Move copy C far away (into another universe)
4. Add GPL code A to BSD code C.
Now, by my logic, and my reading of the GPL, yes, the resulting
product which contains 'A' and 'C' is now under the GPL, and so
copy C automatically falls under the GPL too.
BUT!
Copy B is _NOT_ under the GPL.
This is where we appear to differ. I reject the notion that because
it is possible in some universe to combine BSD + GPL'd code, that it
automatically forces all other copies of the BSD code to fall under
the GPL.
--
Jonathan
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