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Re: What is the simplest 2nd Order Logical statement you can think of?
Posted:
Dec 26, 2011 6:55 AM
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On Dec 26, 3:02 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Dec 26, 4:55 pm, Rupert <rupertmccal...@yahoo.com> wrote:
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> > On Dec 25, 9:16 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
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> > > Mine is:
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> > > FORALL(f): FORALL(x): FORALL(y): f(x)=y <-> f'(y)=x
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> > > Does anyone in SCI.MATH actually know what 2OL is??
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> > > Some examples would help!
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> > > Herc
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> > A second-order language is a language which contains quantifiable
> > variables ranging over subsets or relations or functions on the domain
> > of discourse. But just because you are trying to prove a formula which
> > is most naturally expressed in a second-order language, doesn't mean
> > you have to use the full power of second-order logic to prove it.
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> THERE IS NO 2ND ORDER LANGUAGE OF ANYTHING
> THERE IS NO LANGUAGE OF ARITHMETIC OF THIS OR THAT
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> They are FANTASY CONSTRUCTS INVENTED TO CIRCUMVENT THE FACT
> CARDINALITY ONLY EXISTS IN SYSTEMS OF 2OL.
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> Saying "IN THE 2ND ORDER THEORY OF ARITHMETIC..."
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> is like saying "on the trinity planet of Zeus where integrals contain
> between 4-78 terms".
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> IT's STILL CALCULUS.
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> FORALL(F):X->Y IS 2OL NO MATTER HOW YOU PROVE IT... MORON!
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> Herc
Yet another classic sci.physics post.
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