Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Re: What is the simplest 2nd Order Logical statement you can think of?
Posted:
Dec 26, 2011 6:55 AM


On Dec 26, 3:02 pm, Graham Cooper <grahamcoop...@gmail.com> wrote: > On Dec 26, 4:55 pm, Rupert <rupertmccal...@yahoo.com> wrote: > > > > > > > > > > > On Dec 25, 9:16 pm, Graham Cooper <grahamcoop...@gmail.com> wrote: > > > > Mine is: > > > > FORALL(f): FORALL(x): FORALL(y): f(x)=y <> f'(y)=x > > > > Does anyone in SCI.MATH actually know what 2OL is?? > > > > Some examples would help! > > > > Herc > > > A secondorder 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 secondorder language, doesn't mean > > you have to use the full power of secondorder logic to prove it. > > THERE IS NO 2ND ORDER LANGUAGE OF ANYTHING > THERE IS NO LANGUAGE OF ARITHMETIC OF THIS OR THAT > > They are FANTASY CONSTRUCTS INVENTED TO CIRCUMVENT THE FACT > CARDINALITY ONLY EXISTS IN SYSTEMS OF 2OL. > > Saying "IN THE 2ND ORDER THEORY OF ARITHMETIC..." > > is like saying "on the trinity planet of Zeus where integrals contain > between 478 terms". > > IT's STILL CALCULUS. > > FORALL(F):X>Y IS 2OL NO MATTER HOW YOU PROVE IT... MORON! > > Herc
Yet another classic sci.physics post.



