Re: Aristotles logic decidable?

From: David Costa (hawaii1976lvr_at_hotmail.com)
Date: 03/25/05 

Date: Fri, 25 Mar 2005 15:16:29 +0100

Klaus wrotes:
>
> Your interpretation is a correct intensional semantic for Aristotelian
logic.
>

Klaus, unfortunately this is *not* true. My proposed intensional
semantics.............

A(x,y) : The set y is a subset of the set x
I(x,y) : There is a set z in U such that x and y are both a subset of z
E(x,y) : There is no set z in U such that x and y are both a subset of z
O(x,y) : The set y is not a subset of the set x

where each element of U is a subset of a certain 'universe' set, the
'universe' set is not an element of U and U is closed under the
set-complement operation

............. is *not* correct for Aristotelian logic!

Why? Let's denote the complement of a categorical term y by using the
expression c[y]. Aristotles requires that the following four rules hold:

A obversion: A(x, y) implies E(x, c[y])
I obversion: I(x,y) implies O(x,c[y])
E obversion: E(x,y) implies A(x, c[y])
O obversion: O(x,y) implies I(x, c[y])

Although A and I obversion indeed hold when using the above semantics it can
be shown that E and O obversion does *not* hold when using the above
semantics. So we need to change/refine/improve the above semantics in order
to be able to induce the other two obversions as well. Do you have any idea
how to do this?

David Costa

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