Re: Axioms of Boolean Algebra got from Concept Algebra


Jan Burse wrote:
I didn't say that the structures are the same.
But they have the same signature:
A set S:
^: S x S -> S
v: S x S -> S
~: S -> S
1: S
0: S

I replied
And this signature is bloated and redundant.
Of the six things you have specified here, THE ONLY TWO
you actually NEED are
^ and ~ . Under FOL *with* equality,
the following 3 axioms define boolean algebra:

(x^y)^z = x^(y^z)
xy = yx
~( ~(x^y) ^ ~(x^~y) ) = x
Bertie Reed asked,
Hmm what is xy?

xy was supposed to be x^y, and yx was supposed to be y^x.
Conjunction is like multiplication, which is often represented
by juxtaposition.

Your axioms, and therefore your signature,
don't need v, O, or I , because they are all definable.

This signature is also good because you can do the boolean
ring axioms in it as well; you just have to use a different
definition for the ring's + from the one you use for the algebra's v.
But this axiomatization shows that the "underlying" signature and
axioms are in fact not merely isomorphic BUT IDENTICAL.

Of course, the hard part is figuring out whether you can or cannot
repeat this feat in FOL withOUT identity.

.

Relevant Pages

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  • Re: Torkel Franzen on truth
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  • Re: Axioms of Boolean Algebra got from Concept Algebra
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