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Math system

  1. Aug 18, 2003 #1
    A mathematical system S is defined as S={E,O,A}. A is the set of axioms describing the system. Is the definition of E considered an axiom? For example, if I want E={a,b}, then in the set A, do I write A={...,E={a,b},...}?
    Also, is the definition of O an axiom? Say O={~,V} and then I define ~ to be a function from E to E such that ~={(a,b),(b,a)}. V is a binary operation on E such that V={((a,b),a),((b,a),a),((a,a),a),((b,b),b)}. Then can the property xVy=yVx be derived as a theorem based on the definition of V or must it be considered an axiom? Is the property ~(~(x))=x a theorem based on the properties of ~, or is it better to consider this an axiom? Which is more proper: to establish axioms describing the properties of the operation V and function ~ and from these properties determine their exact definition or to define the functions exactly and derive their properties (for this situation, the fact the V and ~ are of the exact form above is more important than the fact that they have the properties above)?
    I hope my question is clear enough. I'm really not so sure what I'm asking myself.
  2. jcsd
  3. Aug 18, 2003 #2


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    I think I can see a glimmer of what you are talking about. It would help me greatly if you would define all the terms. What is E? What is O? Is O the set of operations? E the set of Theorems? Please provide more fundamental details.
  4. Aug 18, 2003 #3
    This is from my abstract algebra book:
    E is a nonempty set of elements.
    O is a set of operations and relations on E.
    A is a set of axioms concerning the elements of O and E.

    It may help to know that I’m attempting to recreate the theorems from my logic book using the math I learned from my algebra book. The logic book is very simplistic – it doesn’t even mention the concept of theorems and axioms. It just lists “rules” which I must accept as true. I’ve been looking up the axioms of logic for several days. One source listed four. One source listed 19 for prepositional logic alone. This source listed such axioms as (A=B)=(B=A). These are very fundamental axioms. I’m willing to assume the normal axioms which I am probably already assuming by writing the logic in algebraic form. I’m most interested in which axioms are fundamental to logic. The function ~ is the negation function. The operation V is the disjunction operation. If I define V and ~ explicitly (ie, by say ~={…} and V={…}), I can determine associative, commutative, and distributive properties by exhausting all possibilities. Then I can use the properties of V and along with definitions of &, =>, <=> in terms of ~ and V to discover properties of these operations (ie, equivalences). Then when it comes to inference rules such as P, P=>Q infer Q, I can determine if there is an operation R such that xRy=(x&(x=>y))=>y=a for all x and all y (x,y in E). But although I’m confident in a great deal of what I’ve written here, I’m afraid that I’m making too many implicit assumptions. I want to make these assumptions explicit and to do this I want to make sure I’ve established all necessary axioms.
  5. Aug 19, 2003 #4


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    Okay, I get what you are talking about. No, the definition of A and O would not be axioms. Axioms are statements ABOUT the elements of A and O. It might well happen that members of A and O are NOT defined in any specific sense- the are simply listed as "labels" with all of there properties given by the axioms.
  6. Aug 19, 2003 #5
    A={} (so far)

    So to prove ~(~(a))=a, will the following suffice?
    Say a=t. Then ~(a)=f and ~(f)=t so that ~(~(a))=t=a.
    Say a=f. Then ~(a)=t and ~(t)=f so that ~(~(a))=f=a.
    Since a is either t or f, the property is proven.

    I define V by the operation table:
    V t f
    t t t
    f t f

    To prove that aVa=a, will it suffice to show that tVt=t and fVf=f? I initially included this (and other properties) as an axiom. I can also show that aVb=bVa by the operation table through only three instances: tVt=tVt, tVf=fVt, fVf=fVf.
    Again, finding the properties this way saves me an axiom. I like this axiom-avoiding approach. It seems this way that I could develop a great portion of logic without a single axiom. The distribution of V over V can be established by examining 8 instances. Eventually, I can use these properties to prove others and discard this instance by instance examination approach to proof. Is it legitimate to prove properties by examining every possible instance?
  7. Aug 19, 2003 #6


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    Only if you can prove, by some means, that you have examined every possible instance.

    More formally, you need a theorem (or axiom) of the form that:

    ~(a = t) => a = f

    Which could be derived, say, from the axiom that, for all a, a in {t, f}
    Last edited: Aug 19, 2003
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