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I was thumbing through my bro's logic book and got caught far longer than I expected. I got to the point where what they call first-order logic is introduced, but I don't understand why they define constant symbols. It looks to me as if everything that can be done with them could also be done with free variables. Looks a bit pointless to me. Usually I'd ask my bro but he is on the road for a week, so I hope some of you might help me. Thx!


To ask more specific: What can I do or express with a set of sentences with $n$ constant symbols that I can't express with a set of formulas with $n$ free variables but no constants?

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    Constants are not absolutely needed. They do make things more pleasant in many situations, including the construction of models, and in applications of the Compactness Theorem. In various formal theories, such as field theory, or Peano Arithmetic, it is useful to have constant symbols. They make the axioms more natural, more understandable.2012-01-22
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    So it is nothing more than a convenience like "p->q" is for "not p or q"?2012-01-22
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    @André: I disagree. Without constant symbols it is impossible to write down e.g. the first-order theory of groups with an element of infinite order.2012-01-22
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    @Zhen Lin: Certainly true, unless one cheats by introducing additional function symbols.2012-01-22
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    Why? Shouldn't something like: 1. for all x,y: f(x,f(y,z))=f(f(x,y),z) 2. for all x: f(x, e) = x 3. for all x: f(e, x) = x 4. for all x exists y: f(x, y) = e 5. (one for each n>=1) not f^n(c,c) = e 6. not c = e do the trick (c and e are free variables)?2012-01-22
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    @alexis: The usual definition of truth for formulas with free occurrences of variables is to interpret these variables to be *universally quantified*. (Some presentations don't bother to assign truth value to formulas with free variables.)2012-01-22
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    @André: In the book an interpretation J=(S,f) is defined as a structure S and a (variable-)assignment f. When "satisfies" is defined, an unquantified variable x gets the value f(x) which still seems to me to be the same as c^S for a constant c. I haven't read anything yet towards free variables being universally quantified. Please explain further.2012-01-22
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    @alexis: how would you axiomatize, in usual first-order logic, but without constant symbols, a group which has an element that is not of finite order?2012-01-22
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    @CarlMummert As I described in my comment above. Sorry that layout got screwed.2012-01-22
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    @alexis: the set of formulas that you have above would be true in a group that has no elements of infinite order, but does have for each $n$ an element of order $n$. Well, if you tweak them to make them syntactically correct (what is $f^n(c,c)$ when $f\colon G^2\to G$?) and to make them consistent.2012-01-22
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    @Carl Mummert: about that group, what if you let $\Phi_n$ be the axiom: " $\lnot (\forall g)(g=e \vee g^2 = e \vee \cdots \vee g^n = e)$", where $x=e$ is an abbreviation for the formula "$(\forall h)(hx=h)$" and add the axiomscheme $(\Phi_n)_{n\in\mathbb N}$ ?2012-01-22
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    @Myself: that scheme says there is no bound on the orders but not that there is an element of infinite order. The product of all finite cyclic groups has no bound on orders of elements but no element of infinite order.2012-01-23
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    @CarlMummert: I hope that the axioms 1-4 and 6 are correct/clear. But yes, I made a mistake with axiom 5. Let me try again, I hope I get the TeX markup right: $\{ \neg f(c, f(c, f(c, ... f(c,c))...)) = e | \forall n >= 1\ \mbox{ applications of }\ f \}$. Together with the other axioms shouldn't this cause the variable c to be mapped to an element of infinite order?2012-01-24

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