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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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    @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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Variables and constants serve distinct purposes in first order logic, and for that reason behave differently on a syntactic level.

The "scope" of a variable (in a sentence) is only in the formula in which it is defined, or the subformula where it is being quantified over. It serves not to distinguish a specific element, but to represent a set of possible elements. When a variable (in a formula) is interpreted in a structure, it can (and often should) be interpreted in different ways. In a sense, a variable only acquires identity in a particular interpretation inside a structure: in a theory, it is a placeholder. It is this understanding that allows us to quantify over variables.

Constants, however, are much more global. One can reuse a constant in many different formulas, so as to connect them. For example, if I wanted to talk about constant $c$ larger than all natural numbers, I could write $c > 0, c > S(0), \cdots$, an infinite family of formulas. To interpret these formulas in a structure, I would need a number greater than all natural numbers. If $c$ were merely a variable, I would have to write $\exists x ( x > 0), \exists x (x > S(0)), \cdots$: each formula would be disjoint from the other, and I interpret this collection of formulas in $\mathbb{N}$, even though it contains no element larger than all natural numbers. Moreover, certain theories (such as the theory of groups), make use of a distinguished element (such as the identity element). We would like to talk about that distinguished element, rather than treating it as a placeholder, which it is not. We would not want to consider all possible ways of interpreting elements in a group as the identity element: for a fixed group, there is a fixed identity element. As a result, one might say that constants have identity even in a theory, and indeed we introduce new constants to our language if we want to talk about larger models.

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    But doesn't a set of formulas with $n$ free variables and no constant symbols provide the same expressive power as a set of formulas with $n$ constant symbols and no free variables?2012-01-27
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A "pointed set" is a set together with a distinguished element $*$. The first-order language used to describe such a structure has one constant symbol "$*$" and the usual logical symbols. There are no additional axioms.

The first-order language for "real-closed field" has constant symbols $0, 1$, operation symbols (binary function symbols) $+, \times$ and binary relation symbol $\lt$. There are then some axioms describing properties that these must satisfy.

I don't know if these are what you are driving at, or completely off the mark...