6
$\begingroup$

I was reading through the Wikipedia page for first-order logic and all made sense to me, though I got confused at this quote "In other words, the symbol $f$ is associated with the function $I(f)$ which, in this interpretation, is addition." and am unsure of what the I in the last sentence means. Is it a function that maps the function symbol to a function?
Is there a name for it?

Thank you friends

$\endgroup$
1
  • 2
    $\begingroup$ They define $I$ to be an interpretation (a model) which takes anything from $D^n$ to $D$ which is the domain of discourse over an n-arty input. Takes anything from constants, predicates, and functions, and puts them into the model with their own assignments in the universe (or domain) of the model. I personally don't like the way that they're teaching it, but that's the answer for your question. In this case, "I" is supposed to be understood meta-semantics as addition. $\endgroup$
    – Math3147
    Sep 3, 2021 at 3:03

1 Answer 1

4
$\begingroup$

It is defining an interpretation, and I personally don't like that way. Allow me to explain it.
Definition:
An interpretation is a pair ($\mathbf{M}$, $j$), with $j$ being the assignment and $\mathbf{M}$ being the model. j takes elements like functions, relations, and constants from a language $\mathcal{L}$ and "maps" them into elements in the universe of the interpretation (or domain of), and they're usually denoted as: $c^{\mathbf{M}}$, $F^{\mathbf{M}}$, $R^{\mathbf{M}}$ (unimaginatively). I explained in the comment what they exactly mean by turning it into a function that takes the tuples and maps them into elements in the domain of the model. An interpretation may also be called a structure.
Examples:
($i$): The language of group theory has $e$ as a constant which will be assigned to $e^{\mathbf{G}}$, and $\circ$ which is some arbitrary binary operation (predicate constant), which will be assigned as $\circ^{\mathbf{G}}$, and at the end we have the universe (domain) $\mathcal{G}$, and our interpretation looks like this: $\mathbf{G}=(\mathcal{G},\, e^{\mathbf{G}},\,\circ^{\mathbf{G}})$. Now, how is this useful? Fill in the gaps! ($\mathbb{R}$, $\cdot$, $1$) for example. Now you've multiplicative group.
($ii$): Take the language of graphs which has $R$ as its predicate, and so will will have $R^{\mathbf{Gr}}$. Define a domain $\mathcal{R}$, and you're done. The interpretation $\mathbf{Gr}$ = ($\mathcal{R}$, $R^{\mathbf{Gr}}$). You can replace $R^{\mathbf{Gr}}$ with the Cartesian relation for example, and leave the reals as the domain: ($\mathbb{R}$, $\times$), and you've a Cartesian graph.


The comment:

They define $I$ to be an interpretation (a model) which takes anything from $D^n$ to $D$ which is the domain of discourse over an $n$-arty input. Takes anything from constants, predicates, and functions, and puts them into the model with their own assignments in the universe (or domain) of the model. I personally don't like the way that they're teaching it, but that's the answer for your question. In this case, "$I$" is supposed to be understood, in a meta-semantics sense, as addition.

Note that loosely speaking.. this is not exactly correct, and plays around with some words in a misleading way. That's why I suggest seeing the explanation that I wrote.
Lastly an interpretation is of a language, and a model is of a theory. Switching them around is "fine" because they're the "same structure" but it is not recommended, and should only be left for purposes of intuition.

$\endgroup$
3
  • $\begingroup$ It might be an idea to give a concrete example of a basic interpretation to let the OP see what's going on? The wiki explanation has manged to turn a really basic idea into something confusing. For instance, $\begin{array}{lll}\Im:&&\\& \text{Domain:} & \{0, 1, 4\}\\& \text{F:} & \{1, 4 \}\\& \text{G:}& \{0, 1\}\\ & \text{H:} & \emptyset \\ & \text{a:} & 0\\ & \text{b:} & 1 \\& \text{c:} & 4\\\end{array}$ $\endgroup$
    – Ten O'Four
    Sep 3, 2021 at 4:03
  • 1
    $\begingroup$ @TenO'Four Good idea. Will edit. $\endgroup$
    – Math3147
    Sep 3, 2021 at 4:10
  • 1
    $\begingroup$ The examples really ground it now 👍 $\endgroup$
    – Ten O'Four
    Sep 3, 2021 at 5:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.