5
$\begingroup$

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. $\endgroup$
    – user956717
    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. $\endgroup$
    – user956717
    Sep 3, 2021 at 4:10
  • 1
    $\begingroup$ The examples really ground it now 👍 $\endgroup$
    – Ten O'Four
    Sep 3, 2021 at 5:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .