Variables as symbols or objects and interpretation

Question

I've really only started learning a little bit of logic in my education and I'm confused by the treatment of variables sometimes. In standard mathematics nearly all instances of a variable $x$ refer to it's value, of course '$x$' itself it simply a symbol which can be instantiated anywhere on the page.

However in logic we treat every statment with occurences of $x$ as a string and introduce an 'interpretation' to read it as a statement about objects.

What is confusing me is that we use $x+1=2$ and have $x$ in this open formula and here under our intepretations we take $x$ as an object and a statement for which we can quantify or just discuss the truth of, but equally the statement itself is just a string of symbols.

But then we use $x$ as a symbol in $s(x)$ and our notation does not change, $s$ is the assignment function on a symbol '$x$', I asked about this before but my knowledge surrounding the difference between 'symbol' and 'object' was insufficient to grasp the answers, and it was unfair to give excessive follow-up questions to the answerer.

Whats I'm struggling with is if an open formula is just a set of symbols and 'x' is just a set of symbols then $s(x)$ is given the denotation of a number (the assignment function on the symbol $x$) but it just like $x+1$, $s(x)$ is just a string of symbols. This may be related to the distinction between metalanguage and object language, but I'm not sure. However, I find the $s(x)$ notation problematic, yes $x$ is a symbol but it refers to the object in the same way '$5$' does.

Why can we use the same notation to denote 'symbol' and 'object' and what effect does 'interpretation have on this'?

Perhaps we can solve this by stating that in the 'metalanguage' a statement like: $s(x)$ can be used, but in the 'object' language we would find $x+1=2$ or $f(x)$ where $f$ maps in the objects.

This may feel like a duplicate but I feel I have widened the scope of it enough.

See answers to your previous post Meaning of 'argument of a function' for what the argument of a function is as well as a brief introduction to the general concept of function, i.e. functions that are not "algebraic". — Mauro ALLEGRANZA, Oct 27 '22 at 14:20
See also your previous post 'x is a....' statements regarding variables as placeholders. — Mauro ALLEGRANZA, Oct 27 '22 at 14:21
As well as your post 'Equality is a relationship between two expressions' — Mauro ALLEGRANZA, Oct 27 '22 at 14:22
Am I reading it correctly that your question is basically "If our metalanguage itself consists of symbols like $s(...)$, don't we need a metalanguage for the metalanguage and so forth ad infinitum"? — Natalie Clarius, Oct 27 '22 at 14:54
@lemontree Is $x+1=2$ in my object language and $s(x)$ in my metalanguage? In which case writing '$x$' has different denotions in each? In that case, it would appear so — , Oct 27 '22 at 15:25
What math courses have you taken? Any proof-based courses in, say, real analysis or group theory? If not, you may be getting ahead of yourself by studying logic to this depth. — Dan Christensen, Oct 27 '22 at 15:58
The meta language can contain symbols of the object language, after all the meta language talks about the object language (that's what "meta" means). Just think of a grammar book about Chinese written in English. Chinese is the object language and English the meta language, but of course the English text will contain Chinese symbols quoted in it. It's the same with $s(x)$. — Natalie Clarius, Oct 27 '22 at 21:56
@lemontree thanks a lot, that clears it up a good bit for me, — , Oct 28 '22 at 09:20

score 0 · Accepted Answer · answered Oct 28 '22 at 12:10

[turning my comment into an answer for future readers]

Your main confusion seems to be how it can be that $x$ is a symbol of the object language but $s(x) = a$, containing $x$, is an expression of the meta language. The meta language can contain symbols of the object language, after all the meta language talks about the object language (that's what "meta" means). Just think of a grammar book about Chinese written in English. The object language is Chinese and the meta language is English, but of course the English text will contain Chinese symbols quoted in it. It's exactly the same with $s(x)$, except we don't typically put $x$ in quotes or italics.

score -1 · Answer 2 · answered Oct 27 '22 at 23:43

in first order predicate logic a sentence can have a truth value only if all variables are bound.

That is, a variable is in the scope of a quantifier.

The scope of a quantifier is the first formula immediately following the quantifier.

a variable has free occurrence if it is not in the scope of a quantifier.

otherwise it's occurrence is bound.

examples: 1. $(\forall x)(Px \land Py\implies Qx)$. here $x$ has 3 bound occurrences and $y$ one free occurrence.

$Px,\;Qx$ are predicates (functions, formulae, etc).

2. $(\forall x)(\forall y)(Px\land Py\implies Qx) \lor Sx$ here $y$ is bound in it's two occurrences. $x$ is bound in it's 3 occurrences and free in the fourth occurrence because it is not in the scope of it's quantifier.

Your formula $x+1=2$ has no truth value since the occurrence of $x$ is free.

If however the formula reads: $(\exists x)(x+1=2)$

a truth value can be assigned. an interpretation in the domain of natural numbers and '+' as regular addition yields true. $(\exists x)(x+1=2)$ namely, $x=1$.

To summarize a sentence can have a truth value only if all the variables are bound.

Variables as symbols or objects and interpretation

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