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Trying to figure out the logic in which the following formula is expressible: $\forall i\in N: (x_i > y_i)$, which is equivalent to the "infinite" conjunction $\bigwedge_{i\in N} (x_i > y_i)$.

Now a 1st order logic allows arbitrary number of variables $\{x_i,y_i\mid i\in N\}$, but only a finite number of atomic formulas can be composed. (Here $x_i > y_i$ is an atomic formula based on binary predicate "$>$", and so the above formula is a composition of infinite number of atomic formulas.) Thus in the form written, the above doesn't seem to be a formula of 1st-order logic. Also, it's not clear how to rewrite this formula in 1st-order logic if indeed it belongs there.

Next in 2nd-order logic, that allows quantification over predicates (as well as functions), again it is not clear whether the above can be written as formula in 2nd-order logic.

Any insights? Thanks.

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    $\begingroup$ Wouldn't you usually write $x(i)$ and $y(i)$ as functions rather than $x_1, x_2, \dots, y_1, y_2,\dots$ as infinite collections of variables? Then in your expression $x_k$ and $y_k$ would just be alternate notation for $x(k)$ and $y(k)$. $\endgroup$
    – none
    Commented Dec 17, 2015 at 8:14
  • $\begingroup$ You can have infinitely many axioms for a first-order theory. So a list of axioms like $x_0 > y_0$, $x_1 > y_1$, ... is perfectly acceptable. Of course the theory might not be finitely axiomatizable, but I don't see that requirement in your question. $\endgroup$ Commented Dec 17, 2015 at 20:52
  • $\begingroup$ @ToddTrimble True; however, I understood the OP as asking for a logic which could express this as a single sentence. $\endgroup$ Commented Dec 28, 2015 at 1:38
  • $\begingroup$ @NoahSchweber Well, it's likely you're right, but (1) OP has never said so, and ought to; (2) I wonder what the point would be for that (why does OP care, what's the motivation, etc.). $\endgroup$ Commented Dec 28, 2015 at 1:51
  • $\begingroup$ @ToddTrimble That's a very fair point. (Off the top of my head, one reason to care is in case you want this expression to be part of a larger one - e.g. if you want to say something like, "$\forall \overline{x}\exists \overline{y}([x_i>y_i]\wedge ...)$", at which point it's not enough to just express it as a collection of sentences. Of course, you're right that the OP should say this.) $\endgroup$ Commented Dec 28, 2015 at 1:55

2 Answers 2

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You're looking for infinitary logic - probably $\mathcal{L}_{\omega_1\omega_1}$ if you're using countably many distinct free varables. See https://en.wikipedia.org/wiki/Infinitary_logic, as well as https://www.math.wisc.edu/~keisler/kk2.pdf (the latter is a biographical paper about Barwise, but it is also a good source on infinitary logic in general).


Basically, here's how it works: we fix cardinals $\kappa$ and $\lambda$, with $\kappa\ge\lambda$. Then $\mathcal{L}_{\kappa\lambda}$ is (informally) the set of formulas generated by starting with first-order logic and closing under

  • conjunctions and disjunctions over sets of $<\kappa$-many formulas, and

  • quantification over $<\lambda$-many variables.

We can also define proper class sized infinitary logics as $$\mathcal{L}_{\infty\lambda}=\bigcup_{\kappa\in Card} \mathcal{L}_{\kappa\lambda}\quad\mbox{and}\quad \mathcal{L}_{\infty\infty}=\bigcup_{\lambda\in Card}\mathcal{L}_{\infty\lambda}.$$

Of course, the precise definitions are a bit technical, but this is the key idea.

Of special interest is $\mathcal{L}_{\infty\omega}$, which is roughly the "logic of back-and-forth arguments": two structures are $\mathcal{L}_{\infty\omega}$-equivalent iff player II has a winning strategy in the Ehrenfeucht-Fraisse game of length $\omega$ - or, equivalently, if there is some forcing extension of the universe in which they become isomorphic. (The first fact is due to Karp; the second is folklore, but I believe first observed by Barwise.)


Note that we could similarly define infinitary second-order logic, etc. I know less about such things, though.

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  • $\begingroup$ Thanks, that's quite helpful, and I will look into suggested reference. $\endgroup$
    – user84230
    Commented Dec 17, 2015 at 4:00
  • $\begingroup$ Wondering if there is a theory for quantifier elimination on infinitary 1st order logic, as there is one finitary 1st order logic over real closed field due to Tarski. $\endgroup$
    – user84230
    Commented Dec 17, 2015 at 4:02
  • $\begingroup$ @user84230: that's an interesting question! Note that infinitary logic as usually described is not decidable, or is it even clear how to finitely represent a formula without further restrictions. $\endgroup$
    – cody
    Commented Dec 17, 2015 at 18:18
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Using infinitary logic to express your statement is somewhat overkill. The usual formulation of 2nd order logic only allows subsets $X\subseteq\mathbb{N}$ to be quantified over (which can be seen as functions $X\rightarrow {\bf 2}$) rather than arbitrary functions, which would easily allow you to define two sequences $x,y:\mathbb{N}\rightarrow\mathbb{R}$ and say $$ \forall i\in\mathbb{N},\ x(i)>y(i)$$ as mentioned in the comments.

However, the usual formulation of 2nd order logic is quite strong, and functions $\mathbb{N}\rightarrow\mathbb{N}$ and sequences $\mathbb{N}\rightarrow\mathbb{R}$ can easily be coded in it! I'll try to sketch the argument below, but first note that because of all this expressive power, there's no hope of quantifier elimination or decidability in this logic.

  1. Coding rationals: first note that by usual arithmetic tricks that I won't get into here, it's easy to code pairs in $\mathbb{N}\times\mathbb{N}$, triples etc. even finite sequences using just natural numbers. By suitably defining operations and equality, we can code $\mathbb{Z}$ as a subset of $\mathbb{N}\times\mathbb{N}$, and $\mathbb{Q}\subset \mathbb{N}\times \mathbb{Z}$, so $\mathbb{Q}$ is coded as a subset of triples, themselves coded as natural numbers.

  2. Building real numbers. So by 1. to build a function $\mathbb{N}\rightarrow \mathbb{Q}$, it suffices to build functions $\mathbb{N}\rightarrow \mathbb{N}$! But this can be done by describing a function $f$ as a subset $f\subseteq\mathbb{N}\times\mathbb{N}$ with certain properties, as is done in set theory. As before, pairs can be coded as individual numbers, so you can describe these subsets in 2nd order arithmetic.

  3. Now that real numbers are described as sequences of natural numbers (elements of $\mathbb{N}\rightarrow\mathbb{N}$, it's pretty easy to build sequences of real numbers, by noticing that a function $f:\mathbb{N}\rightarrow \mathbb{N}^{\mathbb{N}}$ is just a function $\hat{f}:\mathbb{N}\rightarrow\mathbb{N}\times\mathbb{N}$ (known as the reverse curryfication of $f$, or as Schönfinkel's trick).


To answer your follow-up question, I don't know of any quantifier elimination for ${\cal L}_{\omega_1\omega_1}$, but keep in mind that even quantifier-free formulas are undecidable in that logic, because of it's infinitary nature (which makes it not very interesting, in my opinion).

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  • $\begingroup$ Thanks for your answer. Even 1st order logic allows functions, and so sequences. On the other hand, the quantification over index i is just over a variable, rather a set/predicate/function. So in what way 2nd order logic helping express the sentence? $\endgroup$
    – user84230
    Commented Dec 17, 2015 at 22:39
  • $\begingroup$ Also about quantifer elimination, is there any result for L_{\omega_1,\omega}? $\endgroup$
    – user84230
    Commented Dec 17, 2015 at 22:46
  • $\begingroup$ First order logic allows for fixed functions, so if your sequences $x_i$ and $y_i$ are fixed (you don't want to quantify over them) then you can do it in multi-sorted first order logic with 2 additional function symbols. But if you want to quantify over the sequences, you need something extra. Of course set theory can be specified in first order logic with enough symbols, so it really depends on what you allow. I was taking as basis the 1st order theory of Real Closed Fields vs the 2nd order theory of Arithmetic also called analysis precisely because you can encode sequences of reals. $\endgroup$
    – cody
    Commented Dec 17, 2015 at 22:48
  • $\begingroup$ So the sentence is \forall x,y,i: x(i) > y(i), which is a sentence of 2nd order logic, for qunatification over function. While seq of reals can itself be coded in 2nd order theory of arithmetic, or 1st order theory of real closed field. Thanks. $\endgroup$
    – user84230
    Commented Dec 17, 2015 at 23:41
  • $\begingroup$ Is there any result on QE for such 2nd order logic of real closed field, or 2nd order logic of arithmetic? What is the richest class for which QE result exists? Thanks. $\endgroup$
    – user84230
    Commented Dec 17, 2015 at 23:45

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