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In first-order arithmetic, you can't quantify over sets of numbers. However, you can include sets as free variables. I don't think this is just a meta-linguistic thing, as I've read papers about Peano Arithmetic with first-order sentences containing predicate variables which definitely seem to be used for more than just fill-in-the-blank-with-a-formula. For example, you can prove within Peano Arithmetic that strong induction works, and this paper uses free predicate variables to create a first-order formula stating that a relation is well-ordered.

However, if you really start using free set variables in interesting ways, it seems like you can contradict well known theorems. For example, I could write the formula $(\forall X, n,i)(\Phi(X,n,i)\Leftrightarrow [i=0\land\phi(X,n)\lor i\ge0\land(\exists x)(\forall y)\Phi(\pi(X,x,y),n,i-1)])\Rightarrow\Phi(0,n_0,i_0)$

where

  • $\pi$ is an injection from triples to numbers
  • $\phi$ is a binary predicate like "if the $n^\text{th}$ quantifier-free formula is true given variables $X$"
  • $\Phi$ is a free ternary predicate variable
  • $n_0$ is a free number variable indicating the index of the formula to use
  • $i_0$ is a free number variable indicating the number of existential/universal quantifiers

Assuming I set that up right and assuming you can use free predicate variables like that, this is a formula for the numbers of true arithmetic sentences - in first-order arithmetic. Sure, there's an implicit $(\forall\Phi\in\mathcal{P}(\mathbb{N}))$ in the formula, but according to questions like this, as long as that quantification isn't explicit, it still counts as a first-order formula. But that all contradicts Tarski's Undefinability Theorem, so it can't be right.

It also relates back to questions like this, about why the definitions of addition and multiplication need to be included in the signature and axioms of Peano arithmetic when you could just start any sentence like $(\forall x,z)([\phi_+(x,0,z)\Leftrightarrow x=z]\land(\forall y,z)[\phi_+(x,Sy,Sz)\Leftrightarrow\phi_+(x,y,z)])\Rightarrow...$

And if you can do recursion like that, it kind of makes the whole setup of Godel's $\beta$ function for encoding sequences unnecessary, which I assume it isn't. I've been learning about proof theory for a while and this one basic concept for some reason still confuses me. I keep going back and forth between thinking free set variables add immense power and are the key to everything and that they're just a fancy notation. Is there something I'm missing here? Is the first paper I referenced using "fresh" predicates incorrectly? Does it have to do with the difference between a formula and a sentence? Any explanation would be great. Thanks.

yosho27
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1 Answers1

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Long story short, it's just a meta-linguistic thing. $\mathcal{L}$-formulas for a first-order language $\mathcal{L}$ cannot contain predicate variables, as you can verify by checking their definition in any logic textbook.

Now, some specific proof systems may allow you to use predicate variables (to be used schematically), but the usual semantics of first-order logic does not deal with these at all. This means that such schematic formulas do not "mean" anything under the standard semantic - although their instances do, once you substitute some concrete predicate definable in your language in place of the predicate variable (instantiate the schema with a genuine predicate).

Schematic reasoning with predicate variables in proof systems that support them means precisely that: if you prove a schema containing a unary predicate variable $P$, then you can obtain a proof of any formula obtained by putting an actual (unary) predicate $\varphi$ in place of $P$. But this is a thing that applies to proofs - not to definability, not to satisfaction, not to any other semantic or model-theoretic notion.

Try reading the article you linked more carefully. It does not "use free predicate variables to create a first-order formula stating that a relation is well-ordered (sic)": in fact, it states precisely that there is no formula stating that a relation constitutes a well-ordering (in a non-standard model even $<$ fails to be that). Moreover, if you check carefully, you will see that everything that the paper does falls comfortably under the "fill-in-the-blank-with-a-formula" umbrella.

Z. A. K.
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