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As far as I can tell, there are two mainstream conventions for defining formulas in FOL. Sometimes predicates of arity $0$ are allowed, thus countenancing propositional variables for FOL. Other conventions disallow propositional variables.

I’ve only ever had experience with the first convention, as I think it is much more standard for philosophers. However, mathematicians seem to be adamant that FOL doesn’t contain propositional variables. This is expressed here: Is there no propositional letter in first order logic?.

Is this just for the sake of defining a simpler system, or is there an actual reason to remove propositional variables from FOL? Is there a reason to remove predicates other than $\in$ from set theories? It seems to weaken our expressive capacities, even though adding propositional variables doesn’t do any harm. Further, this convention flies in the face of the “bottom-up” approach for defining new/more expressive logics/theories out of simpler ones. Is there an actual mathematical reason for this convention, or is it just standard practice?

PW_246
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  • Propositional Statement $P="1+2=3"$ might have 0-arity $P()="1+2=3"$ , it might also be Unary $P(X)="1+2=3"$ or Binary $P(X,Y)="1+2=3"$ ETC , where these variables are unused hence redundant ! – Prem Jun 22 '23 at 18:19
  • I've never seen FOL presented as having also propositional variables. Instead, propositional calculus is used to understand the interaction of propositions which themselves can be first-order sentences (or formulas under a fixed assignment), considered in a given structure. – Asaf Karagila Jun 22 '23 at 19:10
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    You seem to be asking two different questions. One about FOL and one about set theory. Please ask one question at a time. – Asaf Karagila Jun 22 '23 at 19:12
  • @AsafKaragila The questions are directly related to each other. – PW_246 Jun 22 '23 at 19:15
  • @Prem not all of FOL is about math, and many things that are reasonable to state in natural languages have difficult first order translations, if any. “Beauty is beautiful” can’t be represented in FOL using a predicate $Bx$ for “x is beautiful”, but it can be represented as the proposition $B$. Since propositional logic is of a lower order than FOL, it stands to reason that it can be used. Again, this is the convention with which I’m familiar. – PW_246 Jun 22 '23 at 19:18
  • Not necessary to involve Math , that was only My Example. With your Example , I might convert that to $P=P()=P(X)=P(X,Y)=\text{"Beauty Is Beautiful"}$ where those variables are unused. Propositions are thus automatically available through Predicates ! – Prem Jun 22 '23 at 19:36
  • @AsafKaragila Proof-theoretic presentations of FOL will have propositional variables (and predicate variables of any arity). Indeed, it rarely makes sense to fix a first-order language $\mathcal{L}$ up front in the proof-theoretic context. – Z. A. K. Jun 22 '23 at 19:56
  • @prem that’s a category error. Beauty can’t be predicated as an object since it is a predicate itself. – PW_246 Jun 22 '23 at 20:13
  • I think you need to bone up on higher-order logic. Predicates as first-class objects have been around since Russell and Whitehead. – Rob Arthan Jun 22 '23 at 21:56
  • I’m not assuming a many-sorted approach in this light, even though I appreciate its existence in other contexts. – PW_246 Jun 22 '23 at 22:41
  • I think you did not get what I am trying to Convey : (A) "1+2 Is 3" & (B) "Beauty Is Beautiful" are Propositions , not made of Predicate variables & not made of Equality & IsBeautiful & ETC : These 2 Propositions are totally unrelated to 2 other Propositions (C) "1+2 Is Not 3" & (D) "Beauty Is Not Beautiful" [[ (A) & (B) are the Negations of (C) & (D) when we make them Predicates , not otherwise ]] – Prem Jun 23 '23 at 04:07

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It depends on what you are doing. For many uses of FOL, sentential variables would simply add clutter without providing any benefit.

For example, in model theory we have structures for FOLs. What would (for example) the first-order theory of groups look like with an addition sentential variable? What assertions about a group would the variable allow you to make? What would happen to the notion of elementary equivalence or elementary substructure? While it would not be hard to come up with answers, it would just be extra work with no payoff. It's a similar story with axiomatic set theory.

Typically definitions and proofs in logic use induction on the complexity of formulas. With sentential variables, you have an extra case to deal with, again with no applications.

In short, leaving sentential variables out of FOL doesn't actually "weaken our expressive capacities" (at least for the things one wants to express in model theory), and it does a (small) amount of harm, namely clutter.

Z.A.K. mentioned that in the proof-theory context, one does include sentential variables. I just checked the classic text Proof Theory by Schütte, and sure enough, his definition of what he calls CP (for "Classical Predicate" calculus) does include sentential variables, along with the sentential constant falsum ($\bot$). On the other hand, the text Proof Theory by Pohlers focusses on number theory, so he does not include them.

My knowledge of formal philosophy doesn't go much beyond Quine, so I don't know why philosophers would routinely include sentential variables. I am puzzled by your "Beauty is beautiful" example. What exactly do you gain by representing this with a single variable $B$? If you wanted to analyze the meaning of that statement, wouldn't you need something with more internal structure? How is having a sentential variable $B$ better than having an individual constant Beauty, a predicate symbol isBeautiful, and writing the formula isBeautiful(Beauty)?

But if it is indeed standard practice in philosophy, there must be good reasons for it.

Michael Weiss
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  • One example I've seen that uses propositional variables (a.k.a. nullary relations): If you add a variable $S_n$ for each $n \in \mathbb{N}$, and axioms $S_n \leftrightarrow \exists x_1 \cdots x_n \bigwedge_{i<j} x_i\ne x_j$, then that forms a conservative extension of FOL with no functions, relations and only equality. That extension has quantifier elimination, which can be used to show it is decidable. Therefore, FOL with no functions, relations and only equality is also decidable. (Or something like that, I might have messed up some details but the idea is sound as far as I recall.) – Daniel Schepler Jun 22 '23 at 21:56
  • @DanielSchepler: it is a standard trick to give "quantifier elimination" processes for certain theories that reduce any formula to an equivalent "quantifier-free" formula involving sentences like your $S_n$ that hold in the theories of interest. In the case in point, you get that FOL with only equality over an infinite universe is decidable. FOL with only equality is not decidable without some restrictions on the size of the universe: consider $\forall x, y(x = y)$. – Rob Arthan Jun 22 '23 at 22:13
  • Oh, right, I guess I was thinking that the question of whether a sentence is a tautology in bare FOL is a decidable problem. – Daniel Schepler Jun 22 '23 at 22:17
  • @DanielSchepler, RobArthan. Interesting example. The proof that comes to mind for me of that result is that the theory is obviously $\aleph_0$-categorical, hence complete, hence decidable. But it does suggest why, in a proof-theory context, one would want sv’s. – Michael Weiss Jun 22 '23 at 22:33
  • We’d have to equate the object “beauty” $b$ and the predicate “is beautiful” $Bx$ in our semantics in that case. – PW_246 Jun 22 '23 at 22:40
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    I can see how one might regard this as an issue, but not with the FOL syntax. If it’s illegitimate to apply the isBeautiful predicate to Beauty, then the natural language sentence “Beauty is beautiful” is already illegitimate. Moreover, just representing the sentence by B hides the issue, while writing it as isBeautiful(Beauty) gives you a handle for discussion. So I still don’t see this example as an argument for sv’s. – Michael Weiss Jun 23 '23 at 00:29
  • @MichaelWeiss: You wrote "But if it is indeed standard practice in philosophy, there must be good reasons for it.". This is a false assumption. There are many 'standard' things in philosophy with absolutely no good reason for them. This is also why ChatGPT can pretend well to be a philosopher but cannot pretend to be a mathematician. – user21820 Jun 30 '23 at 09:42