syntactic implication without appealing to completeness

Question

At my university, the course in logic and set theory develops propositional logic while assuming a notion of a set, which I find dissatisfying because we only the naive notion of a set at that point, which has serious problems. So I'm trying to write my own notes about it, and I can state the axioms of predicate logic without a notion of a set. The problem is I'm finding it hard to make many syntactic deductions from the axioms because I can't appeal to valuations and the completeness theorem. To be clear, the axioms I'm using are

For any formulae $p$ and $q$, $p\implies(q\implies p)$.
For any formulae $p$, $q$ and $r$, $(p\implies(q\implies r))\implies((p\implies q)\implies(p\implies r))$.
For any formula $p$, $\neg(\neg p))\implies p$.

There are also the other axioms that deal with variables, but they aren't relevant for now. The deduction rule I'm using is modus ponens. There's also generalisation, but it isn't relevant for what I'm trying to do at the moment. So I've managed to prove that $\{p\implies q,q\implies r\}\vdash p\implies r$. You might say that I'm using a notion of a set here, but I'm really not; the notation is just short-hand for saying "I can write a proof of $p\implies r$ from the axioms along with the formulae $p\implies q$ and $q\implies r$". I've also proven that $\vdash p\implies p$ and a special case of the deduction theorem: $\vdash p\implies q$ if and only if $\{p\}\vdash q$.

Now I want to prove the normal properties one would expect the logical system to have, such as $\{p\}\vdash \neg(\neg p)$, and $\{a\implies b\}\vdash\{(\neg b)\implies(\neg a)\}$, and $\{a,b\}\vdash a\land b$, but I am stuck. As I mentioned in the beginning, I can't appeal to valuations since I can't define what a function is, so the completeness theorem is out, and I just need to prove syntactic implication directly. I would appreciate any help.

I suppose I should be explicit about my definitions of $\neg$ and $\land$ in case there are multiple conventions. I'm defining $\neg a$ to be $a\implies\bot$ and $a\land b$ to be $\neg(a\implies(\neg b))$.

On the specific issue, see Set theoretic concepts in first order logic as well as When does the set enter set theory?. — Mauro ALLEGRANZA, Dec 19 '22 at 14:48
Regarding $⊢$, you do not need valuation: you have to provide derivations. — Mauro ALLEGRANZA, Dec 19 '22 at 14:50
@MauroALLEGRANZA my point about valuations is that ordinarily one would be able to prove semantic implication and then appeal to the completeness theorem to deduce syntactic implication, but I cannot do that since I can't define a valuation, so I need to prove syntactic implication directly — jeff honky, Dec 19 '22 at 14:52
"to prove syntactic implication directly" is exactly to provide a derivation (a sequence of formulas) starting with premises and ending with conclusion. — Mauro ALLEGRANZA, Dec 19 '22 at 14:58
@MauroALLEGRANZA I know. My point is that one would ordinarily be able to avoid providing an explicit derivation by proving semantic implication and then appealing to the completeness theorem, and I cannot do that. Therefore I am attempting to provide an explicit derivation, but I am unable to do it — jeff honky, Dec 19 '22 at 14:59
@MauroALLEGRANZA I only mentioned it in my question so that nobody tells me to use a valuation and appeal to the completeness theorem — jeff honky, Dec 19 '22 at 15:00
Two ways: either syntactical implication ($\vdash$) or semantical implication ($\vDash$). It is your choice... — Mauro ALLEGRANZA, Dec 19 '22 at 15:01
@MauroALLEGRANZA I don't understand what you're saying. I am trying to prove syntactic implication. Semantic implication makes no sense in my system because I cannot define what a valuation is — jeff honky, Dec 19 '22 at 15:03

Bram28 · Accepted Answer · 2022-12-19T15:25:50.563

For those initial proofs you really need to appeal to the definition of the $\neg$. If you try to work directly with the $\neg$, you probably won't success for many of those proofs, since if $\neg$ is treated as a primitive operator,your system would not be complete. Also, it really helps to make use of the Deduction Theorem!

OK, so let's do $p \vdash \neg \neg p$

Given the definition of the $\neg$, this really means that we need to show $p \vdash (p \to \bot) \to \bot$

Well, this is pretty easy! We know that $p, p \to \bot \vdash \bot$. So, given the Deduction Theorem, this means that $p \vdash (p \to \bot) \to \bot$. Done!

One proof you may want to do soon is $\vdash (\neg p \to \neg q) \to (q \to p)$. You'll again have to use the definition of the $\neg$ again to prove that one. But, once you have proven it, you can work with the $\neg$ as if it were a primitive. This one, together with your first two axioms forms a complete system for $\neg$ and $\to$.

thanks. I will need to slightly extend the case of the deduction theorem that I've proven, but that'll be fine. And then hopefully I can prove the other propositions as well — jeff honky, Dec 19 '22 at 15:25
@esechanota Yeah, shouldn't be so bad! Your system is Church's system (complete for $\to$ and $\bot$), and if you replace the third axiom with $(\neg p \to \neg q) \to (q \to p)$ you have Lukasiewicz' system, which is complete for $\to$ and $\neg$. So you can do searches for proofs in those systems if you get stuck. — Bram28, Dec 19 '22 at 15:29

syntactic implication without appealing to completeness

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