1

If $\bot$ could be defined in positive intuitionistic logic, then minimal logic would be positive intuitionistic logic: see here for the axioms.

Now, how to proof that $\bot$ is not definable in positive intuitionistic logic?

My attempt: we can model implication of positive intuitionistic logic on real line $(0,1]$ with usual order. We can be define $v$ for implication this way:

$$v(q) \leq v(p) \implies v(p \to q) = 1 $$ $$v(q) > v(p) \implies v(p \to q) = v(q)$$

But if $\bot$ would be definable, then for every $p$, $v(\bot) \leq v(p)$, and this would imply that $v(p \to \bot) = 1$, for every $p$. But this property doesn't hold on this model.

Is this right?

Sarvesh Ravichandran Iyer
  • 77,817
Lost definition
  • 1,087

1 Answers1

4

I.

First of all, it is correct to say that "minimal logic is exactly equivalent to IPC with $\bot$ interpreted as an extra propositional variable". If you disagree with that sentence, you probably misunderstood the sentence, or misunderstood something fundamental about minimal logic.

Minimal logic has the exact same formulae as intuitionistic logic, but it does not have the same tautologies. Instead, given a formula $\varphi$ of intuitionistic logic, we have an equivalence between the following statements:

  • the formula $\varphi$ constitutes a tautology of minimal logic;
  • the formula $\varphi[\bot:=P]$, obtained by replacing all occurrences of $\bot$ with a fresh variable symbol $P$ (one which does not occur in $\varphi$) constitutes a tautology of intuitionistic logic.

As far as minimal logic is concerned, $\bot$ really just denotes an arbitrary propositional variable using a funny symbol, and $\neg P$ just denotes the implication $P \rightarrow \bot$.

On the other hand, in intuitionistic logic $\bot$ has a special property: it obeys the principle of explosion, $\bot \rightarrow \varphi$ constitues an intuitionistic tautology for any formula $\varphi$.

Unlike minimal logic, positive intuitionistic logic has a different set of formulae from intuitionistic logic. Certain formulae of minimal logic, such as $\bot$, $\neg P$ or $\bot \wedge (A \rightarrow \bot)$ are not formulae of positive intuitionistic logic.

II.

You ask if $\bot$ is definable in positive intuitionistic logic.

Alas, turning this into a well-posed question would require much more context, context you have not provided. You would need to explain what you mean by "definable", and even what you mean by $\bot$: neither term has a fixed meaning in this setting. In particular, positive intuitionistic logic simply does not include the symbol $\bot$.

Some moderators might say this question should get closed for missing context at this point. Instead of voting to close, I will instead try to guess your intent.

As mentioned above, in intuitionistic logic $\bot$ has the special property that $\bot \rightarrow \varphi$ constitues a tautology for any formula $\varphi$.

So perhaps your real question concerns whether one can replicate that kind of behavior inside positive intuitionistic logic, without using $\bot$ directly. More precisely, I think you wish to ask:

Does positive intuitionistic logic contain a formula with this explosion property, i.e. $\psi$ such that $\psi \rightarrow \varphi$ constitutes a positive intuitionistic tautology for any other formula $\varphi$?

I will give a negative answer to this question below.

III.

There is no such formula $\psi$. In this part I briefly sketch a proof-theoretic argument, which I'll turn into a straightforward semantic argument in Part IV. If you have yet to study proof theory, you should just read Part IV now.

Assume for a contradiction that we found such a positive formula $\psi$. Then for any positive formula $\varphi$, the sequent $\psi \vdash \varphi$ would have a cut-free proof.

Take any propositional letter $Q$ that does not occur in $\psi$. I claim that there is no cut-free proof of $\psi \vdash Q$. By the subformula property, each branch of the proof tree would have to terminate in an application of the identity axiom $\overline{P \vdash P}$ where $P$ occurs as some subformula of $\psi$. But no matter what rules we use, some branch of the tree will always have $Q$ on the right of the turnstile, and that branch cannot terminate in an application of the identity axiom, since $Q$ is not a subformula of $\psi$. Thus, as I claimed, a cut-free proof of $\psi \vdash Q$ cannot exist.

But by assumption, for any positive formula $\varphi$, the sequent $\psi \vdash \varphi$ has a cut-free proof. The assumption that such a $\psi$ exists leads to a contradiction.

IV.

We can turn the proof-theoretic observation of Part III into a straightforward (albeit perhaps less insightful) semantic proof.

By the usual Heyting-valued semantics of intuitionistic logic, if some positive formula $\psi$ was explosive, then for any other positive formula $\varphi$, the equality $v(\psi \rightarrow \varphi) = 1$ would hold for any valuation $v$ taking values in a Heyting algebra $(H,\wedge,\vee,\rightarrow,0,1)$ (see e.g. a related question about lattice-like semantics for minimal logic).

Consider the two-element Boolean algebra $B$ as a Heyting algebra. Observe that

$$ 1 \vee 1 = 1 \wedge 1 = 1 \rightarrow 1 = 1 $$

holds in $B$.

Take any positive formula $\psi$. Take a propositional letter $Q$ that does not occur in $\psi$.

Consider the $B$-valued valuation $v$ that assigns the value $v(P) = 1$ to every variable $P$ that occurs in $\psi$, but assigns $v(Q) = 0$. Using the observation above, inducting on the length of the formula $\psi$, one gets that $v(\psi) = 1$. But then $$v(\psi \rightarrow Q) = v(\psi) \rightarrow v(Q) = 1 \rightarrow 0 = 0 \neq 1$$ so $\psi$ is not explosive.

Since $\psi$ was an arbitrary positive formula, this proves that no positive formula is explosive.

Z. A. K.
  • 13,310
  • 2
    I enjoyed reading this, thanks very much. – MJD Apr 30 '25 at 15:36
  • About I: forget bottom. I think this is a counter example to the equivalence you proposed. In the wiki page I cited, in the formulation with only negation there is explicitly an extra axiom in IPC. So $\phi =(p \to \neg p) \to q$ is a tautology of IPC with only variables with no bottom and is not a tautology of minimal logic. – Lost definition May 01 '25 at 13:03
  • I would accept this: “ minimal logic is exactly equivalent to positive IPC with ⊥ interpreted as an extra propositional constant”. – Lost definition May 01 '25 at 13:07
  • 1
    @Lostdefinition: As explained, $\neg p$ denotes $p \rightarrow \bot$ in minimal logic, i.e. the formula you write clearly has $\bot$. – Z. A. K. May 01 '25 at 13:07
  • Take the description of minimal logic without bottom that it is on the wiki, with only conjunction, disjunction, implication and negation. You could define bottom after, the way you said. But still, what I gave you is a tautology of IPC and not of minimal logic. Correct? – Lost definition May 01 '25 at 13:21
  • I also don’t understand why you use variable and not constant. All the languages you cited have an enumerable number of variables. Adding one more doesn’t change the expressive power of them. Right? – Lost definition May 01 '25 at 13:34
  • @Lostdefinition: We (I and the answer you quoted in your now-deleted edit 1) use the term variable here because the point being made is that $\bot$ behaves like a propositional variable symbol in minimal logic, i.e. it satisfies substitution. This is the content of the observation made in Part I. Minimal logic is of course not more expressive than positive intuitionistic logic, since the $\bot$ of minimal logic behaves just like any propositional variable symbol, and as you state, we already have more than enough propositional variable symbols. – Z. A. K. May 01 '25 at 14:12
  • @Lostdefinition: You ask "You could define bottom after [..]?". Since define does not have any standard meaning here, can you provide some context on what you mean by define? – Z. A. K. May 01 '25 at 14:14
  • For the claim made in Part III of my answer, the exact choice of syntax used to formulate minimal logic makes no difference, as no formulation includes an explosive formula. But there's a good chance that my answer does not answer the actual question you wanted to ask (see my guesswork in Part II to arrive at the question). If so, try to pin down the exact question so it can get answered without guesswork. – Z. A. K. May 01 '25 at 14:25
  • Let us continue this discussion in chat. – Lost definition May 01 '25 at 15:54
  • I was just saying I don't think there is a equivalence between the items that you propose in part I). I will try to be more precise. Suppose minimal logic and IPC share the same language with conectives ${ \land, \lor, \to, \neg}$. Now, suppose you introduces a new variable $\bot$ in IPC (now the languages are different). Let $\phi = (\bot \to \neg \bot) \to p $. Then $\phi[\bot := q] = (q \to \neg q ) \to p$ is a tautology in IPC but not in minimal logic. So the implication from the second bullet to the first isn't valid. – Lost definition May 01 '25 at 16:10
  • Ok, I think I understand now what you are saying. You are saying that, in a language without negation, defining negation in terms of bottom, and accepting all axioms of IPC in this way, you would have that equivalence you propose. But that is not trivial. You would have to give me a proof of that. – Lost definition May 01 '25 at 20:54
  • @Lostdefinition: You claim above that $(q \rightarrow \neg q) \rightarrow p$ is an IPC tautology. It is very clearly not. It's not even a classical tautology! With that in mind, I'll have to step away from our exchange at this point: the gap between truth tables and any nontrivial proof about the expressive power of different logics is simply too large to be bridged in a short discussion. – Z. A. K. May 01 '25 at 23:58
  • @Z.A.K., the formula is $\neg q \to ( q \to p)$. I did just a wrong parenthesis placing. A error you could spot easily if you were willing to talk to me. This is classical tautology (you can do the truth table), and also intuitionitic one simply because it is a axiom in the link a posted. I find rude your assumption about my education, about what courses I have take and about the gap you suppose there is in my education. All I did is ask a simple question here, that could be a start of fruitful exchange in a good willing talk. – Lost definition May 05 '25 at 15:24
  • @Z.A.K. I think I understood your proof-theoretic proof (and no, I didn't take a course on proof-theory to be able to understand it). Yes, I think it's an answer to the first question I posed. About definability, you could understand as you did, but the definition could be understood more abstractly by means of translations and definitional extentions of $IPC^+$ to IPC. But I don't understand why people were so defensive about this simple question here. I did the same question in another place and it was simple answered and considered as a simple exercise. – Lost definition May 05 '25 at 20:24