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The existence of intermediate logics between intuitionistic logic and classical logic is well known (for example, adding $(A \to B) \lor (B \to A)$ as an "axiom of linearity" to Heyting's logic defines Gödel-Dummett logic, a logic stronger than intuitionistic logic but weaker than classical logic). But is there an argument to exclude the possibility of an intermediate logic between minimal logic and intuitionistic logic or, on the contrary, an argument to assert that such an intermediate system is possible?

Reminder: in intuitionisitic logic, we have $$\lnot \lnot A \nvdash A $$ contrarily to classical logic, and in minimal logic, we have $$\bot \nvdash C$$ contrarily to intuitionistic logic.

Here is what should be a clue to answer my question. In intuitionistic logic, the assumption of the classical theorem called "Dummett's formula", that is,

$$(A \to B) \lor (B \to A)$$

does not entail the law of elimination of double negation: \begin{equation*} \tag{1} (A \to B) \lor (B \to A) \nvdash_{i} \lnot \lnot A \to A \end{equation*}

We can therefore add $(A \to B) \lor (B \to A)$ to the system of intuitionistic logic to obtain an intermediate system.

Is there a formula that is provable in intuitionistic logic, but not in minimal logic, and that is such that its assumption in minimal logic does not entail the explosion rule? If there is such a formula, adding this formula to minimal logic would provide an intermediate logic. On the contrary, if it is provable that such a formula does not exist, it must be admitted that there is no intermediate logic between minimal logic and intuitionistic logic.

Contrarily to the first reply made to this question, my intuition is "No, there is no intermediate logic between minimal and intuitionistic logic". Here is a sketch of argument: the proof of any formula F that is provable only thanks to the explosion rule contains a contradiction of two literals, say $\lnot A$ and $A$, to get an atomic formula B via the explosion rule. The assumption of F in minimal logic must therefore entail this instance of explosion rule. For example, this sequent: \begin{equation*}\tag{2} ( (A \lor B) \land \lnot A) \to B \vdash (\lnot A \land A) \to B \end{equation*} is provable in minimal logic, by contrast with (1) where the assumption of the Dummett formula does not entail the provability of an instance of the rule of Double Negation Elimination. That is why I suspect there is no intermediate logic between minimal logic and intuitionistic logic, at least not like in the well-known case of intermediate logics between intuitionistic logic and classical logic.

  • Below, a specific logic, B = ML + $\bot \vee (\bot \rightarrow y)$ was given. This B is intermediate between ML and IL since 1. B proves e.g. $\bot \vee (\bot \rightarrow C)$, which ML does not prove; 2. anything B proves is also provable in IL; 3. B fails to prove e.g. $\bot \rightarrow C$, which IL does prove. Do you disagree with any of these claims, for example think that in fact there is a derivation of $\bot \rightarrow C$ in B, or else a derivation of $\bot \vee (\bot \rightarrow C)$ in minimal logic? If so, please share those derivations so we can take a look. – Z. A. K. Feb 27 '25 at 12:27
  • Your sketched argument does not seem to lead anywhere? Logics intermediate between ML and IL prove plenty of instances of explosion - in fact, ML itself proves instances, e.g. ML proves $\bot \rightarrow (A \rightarrow \bot)$. They just don't prove all instances, that's what makes them intermediate. Similarly, logics intermediate between IL and CL prove plenty of instances of DNE, (e.g. $\neg\neg\neg A \rightarrow \neg A$). IL + the Dummett axioms entails the provability of plenty of new instances as well, such as $\neg\neg((A \to B) \lor (B \to A)) \rightarrow ((A \to B) \lor (B \to A))$. – Z. A. K. Feb 27 '25 at 12:33
  • @Z.A.K. $\bot \to (A \to \bot)$ is certainly not an instance of explosion. It is just one of the paradoxes for "relevance logic" i.e. $B \to (A \to B)$. Similarly, $\lnot \lnot \lnot A \to \lnot A$ is nothing else than $(((A \to B) \to B) \to B) \to (A \to B)$ and would be better called "Triple Negation Elimination" than DNE. – Joseph Vidal-Rosset Feb 27 '25 at 12:56
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    The principle of explosion is the schema $\bot \rightarrow y$. Substituting any $y$ gives an instance of explosion: that's what an instance means, a substitution instance. Some instances will be provable in minimal logic, other instances not. As I said above, $\bot \rightarrow (A \rightarrow \bot)$ is an exp. instance that is provable in minimal logic; so is $\bot \rightarrow (A \rightarrow A)$. Other instances, such as $\bot \rightarrow (A \rightarrow B)$, are not provable in ML. Similarly for $\neg\neg y \rightarrow y$: taking $y:=\neg A$, we just happen to get an IL-provable instance. – Z. A. K. Feb 27 '25 at 13:32
  • Every single intermediate logic (in the usual sense, intermediate between intuitionistic and classical) proves more instances of double-negation elimination than intuitionistic logic itself. Classical logic is the only extension of IL that proves all instances of double-negation elimination. – Z. A. K. Feb 27 '25 at 13:34

1 Answers1

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We can construct infinitely many logical systems stronger than minimal and weaker than intuitionistic logic.

Recall that any minimal logic proof translates into an intuitionistic proof by choosing a variable not present in the proof and replacing every occurrence of $\bot$ in the proof with that variable symbol. This works because $\bot$ has no special rules governing it in minimal logic.

The idea is to now extend minimal logic by asserting that $\bot$ obeys some law $L(p)$ such that

  1. both $\bot$ and $\top$ obey the chosen law in intuitionistic logic (i.e. $\vdash L(\bot)$ and $\vdash L(\top)$ in intuitionistic logic),
  2. but an arbitrary proposition need not (i.e. $\not\vdash L(p)$ in intuitionistic logic),
  3. and the symbol $\bot$ itself does not occur in $L(p)$.

Then adding the axiom $L(\bot)$ to the usual Hilbert system of minimal logic results in a system B that interpolates between minimal and intuitionistic.

Since intuitionistic logic proves $L(\bot)$, one can translate any B-proof of a formula into an intuitionistic proof of the same formula. To complete the construction, one only has to check that B does not prove explosion. But since intuitionistic logic also proves $L(\top)$, any B-proof remains a valid intuitionistic proof if one replaces the symbol $\bot$ with $\top$ everywhere. Since $\not \vdash \top \rightarrow C$ in intuitionistic logic, B in turn does not prove $\bot \rightarrow C$.

I still need to justify that infinitely many different suitable choices exist for such a law $L$. Fortunately, the Rieger-Nishimura lattice contains an infinite subposet of formulae that both $\bot$ and $\top$ satisfy in intuitionistic logic:

a subposet of the RN lattice consisting of formulae that bottom and top obey

Since none of these laws imply each other even in intuitionistic logic, choosing any one of them as our basis of $L(p)$ yields a distinct intermediate logic: one just has to eliminate the original occurrences of $\bot$, hidden in the negation: but that's easily done by replacing the original occurrences with any fixed fresh variable symbol in positive, and with $p$ itself in negative positions. This still leaves us with infinitely many different laws.

For example, choosing the formula $p \vee \neg p$, we get that the logic "minimal logic + $\bot \vee (\bot \rightarrow y)$" is intermediate between minimal and intuitionistic.

Z. A. K.
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  • It is known that there exist continuum many intermediate logics (between intuitionistic logic and classical logic). What about for logics between minimal logic and intuitionistic logic? (It may be easily deducible from what you've written, but having not studied this stuff in a while, I haven't tried to see whether that is the case.) – Dave L. Renfro Feb 27 '25 at 01:11
  • @DaveL.Renfro Yes, that's a good followup question. I believe the answer should be positive. Your intuition is correct that there will not be a short demonstration of this based on the argument above: you need something that allows you to choose sets of infinitely many "logically independent" laws to adapt the usual methods that construct continuum many LJ-LK intermediates, and the method presented above won't let you choose even 3. I don't think it'd be very hard, but the question only asked for one example, so I'll let someone else go further and establish that there are continuum-many. – Z. A. K. Feb 27 '25 at 01:38
  • Your intuition is correct that there will not be a short demonstration of this based on the argument above -- Actually, this wasn't my intuition but instead I was just being careful in case it was obvious there was some kind of bisection method (such as can be used to show there are continuum many nodes in an infinite binary tree, or the are continuum many points in the Cantor set) using what you have already given, which I haven't tried to follow. – Dave L. Renfro Feb 27 '25 at 01:55
  • @Z.A.K. You wrote "Then adding the axiom L(⊥) to the usual Hilbert system of minimal logic results in a system B that interpolates between minimal and intuitionistic." . I do not understand. What do you call "axiom L(⊥)" exactly? – Joseph Vidal-Rosset Feb 27 '25 at 12:00
  • $L(\bot)$ is the formula you get when you substitute the symbol $\bot$ for the variable $p$ in the formula $L(p)$. Adding the axiom $L(\bot)$ to the usual Hilbert system means taking the Hilbert system in which all the axioms of minimal logic are axioms and also every substitution instance of the formula $L(\bot)$ is an axiom. I give an explicit example of such a system at the end of the answer, where the formula $L(\bot)$ happens to be $\bot \vee (\bot \rightarrow y)$. – Z. A. K. Feb 27 '25 at 12:08
  • @Z.A.K. Thanks. Your expression was confusing with rule $L\bot$ in G3ip. Can you give us an example of a formula that is unprovable with minimal logic but that you can prove with adding this axiom to minimal logic? – Joseph Vidal-Rosset Feb 27 '25 at 12:32
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    @JosephVidal-Rosset: For example, $\bot \vee (\bot \rightarrow C)$ itself is not provable in minimal logic, but has a one line proof using the axiom. – Z. A. K. Feb 27 '25 at 12:34
  • @Z.A.K. Do you mean that adding this formula as axiom to minimal logic, you can "prove" this formula in this new system? That's it? – Joseph Vidal-Rosset Feb 27 '25 at 12:56
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    @JosephVidal-Rosset: I'm not sure what else you could expect it to be. If you want something less trivial, you can take the formula $(C \rightarrow \bot) \vee (\bot \rightarrow C)$: to prove this in B you have to use the axiom and also do some additional work. It is also not provable in Minimal Logic. – Z. A. K. Feb 27 '25 at 13:27