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If $A$ is a theorem of classical logic, is it the case that $\neg \neg A$ is a theorem of intuitionistic logic? And secondly, if $A \to B$ is a theorem of classical logic, is $\neg B \to \neg A$ a theorem of intuitionistic logic. Third, is $\neg A \vee \neg \neg A$ a theorem schema of intuitionistic logic?

Bruno Bentzen
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user107952
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  • I don't know anything about intuitionistic logic, but these all seem unlikely to me. Especially the last one. For $\neg A\vee\neg\neg A$ to be intuitionistically valid, wouldn't either $\neg A$ or $\neg\neg A$ have to be intuitionistically valid? – bof Jun 07 '15 at 01:10
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    @user107952: Could you please improve the question a little by editing it to give some background of where you encountered it or came up with it? On this site, questions that have no context are often closed. – Carl Mummert Jun 07 '15 at 01:26
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    For the second and third questions, you should look into the"topological interpretation" of intuitionistic logic, which will give you the means to answer them and many similar questions. One reference is the paper Intuitionistic Logic by Dirk van Dalen from The Blackwell Guide to Philosophical Logic. http://www.phil.uu.nl/~dvdalen/articles/Blackwell%28Dalen%29.pdf – Carl Mummert Jun 07 '15 at 01:27
  • Note that $¬A∨¬¬A$ is only the excluded middle disguised: let $\alpha \equiv \neg A$. – Bruno Bentzen Jun 07 '15 at 10:27

2 Answers2

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The answer to the first question,

If $A$ is a theorem of classical logic, is it the case that $\lnot \lnot A$ is a theorem of intuitionistic logic?

is "yes" when we are talking about propositional logic. This is a very classical result of Glivenko.

For first-order logic, it is not enough to merely preface the formula with two negations. However, there are a number of so-called negative translations that accomplish a similar thing in a more complicated way: to each first-order formula $\phi$ is associated another formula $\phi^*$ such that if $\phi$ is provable in classical first-order logic then $\phi^*$ is provable in intuitionistic first-order logic.

This use of negative translations is very well studied in proof theory, and is a standard method that has become commonly used. However, there is no truly accessible literature on proof theory. The Wikipedia article linked above has some graduate-level references.

Carl Mummert
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  • Does the stuff about "negative translations" allude to the fact (if such it be) that the symbols $\wedge,\neg,\forall$ have the same meaning in intuitionistic as in classical logic, wherefore a sentence with no connectives other than $\wedge,\neg$ and no quantifiers other than $\forall$ is intuitionistically valid if and only if it is classically valid? Is this view oversimplified, or just wrong? – bof Jun 07 '15 at 01:38
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For an intuitionistically valid proof of $A \to B \vdash \lnot B \to \lnot A$ with Natural Deduction :

1) $A \to B$ --- premise

2) $\lnot B$ --- assumed [a]

3) $A$ --- assumed [b]

4) $B$ --- from 1) and 3) by $\to$-elimination

5) $\bot$ --- from 2) and 4) by $\to$-elimination : $\lnot B$ is an abbreviation for : $B \to \bot$

6) $\lnot A$ --- from 3) and 5) by $\to$-introduction, discharging [b]

7) $\lnot B \to \lnot A$ --- from 2) and 6) by $\to$-introduction, discharging [a].

Nowhere in the derivation we have used Exluded Middle, nor Double Negation; thus, the proof is intuitionistically valid.

Now the issue is :

if $\vdash_K A \to B$, then $\vdash_J ¬B \to ¬A$ or not ?

Whe can apply the Gödel–Gentzen translation to $⊢_K A \to B$ and we have that : $⊢_J ¬¬A \to ¬¬B$.

But this is simply a "double contraposition"; thus, it seems that $⊢_J ¬B \to ¬A$.

Regarding $\lnot A \lor \lnot \lnot A$, it is not intuitionistically provable.

The counter-example with Kripke semantics is a structure with three nodes : $0,1,2$ such that $0 \le 1$ and $0 \le 2$, where $A$ holds only in $1$ (i.e. $1 \Vdash A$).

We have that $2 \Vdash \lnot A$, and thence, by the semantical clause : $k \Vdash \lnot A$ iff for all $k' \ge k : (k' \nVdash A)$ :

$0 \nVdash \lnot \lnot A$.

But since $1 \Vdash A$, also $0 \nVdash \lnot A$, and thus :

$0 \nVdash \lnot A \lor \lnot \lnot A$.

Mauro ALLEGRANZA
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  • @StevenTaschuk - Good point ... $p \to (q \lor \lnot q)$ is a tautology that is not intuitionsitically valid; it seems to me that neither $\lnot (q \lor \lnot q) \to \lnot p$ is intuitionsitically valid. If so, we have a counterexample to the non-trivial case of : if $\vdash_K A \to B$ then $\vdash_J \lnot B \to \lnot A$. – Mauro ALLEGRANZA Jun 07 '15 at 19:19