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I am following a course in intuitionistic mathematics and I have been given an exercise about intuitionistic propositional logic.

The problem

Find four propositions $X_0, X_1, X_2, X_3$, such that the following hold:
(i) For each $i < 4$, the basic formulas of $X_i$ are among $A, B, \bot$
(ii) For each $i < 4$, $X_i \vdash_{NK} A\lor B$ and $A\lor B \vdash_{NK} X_i$
(iii) For each $i<4$, for each $j < 4$, if $i \neq j $ then $X_i \not\vdash_{NI} X_j$

Here $NK$ denotes the natural deduction scheme in classical logic and $NI$ the natural deduction scheme in intuitionistic logic.

Attemps

I am at a point where I have four propositions but recently I have found that (iii) does not hold.

  • $X_0 = ((\neg\neg A \to A) \land A) \lor ((\neg\neg B\to B) \land B)$
  • $X_1 = \neg A \to B$
  • $X_2 = \neg B \to A$
  • $X_3 = \neg\neg A \lor \neg\neg B$

Here it can be shown that $X_0 \vdash_{NI} X_3$ and for all the other combinations, I can find Kripke models to show the statements are intuitionistically not valid. Besides these four, there were other candidates that are classically equivalent to $A\lor B$ which were considered such as $A\lor B, \neg\neg(A\lor B)$ and $((A\to B)\to B) \land ((B\to A) \to A)$. Yet combining these have failed to give a satisfying result.

My question therefore is, are there other propositions that are classically equivalent to $A\lor B$ which I have not considered and cannot intuitionistically be proven assuming that proposition?

Tungsten
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  • Hint: can you express $A \vee B$ in classical logic by a simple formula which only uses implication? – Pilcrow Dec 19 '21 at 18:28
  • I can only think of either $(A\to B)\to B$ and $(B\to A)\to A$. Are these what you mean or are there different propositions? – Tungsten Dec 19 '21 at 21:56
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    Yes, that's what I had in mind. Now you can try comparing these with what you have. Also, note that in addition to $\neg \neg A \vee \neg \neg B$ you have, in the same spirit, $\neg \neg A \vee B$ and $A \vee \neg \neg B$. – Pilcrow Dec 19 '21 at 22:02

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