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  1. $$\neg\neg(p\wedge q)\to (\neg\neg p\wedge q)\vee (p\wedge\neg\neg q)$$

  2. $$\neg\neg(p\vee q)\to (\neg p\to q)\vee (\neg q\to p)$$

I want to prove both in intuitionistic logic. Unless, i want to find the counter example of a Heyting algebra.

The converses are easily provable in IPC. I've shown that $$\neg\neg(p\wedge q)\leftrightarrow\neg\neg p\wedge \neg\neg q$$ $$\neg\neg p\vee\neg\neg q\to (\neg p\to q)\vee(\neg q\to p)$$ So, the last says that the (2.) can be provable in KC.

Z. A. K.
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homology
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    I must say, (2) is quite a sadistic exercise to assign to a student whose only tool is Heyting algebra semantics: while (2) is not provable, the smallest Heyting counter-model has 10 elements, and the countermodel over $\mathbb{R}$ is similarly non-obvious. – Z. A. K. Jul 23 '25 at 12:38
  • @Z.A.K. Brutal... I was actually thinking it was true because I could not find any counter example. – P. Quinton Jul 23 '25 at 13:30
  • thanks a lot! I've got a more question. If i choose the (2.) as axiom, then, KC could be implied? that is, $$IPC+(2.) \vdash \neg p\vee\neg\neg p$$. i think that might be false. How can i find the counterexample? Could you recommend some key word to find them? – homology Jul 23 '25 at 17:09
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    @homology: This is indeed false. Consider the 5-element Heyting algebra obtained by adjoining ${0,1,2}$ to the powerset algebra of ${0,1}$. You can check that in the resulting HA we have $\neg {0} = {1}$ and $\neg\neg {0} = {0}$, so $\neg {0} \cup \neg\neg {0} = {0,1} \neq \top$. But this algebra has fewer than ten elements, so it validates (2). – Z. A. K. Jul 23 '25 at 18:00
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    I changed the title from the nondescript "Intuitionistic logic" to "Heyting countermodels to ¬¬(p∧q)→(¬¬p∧q)∨(p∧¬¬q) and ¬¬(p∨q)→(¬p→q)∨(¬q→p)". – Z. A. K. Jul 24 '25 at 04:03

3 Answers3

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As @P. Quinton shows above, (1) is not provable in intuitionistic logic, since $\neg\neg(p\wedge q)\to (\neg\neg p\wedge q)\vee (p\wedge\neg\neg q) = \top$ is not an identity in the 3-element Heyting algebra.

Unfortunately, a similar counterexample to equation (2) requires a much larger, 10-element Heyting algebra (Kripke semantics fares somewhat better, as @Bumble's excellent answer shows; proof-theoretic arguments are easier still).

So let us construct an explicit counterexample to (2) as a Heyting algebra.

Consider the set of numbers $S = \{1,2,3,4,6,9,12,18,36,72\}$. This set forms a Heyting algebra with

  • Conjunction $\wedge$ given by greatest common divisors,
  • Disjunction $\vee$ given by least common multiples,
  • Top element $\top = 72$, and
  • Bottom element $\bot = 1$.

As usual, we can calculate the Heyting implication $\Rightarrow$ using the formula

$$ y \Rightarrow z = \bigvee \left\{x \in S \mid (x \wedge y) \vee z = z \right\}. $$

This means that $x \wedge y$ divides $z$. For convenience I provide a full implication table below, but we only really need to calculate the negation table (first column) and two more implications to show that equation (2) fails in this algebra.

 => | 1, 2, 3, 4, 6, 9,12,18,36,72
----+------------------------------
  1 |72,72,72,72,72,72,72,72,72,72
  2 | 9,72, 9,72,72, 9,72,72,72,72
  3 | 4, 4,72, 4,72,72,72,72,72,72
  4 | 9,18, 9,72,18, 9,72,18,72,72
  6 | 1, 4, 9, 4,72, 9,72,72,72,72
  9 | 4, 4,12, 4,12,72,12,72,72,72
 12 | 1, 2, 9, 4,18, 9,72,18,72,72
 18 | 1, 4, 3, 4,12, 9,12,72,72,72
 36 | 1, 2, 3, 4, 6, 9,12,18,72,72
 72 | 1, 2, 3, 4, 6, 9,12,18,36,72

Now, let us show that equation (2) fails in this algebra, i.e. that we can choose $p,q \in S$ so that

$$\neg\neg(p\vee q)\Rightarrow ((\neg p\Rightarrow q)\vee (\neg q\Rightarrow p)) \neq 72.$$

Set $p = 2$ and $q = 3$. Then $p \vee q = 6$, so $\neg\neg (p \vee q) = \neg\neg 6 = \neg 1 = 72$. Since $72$ is the top element, this reduces the left-hand side of equation (2) to $((\neg 2\Rightarrow 3)\vee (\neg 3\Rightarrow 2))$.

We now compute $\neg 2\Rightarrow 3$ as $\neg 2\Rightarrow 3 = 9 \Rightarrow 3 = 12$ and $\neg 3\Rightarrow 2$ as $\neg 3 \Rightarrow 2 = 4 \Rightarrow 2 = 18$. But the least common multiple of $12$ and $18$ is $12 \vee 18 = 36 \neq 72$.

This shows that $S$ does not satisfy the equational form of (2), and therefore (2) does not have a proof in intuitionistic logic.

Z. A. K.
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The totally ordered set $\{ 0, 1/2, 1\}$ is a counter example to (1) with $p=q=1/2$. \begin{align*} \neg\neg(p\land q)&=\neg\neg(1/2)=1\\ \neg\neg p\land q=p\land\neg\neg q&=1/2\land 1=1/2\\ \neg\neg(p\wedge q)\to (\neg\neg p\wedge q)\vee (p\wedge\neg\neg q)&=1\to1/2=1/2 \end{align*}

P. Quinton
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This doesn't properly answer your question, because you asked for Heyting countermodels. But if you were to use Kripke semantics, there are countermodels as follows:

  1. Nodes: $\{0, 1, 2\}$ Accessibility relations: $\{0 \leq 1, 0 \leq 2\} $ Interpretation function: $$I(0, p) = \bot, \ I(1, p) =\top, \ I(2, p) = \bot,$$ $$I(0, q) = \bot, \ I(1, q) =\bot, \ I(2, q) = \top$$

  2. Nodes: $\{0, 1, 2, 3, 4\}$ Accessibility relations: $\{0 ≤ 1, 1 ≤ 2, 0 ≤ 3, 3 ≤ 4\}$ Interpretation function: $$I(0, p) = \bot, \ I(1, p) = \bot, \ I(2, p) = \bot, \ I(3, p) = \bot, \ I(4, p) = \top,$$ $$I(0, q) = \bot, \ I(1, q) = \bot, \ I(2, q) = \top, \ I(3, q) = \bot, \ I(4, q) = \bot$$

Bumble
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  • You could always convert a Kripke countermodel into a Heyting algebra countermodel by taking the Heyting algebra of upward-closed subsets. – Daniel Schepler Jul 25 '25 at 21:14
  • Unless I'm missing something, I don't think (1) works. No node satisfies $p\land q$, so every node satisfies $\lnot(p \land q)$, and no node satisfies $\lnot\lnot(p \land q)$, meaning the model should vacuously satisfy the implication. A possible countermodel which should work: $0 \le 1 \le 2$, with $p$ starting to be true at 1 and $q$ true only at 2. – Daniel Schepler Jul 25 '25 at 21:26
  • @DanielSchepler I think you're right. A simpler solution would be $ 0 \leq 1, I(0, p) = \bot, I(1, p) = \top, I(0, q) = \bot, I(1, q) = \top. $ – Bumble Jul 25 '25 at 23:08