Co-Leibniz Identity for Decidability

Heyting algebras are structures that can be used to give a semantics to intuitionistic propositional logic, and as it turns out, they can easily be dualised, yielding the aptly named co-Heyting algebras.

Co-Heyting algebras come equipped with three binary operations \((\land, \lor, \leftharpoondown)\), where \(\leftharpoondown\) is the dual of implication \(\to\) and usually called subtraction or exclusion.

I will not go into too much detail on co-Heyting algebras here, but let me at least give you an intuition for why the name subtraction makes sense. In a Heyting algebra we have the following identity:

If you imagine the \(\leq\) to be another \(\to\), then the above reads \(a \land b \to c \iff a \to (b \to c)\) which should look like a familiar logical principle / currying of functions / adjunction of functors. The dualized identity in a co-Heyting algebra looks like this:

And if you think of \(\lor\) as \(+\), then the above tells us that we can subtract \(b\) on both sides of the left equation without breaking the inequality.

Let’s now move on to the main definition in co-Heyting algebras that I want to highlight here: the boundary of an element \(s\), which is defined by

The interesting thing about this definition of a boundary is that it satisfies the Leibniz rule familiar from calculus:

What surprised me was that nLab does not mention any comparable definition for Heyting algebras, even though we have this strong duality between the two. And indeed, if we straight up dualise the definition, by flipping \(\land\) and \(\lor\) as well as the negation, we get an expression that looks oddly familiar

In constructive logic, \(\varphi \lor \neg \varphi\) is often referred to as the “decidability” of a statement \(\varphi\).

Accordingly, we can indeed state and prove a dualised version of the Leibniz rule for this decidability operator:

This can be verified for any Heyting algebra, but below I give a quick verification of this fact by using the usual definition of decidability in the Coq proof assistant:

Definition iffT (X Y: Type) : Type := (X -> Y) * (Y -> X).
Notation "X <=> Y" := (iffT X Y) (at level 95).

Definition dec (P : Type) : Type := P + (P -> False).

Lemma dec_sum A B :
  dec (A + B) <=> (dec A + B) * (A + dec B).
Proof.
  split; intros H; unfold dec.
  - destruct H as [|]; split; try tauto.
  - destruct H as [[[]|] [|[]]]; try tauto.
Qed.

While I don’t have a good intuitive grasp on why the equivalence holds, I can give a good pictorial view on how to think about both \(\partial\) and \(\delta\).

Consider a Venn-diagram showing overlapping sets \(A\) and \(B\). We can then think of the boundary \(\partial A\) as the line that we would use to outline the set \(A\), and likewise for the boundary of other sets. The Leibniz identity then simply reflects a way to compute the boundary \(\partial (A \cap B)\) based on the boundaries of \(A\) and \(B\). A similar visual explanation holds up for \(\delta A\); it consists of everything in the picture except the boundary \(\partial A\).

Apart from the connection to decidability I showed above, I have not yet encountered the co-Leibniz identity elsewhere “in the wild”, and the same goes for people I have asked so far. So if you have, I would be interested to hear about it!