Observable implication and the Heyting rule

We are continuing the open-set interpretation. Interior means the largest open subset. Open-set negation takes the interior of the complement each time that connective is evaluated. Earlier classical negation used the complement itself. Keep that distinction while we introduce the corresponding open implication. Given open events U and V, seek an open condition W such that W together with U guarantees V. Among all such open conditions, choose the largest by inclusion. This is the least restrictive sufficient open condition: it allows as many points as possible while retaining the guarantee. Without the openness requirement, a point is allowed whenever it lies outside U or inside V. This is the classical implication region. But that region may not be open in the selected observation topology. Taking the interior supplies the largest open subset of that classical candidate. It is therefore the largest open W whose intersection with U lies in V. The operation is called Heyting implication. Here is the defining equivalence in words. For any open W, W is contained in the Heyting implication from U to V if and only if the intersection of W with U is contained in V. The openness of W is part of this rule. The two sides express the same guarantee in different positions. Use the positive four-state topology. On this four-card carrier, the fixed X,Y values are A: one-zero, B: one-one, C: zero-one and D: zero-zero. Only positive X and Y are selected. The smallest neighborhoods are A and B from A; B alone from B; B and C from C; and the whole carrier from D. One means true and zero means false. U is the X region and V is the Y region. The classical implication candidate is the complement of U together with V, the set B, C and D. Its largest open subset is B and C. D cannot be included in any open subset of the candidate because D's smallest neighborhood is the whole carrier, including A outside the candidate. The interior removes exactly that obstruction here. The implication itself is an open condition satisfying the defining guarantee. Intersect it with U and the result is included in V. This is open-set Modus Ponens, obtained directly from the largest-condition property. In our example the intersection is B, which lies in V. The visible calculation illustrates the general inclusion; the defining property, rather than this one four-point check, explains why the rule always holds for open inputs. Combine the implication with the open negation of V. If this open condition also met U, Modus Ponens would put the intersection in V, while the negation keeps it disjoint from V. The intersection with U must be empty. Being an open condition disjoint from U, it is included in the largest such open condition, Heyting negation of U. This derives the open-set Modus Tollens inequality without replacing Heyting negation by an arbitrary complement. Why can we not add D to the computed Heyting implication while keeping it open and retaining the guarantee? Use D's smallest neighborhood rather than the visual position of its card. Every open containing D contains the whole carrier in this positive topology. It would therefore include A, which lies in U but outside V. The guarantee fails. The largest allowed open is B and C, with no D. The opens with intersection, union and these implication and negation operations form a Heyting algebra. In a discrete topology the interiors do nothing and these operations recover the classical set forms. Keep the two semantics explicitly labelled in every comparison.