Heyting

Vikraman Choudhury April, 2018

1 Heyting Duality

Pairs of “equivalent” concepts or phenomena are ubiquitous in , and dualities relate such two different, even opposing concepts. Stone’s Representa- tion theorem (1936), due to Marshall Stone, states the duality between Boolean algebras and topological spaces. It shows that, for classical propositional logic, the Lindenbaum-Tarski algebra of a set of propositions is isomorphic to the clopen subsets of the set of its valuations, thereby exposing an algebraic viewpoint on logic. In this essay, we consider the case for intuitionistic propositional logic, that is, a duality result for Heyting algebras. It borrows heavily from an exposition of Stone and Heyting duality by van Schijndel and Landsman [2017].

1.1 Preliminaries Definition (). A lattice is a poset which admits all finite meets and joins. Categorically, it is a (0, 1) − 푐푎푡푒푔표푟푦 (or a thin category) with all finite limits and finite colimits. Alternatively, a lattice is an algebraic structure in the signature (∧, ∨, 0, 1) that satisfies the following axioms.

• ∧ and ∨ are each idempotent, commutative, and associative with respective identities 1 and 0.

• the absorption laws, 푥 ∨ (푥 ∧ 푦) = 푥, and 푥 ∧ (푥 ∨ 푦) = 푥.

Definition (). A distributive lattice is a lattice in which ∧and ∨ distribute over each other, that is, the following distributivity axioms are satisfied. Categorically, this makes it a distributive category.

1 • 푥 ∨ (푦 ∧ 푧) = (푥 ∨ 푦) ∧ (푥 ∨ 푧)

• 푥 ∧ (푦 ∨ 푧) = (푥 ∧ 푦) ∨ (푥 ∧ 푧) Definition (Complements in a lattice). A complement of an element 푥 of a lattice is an element 푦 such that, 푥 ∧ 푦 = 0 and 푥 ∨ 푦 = 1. Complements need not be unique. If it is unique, we denote the complement of 푥 by ¬푥. Definition (Boolean algebra). A Boolean algebra 픅 is a distributive lattice with unique complements.

1.2 Stone Duality We first state the Stone Representation Theorem, which establishes the duality of Boolean algebras and Stone spaces. Definition (). A (푋, 휏) is totally disconnected if 휏 has a basis of clopen sets. A Stone space is a totally disconnected compact . Definition (푃퐹). For 픏 a bounded distributive lattice, let 푃퐹(픏) be the set of prime filters. For 푎 ∈ 픏, we define

퐹 ∶ 픏 → 푃퐹(픏)

푎 ↦ 퐹푎 ≔ { 푃 ∈ 푃퐹(픏) ∣ 푎 ∈ 푃 } Note that 푃퐹(픏) is a also a poset in its own right, with respect to inclusion. It is easy to check that it is also a bounded distributive lattice. Definition (퐶푙). For 푋 a topological space, let 퐶푙(푋) be the set of clopen sets of 푋. Definition (퐶푈). For 푋 a topological space, let 퐶푈(푋) the set of clopen up-sets of 푋, ordered by inclusion. Lemma. The map 퐹 ∶ 픏 → 푃퐹(픏) is a homomorphism. It preserves joins, meets, 0, 1, and complements.

Proof. 퐹0 = ∅ because no prime contains 0. 퐹1 = 푃퐹(픏) because every prime filter contains 1. 퐹푎 ∪ 퐹푏 ⊆ 퐹푎∨푏 because filters are up-sets and 푎 ∈ 푃 or 푏 ∈ 푃 implies 푎 ∨ 푏 ∈ 푃. Moreover, 퐹푎∨푏 ⊆ 퐹푎 ∪ 퐹푏 because 푃 ∈ 퐹푎∨푏 is prime, hence 푎 ∨ 푏 ∈ 푃 implies 푎 ∈ 푃 or 푏 ∈ 푃. The case for meets follows dually because filters are closed under finite meets, and are up-sets.

2 ′ If 픏 has complements, we want to prove that ¬퐹푎 = 퐹푎 = 퐹¬푎. Since 푃 ∈ 푃퐹(픏) is proper, it cannot contain both 푎 and ¬푎. Since 푃 is prime, and hence an ultrafilter (using the ultrafilter lemma), it contains either 푎 or ¬푎. This means that the set of prime filters containing ¬푎 is precisely the set of prime filters not containing 푎.

Definition (Topology τ on 푃퐹(픏)). We equip 푃퐹(픏) with a topology 휏 as follows. ′ For 푎, 푏 ∈ 픏, let 푆 ≔ { 퐹푎 ∣ 푎 ∈ 픏 } ∪ { 퐹푏 ∣ 푏 ∈ 픏 } be a subbasis. Then 푇 ≔ ′ { 퐹푎 ∩ 퐹푏 ∣ 푎, 푏 ∈ 픏 } is a basis of 휏. Lemma. If 픏 is a distributive lattice, then 푃퐹(픏) is a Stone space.

′ Proof. Let 퐴 be in 푇, the basis of 휏, there are 푎, 푏 ∈ 픏 such that 퐴 = 퐹푎 ∩ 퐹푏. Since ′ ′ ′ ′ 퐴 is in the basis of 휏, 퐴 is open. Now, 퐴 = (퐹푎 ∩ 퐹푏) = 퐹푎 ∪ 퐹푏 by de Morgan’s ′ ′ laws. Since 퐹푎 and 퐹푏 are in the basis, their union 퐴 is open, so 퐴 is closed. Let 푃, 푄 be prime filters with 푃 ≠ 푄. Then either 푃 ⊈ 푄 or 푄 ⊈ 푃. If 푃 ⊈ 푄, ′ there is an 푎 ∈ 픏 with 푎 ∈ 푃 and 푎 ∉ 푄, so 푃 ∈ 퐹푎 and 푄 ∈ 퐹푎. Since 푃 and 푄 are separated by disjoint open sets, 푃퐹(픏) is Hausdorff. To show that 푃퐹(픏) is compact, we need to show that any open cover 풰 ⋃ ′ of 푃퐹(픏) has a finite subcover. If 푈 ∈ 풰, then 푈 = 푎,푏∈퐴 { 퐹푎 ∩ 퐹푏 }, for some 퐴 ⊆ 픏. Since 풰 is a cover, for fixed 퐴1, 퐴2 ⊆ 퐴,

′ ′ 푃퐹(픏) ⊆ ﾌ { 퐹푎 ∩ 퐹푏 } ⊆ ﾌ { 퐹푎 } ∪ ﾌ { 퐹푏 } 푎,푏∈퐴 푎∈퐴1 푎∈퐴2 which is also an open cover of 푃퐹(픏). Hence, ⋂ 퐹 ⊆ ⋃ 퐹 . 푏∈퐴2 푏 푎∈퐴1 푎 Let 퐼 be the generated by 푎 ∈ 퐴1 and 퐺 the filter generated by 푏 ∈ 퐴2.A proof by contradiction and the use of the prime ideal theorem shows that 퐺∩퐼 ≠ ∅.

If 푥 ∈ 퐺 ∩ 퐼, 푥 ≥ 푏∗ ≔ 푏1 ∧ … ∧ 푏푛 for some 푏푖 ∈ 퐴2. Also, 푥 ≤ 푎∗ = 푎1 ∨ … ∨ 푎푚 for some 푎 ∈ 퐴 . Because 퐹 s are prime filters, 퐹 ∩…∩퐹 ⊆ 퐹 ⊆ 퐹 ⊆ 퐹 ∪…∪퐹 . 푖 1 푖 푏1 푏푛 푏∗ 푎∗ 푎1 푎푚 This gives a covering of 푃퐹(픏) which makes it compact.

Lemma. If 푋 is a Stone space, then 퐶푙(푋) is a Boolean algebra.

Proof. This follows by a tedious checking of the axioms of a Boolean algebra, using the fact that clopenness is preserved under intersection, union and complement.

We now claim that 푃퐹 and 퐶푙 establish an equivalence, which proves the Stone representation theorem.

3 Theorem (Stone Representation Theorem). The map 푓 ∶ 픅 → 퐶푙(푃퐹(픅)) given by 푏 ↦ 퐹푏 is an isomorphism of Boolean algebras. The map 푔 ∶ 푋 → 푃퐹(퐶푙(푋)) given by 푥 ↦ 퐹푈 ≔ { 푈 ∈ 퐶푙(푋) ∣ 푥 ∈ 푈 } is a . Proof. First, we notice that 퐶푙(푃퐹(픅)) is a Boolean algebra by the above lemma, hence a subalgebra of the of 푃퐹(픅). Since 퐹푏 is an element of the ′ subbase 푆 of 휏, it is open. Since (퐹푏) is also an element, 퐹푏 is closed, hence clopen. To show that 푓 is injective, consider 푎 ≠ 푏. We want to show that 퐹푎 ≠ 퐹푏. Either 푎 ≰ 푏 or 푏 ≰ 푎. Using 푎 ≰ 푏, let 퐺 =↑ 푎 be the filter generated by 푎 and 퐼 =↓ 푏 be the ideal generated by 푏, so that 퐺 ∩ 퐼 = ∅. Then there is a prime filter

푃 with 퐺 ⊆ 푃 and 푃 ∩ 퐼 = ∅. Hence, 푎 ∈ 푃 and 푏 ∉ 푃, so 푃 ∈ 퐹푎 and 푃 ∈ 퐹푏, so that 퐹푎 ≠ 퐹푏. To show that 푓 is surjective, consider 퐶 a clopen subset of 푃퐹(픅), we need to show that 퐶 is in the of 푓. Since 퐶 is open, for 퐵1, 퐵2 ⊆ 픅,

′ 퐶 = ﾌ { 퐹푏 ∩ 퐹푐 } = ﾌ { 푃 ∈ 푃퐹(픅) ∣ 푏 ∈ 푃, 푐 ∉ 푃 }

푏∈퐵1,푐∈퐵2 푏∈퐵1,푐∈퐵2 Since 푃 is also an ultrafilter, 푐 ∉ 푃 implies 푐′ ∈ 푃, and since 푃 is closed under meets, 푏 ∧ 푐′ ∈ 푃. Hence,

′ 퐶 = ﾌ { 푃 ∈ 푃퐹(픅) ∣ 푏 ∧ 푐 ∈ 푃 } = ﾌ { 푃 ∈ 푃퐹(픅) ∣ 푑 ∈ 푃 } = ﾌ 퐹푑

푏∈퐵1,푐∈퐵2 푑∈퐵3 푑∈퐵3

Since 퐹푑 is open for each 푑, { 퐹푑 ∣ 푑 ∈ 퐵3 } is an open cover of 퐶. Since 퐶 is closed and 푃퐹(픅) is compact, 퐶 is also compact. Hence, 퐶 = 퐹 ∪ … ∪ 퐹 = 퐹 푑1 푑푛 푑1∨…∨푑푛 which is in the image of 푓. To see that 푔 is well-defined, for any 푥 ∈ 푋, 푔(푥) should be a prime filter of 퐶푙(푋). Let 푉 ∈ 퐶푙(푋) with 푉 ∈ 푔(푥) and 푉 ⊆ 푊. Then 푥 ∈ 푊 and 푊 ∈ 푔(푥), hence 푔(푥) is an up-set of 퐶푙(푋). If 푈, 푉 ∈ 푔(푥), then 푥 ∈ 푈 and 푥 ∈ 푉, so 푥 ∈ 푈 ∩ 푉. Thus 푈 ∩ 푉 ∈ 푔(푥) and 푔(푥) is a filter. Suppose 푈 ∪ 푉 ∈ 푔(푥), then 푥 ∈ 푈 ∪ 푉, so 푥 ∈ 푈 or 푥 ∈ 푉, hence 푔(푥) is a prime filter.

Let 휏1 be the topology of 푃퐹(퐶푙(푋)). To check that 푔 is continuous, it suffices to show that the inverse image of every set in the basis of 휏1 is open. If 퐴 is an element in the basis of 휏1, since 퐹 is a homomorphism, ′ { 퐹푈 ∩ 퐹푉 ∣ 푈, 푉 ∈ 퐶푙(푋) } = { 퐹푈∩푉′ ∣ 푈, 푉 ∈ 퐶푙(푋) }

So 퐴 = 퐹푊 for some 푊 ∈ 퐶푙(푋). −1 푔 (퐴) = { 푥 ∈ 푋 ∣ 푔(푥) ∈ 퐹푊 } = { 푥 ∈ 푋 ∣ 푊 ∈ 푔(푥) } = 푊

4 Since 푊 is a clopen subset of 푋, the inverse image of 퐴 is open for every 퐴 in the basis of 휏1. Thus, 푔 is continuous. To show that 푔 is injective, let 푥, 푦 ∈ 푋 with 푔(푥) = 푔(푦). Every 푎, 푏 ∈ 푋 with 푎 ∉ 푏 can be separated by open sets, because 푋 is Hausdorff. But 푋 also has a basis of clopen sets, thus we can separate 푎, 푏 by clopen sets. Hence,

ﾎ 푔(푥) ≔ ﾎ { 푈 ∈ 퐶푙(푋) ∣ 푥 ∈ 푈 } ⊆ { 푥 }

If 푔(푥) = 푔(푦), then ⋂ 푔(푥) = ⋂ 푔(푦), hence { 푥 } = { 푦 }, which implies 푥 = 푦. Lastly, we check that 푔 is surjective. Let 푃 ∈ 푃퐹(퐶푙(푋)). Since 푃 is a proper filter, it is closed under intersection and does not contain ∅. It has the finite intersection property and is compact because it is a Stone space. Hence, 푃 has a non-empty intersection. Let 푥, 푦 ∈ 푃 with 푥 ≠ 푦. By separation, there is a clopen set 푈 with 푥 ∈ 푈 and 푦 ∈ 푋 ∖ 푈. Since 푃 is also an ultrafilter, either 푈 ∈ 푃 or 푋 ∖ 푈 ∈ 푃. If 푈 ∈ 푃, then 푦 ∉ ⋂ 푃, and if 푋 ∖ 푈 ∈ 푃, then 푥 ∉ ⋂ 푃. So ⋂ 푃 = { 푧 } for some 푧 ∈ 푋 and 푃 ∈ 푔(푧). Since 푃 and 푔(푥) are both prime, maximal filters on 퐶푙(푋), 푃 = 푔(푧). Thus, every 푃 ∈ 푃퐹(퐶푙(푋)) is in the image of 푔.

1.3 Heyting Duality A Boolean algebra can be generalized to a Heyting algebra by weakening the com- plement. We say that 푎∗ ≔ 푚푎푥 { 푏 ∈ 픏 ∣ 푏 ∧ 푎 = 0 } is the pseudo-complement of 푎. We introduce an implication operation, such that 푎∗ = 푎 → 0. Definition (Heyting algebra). A Heyting algebra ℌ is a bounded, distributive lattice with a binary implication operation →, such that 푐 ≤ 푎 → 푏 iff 푎 ∧ 푐 ≤ 푏. Definition (Heyting space). Let (푋, 휏, ≤) be a Stone space with a partial order ≤ defined on 푋. For 푥, 푦 ∈ 푋 with 푥 ≰ 푦, if there is a clopen up-set 푈 with 푥 ∈ 푈 and 푦 ∉ 푈, then (푋, 휏, ≤) satisfies the Priestley separation axiom. Furthermore, iffor every clopen 푈 ⊆ 푋 the set ↓ 푈 is clopen, then (푋, 휏, ≤) is a Heyting space, or an Esakia space. Lemma. If ℌ is a Heyting algebra, then (푃퐹(ℌ), ⊆) is a Heyting space. Proof. Since ℌ is a distributive lattice, 푃퐹(ℌ) is a Stone space and compact. Let

푃, 푄 ∈ 푃퐹(ℌ) with 푃 ⊈ 푄. Then there exists a 푥 ∈ 푃 with 푥 ∉ 푄, so that 푃 ∈ 퐹푥 but 푄 ∉ 퐹푥. We know that 퐹푥 is clopen and an up-set of 푃퐹(ℌ). If 푅 ∈ 퐹푥 and 푅 ⊆ 푆, then 푥 ∈ 푆, so 푆 ∈ 퐹푥. So 푃퐹(ℌ) satisfies the Priestley separation axiom.

5 We need to verify that if 푈 ⊆ 푃퐹(ℌ) is clopen, then ↓ 푈 is clopen. Since 푈 is open, it is a union of elements in the basis of the topology of (푃퐹(ℌ), ⊆), which forms an open cover. Since 푈 is closed and 푃퐹(ℌ) is compact, 푈 is compact, so the open cover has a finite subcover. Thus,

↓ 푈 =↓ ( ﾌ 퐹 ∩ 퐹′ ) = ﾌ ↓ (퐹 ∩ 퐹′ ) 푎푖 푏푖 푎푖 푏푖 푖=1…푛 푖=1…푛

If we can show that ↓ (퐹 ∩퐹 ′ ) = 퐹 , then 푎 → 푏 ∈ ℌ, so 퐹 is clopen. Since 푎푖 (푏푖) 푎→푏 푎→푏 clopen sets are closed under finite union, we can conclude that ↓ 푈 is clopen.

Let 푎, 푏 ∈ ℌ, then 푎 ∧ (푎 → 푏) = 푎 ∧ 푏 ≤ 푏. Let 푃 ∈ 퐹푎 ∩ 퐹푎→푏. Now, 푃 is a filter and an up-set, so 푎 ∧ (푎 → 푏) ∈ 푃 and 푏 ∈ 푃, hence 푃 ∈ 퐹푏. Since 퐹푎 ∩ 퐹푎→푏 ⊆ 퐹푏, ′ ′ ′ ′ ′ 퐹푎 ∩ 퐹푏 ⊆ 퐹푎→푏. But 퐹푎→푏 is a down-set, so that ↓ (퐹푎 ∩ 퐹푏) ⊆ 퐹푎→푏. ′ ′ Let 푃 ∈ 퐹푎→푏, we want to show that 푃 ∈↓ (퐹푎 ∩ 퐹푏). Therefore, we need a prime filter 푄 of ℌ with 푎 ∈ 푄 but 푏 ∉ 푄, and 푃 ⊆ 푄. It suffices to show that 푎 → 푏 ∉ 푄 rather that 푏 ∉ 푄. Such a 푄 exists if the filter 퐺 generated by 푃 ∪ { 푎 } does not contain 푎 → 푏. Suppose, towards a contradiction that 푎 → 푏 ∈ 퐺. Then there exists an 푦 ∈ 푃 ∪ { 푎 } with 푦 ≤ 푎 → 푏. Since 푎 → 푏 ∉ 푃, 푦 = 푎 ∧ 푥 for some 푥 ∈ 푃. Since 푎 ∧ 푥 ≤ 푎 → 푏, (푎 ∧ 푥) ∧ 푎 ≤ 푏, or 푎 ∧ 푥 ≤ 푏, which means that 푥 ≤ 푎 → 푏. But, since 푥 ∈ 푃, 푎 → 푏 ∈ 푃 which is a contradiction! ′ ′ Hence, 퐹푎→푏 =↓ (퐹푎 ∩ 퐹푏), and ↓ 푈 is clopen. Lemma. If (푋, ≤) is a Heyting space, then 퐶푈(푋, ≤) is a Heyting algebra, where 푈 → 푉 = (↓ (푈 ∩ 푉 ′))′. Proof. To see that implication is well-defined, we note that (↓ (푈 ∩ 푉 ′))′ is a clopen up-set for clopen sets 푈 and 푉. 퐶푈(푋, ≤) is a bounded distributive lattice by using the properties of the subsets of the powerset. The implication follows the correct properties by a routine application of set-theoretic identities and de Morgan’s laws. We now establish the isomorphism to complete the Heyting duality.

Theorem (Heyting duality). The map 푓 ∶ ℌ → 퐶푈(푃퐹(ℌ)) given by ℎ ↦ 퐹ℎ ≔ { 푃 ∈ 푃퐹(ℌ) ∣ ℎ ∈ ℌ } is an isomorphism of Heyting algebras. The map 푔 ∶ (푋, ≤ ) → 푃퐹(퐶푈(푋, ≤), ⊆) given by 푥 ↦ { 푈 ∈ 퐶푈(푋, ≤) ∣ 푥 ∈ 푋 } is an isomorphism of Heyting spaces. Proof. For 푓 to be a homomorphism, we just need to check that it preserves ′ ′ ′ ′ implication. Let ℎ, 푘 ∈ ℌ. Then 푓(ℎ → 푘) = 퐹ℎ→푘 = (퐹ℎ→푘) = (↓ (퐹ℎ ∩ 퐹푘)) = 퐹ℎ → 퐹푘.

6 Injectivity of 푓 follows by the same argument as in the case of Stone duality. To show that 푓 is surjective, let 푈 be a clopen up-set of 푃퐹(ℌ). Let 푃 ∈ 푈, ′ 푄 ∈ 푈 , then 푃 ⊈ 푄. So there exists some 푎푃푄 ∈ ℌ such that 푎푃푄 ∈ 푃 and 푎푃푄 ∉ 푄. Hence, 푃 ∈ 퐹 , 푄 ∈ 퐹′ , and the 퐹′ s cover 푈′. Since 푃퐹(ℌ) is compact, 푎푃푄 푎푃푄 푎푃푄

′ ′ ′ 푈 ⊆ ﾌ (퐹푎 ) = 퐹푎 푃푄푖 푃 푖=1…푛 where 푎 = 푎 ∧ … ∧ 푎 . Since 푃 ∈ 퐹 for all 푎 , we have 푃 ∈ 퐹 ⊆ 푈. 푃 푃푄1 푃푄푛 푎푃푄푖 푃푄푖 푎푃 Since 푈 is the union of the various 퐹 , they form an open cover of 푈. But 푈 푎푃 is closed and 푃퐹(ℌ) is compact, so 푈 is a finite union

ﾌ 퐹푎 = 퐹푎 푃푖 푖=1…푚 where 푎 = 푎 ∨ … ∨ 푎 ∈ ℌ. Hence 푈 is in the image of 푓. 푃1 푃푚 푔 is well-defined by a similar argument as in the case of Stone duality. 푔 preserves order because 푈 is an up-set. If 푥, 푦 ∈ 푋 with 푥 ≤ 푦, 푈 ∈ 푔(푥) implies 푥 ∈ 푈, hence 푦 ∈ 푈, so 푈 ∈ 푔(푦) and 푔(푥) ⊆ 푔(푦). Since 푋 and 푃퐹(퐶푈(푋, ≤), ⊆) are both Heyting spaces, 푋 is compact and 푃퐹(퐶푈(푋, ≤), ⊆) is Hausdorff, so 푔(푋) is closed in 푃퐹(퐶푈(푋, ≤), ⊆). Suppose towards a contradiction that 푔 is not surjective, then there is a prime filter 푃 of 퐶푈(푋, ≤) with 푃 ∉ 푔(푋). So, 푃 and 푔(푋) are closed and disjoint, and there are disjoint open sets 푈 and 푊 with 푔(푋) ⊆ 푈 and 푃 ∈ 푊. Since 푊 is a union of clopen sets, there is a clopen ′ set 푉 disjoint with 푔(푋) but 푃 ∈ 푉. Now, 푉 is compact. Let 푉 = 퐹푆 ∩ 퐹푇 for some 푆, 푇 ∈ 퐶푈(푋, ≤). Then,

−1 −1 ′ ′ ∅ = 푔 (푉) = 푔 (퐹푆 ∩ 퐹푇) = 푆 ∩ 푇

Thus, 푆 ⊆ 푇, which means 푉 = 퐹푆 ∩ 퐹푇 = ∅, so 푃 ∉ 푉 which is a contradiction. Hence, 푔 is surjective. Finally, we have to check that for every 푥 ∈ 푋 and 푃 ∈ 푃퐹(퐶푈(푋, ≤), ⊆) with 푔(푥) ⊆ 푃, there is a 푦 ∈ 푋 with 푥 ≤ 푦 and 푔(푦) = 푃. Since 푔 is surjective, there is a 푦 ∈ 푋 with 푃 = 푔(푦). Since 푔(푥) ⊆ 푔(푦) and 푔 preserves order, we have 푥 ≤ 푦. As a consequence, if ℌ and 픎 are isomorphic Heyting algebras, then 푃퐹(ℌ) and 푃퐹(픎) are isomorphic Heyting spaces. If 푋 and 푌 are isomorphic Heyting spaces, then 퐶푈(푋) and 퐶푈(푌) are isomorphic Heyting algebras.

7 1.4 Completeness of IPL Using Heyting duality, we get a topological semantics for Intuitionistic Propo- sitional Logic. We can use this to state a completeness theorem for IPL. To do so, we define the Lindenbaum-Tarski algebra by quotienting out the free algebra with the following congruence relation,

휑 ⟷ 휓 ≔ (휑 → 휓) ∧ (휓 → 휑)

By using,

∥ 휑 ∥≔ { 휓 ∣ ⊢ 휑 ↔ 휓 }

we define a Heyting algebra ℌ as follows.

0 ≔∥ ⊥ ∥ 1 ≔∥ ⊤ ∥ ∥ 휑 ∥ ∧ ∥ 휓 ∥ ≔∥ 휑 ∧ 휓 ∥ ∥ 휑 ∥ ∨ ∥ 휓 ∥ ≔∥ 휑 ∨ 휓 ∥ ∥ 휑 ∥→∥ 휓 ∥ ≔∥ 휑 → 휓 ∥ It is routine to check that this indeed satisfies the axioms of a Heyting algebra. The valuation function is defined as 픳(휑) ≔∥ 휑 ∥. Hence, if 픳(휑) = 1, we have that ⊢ 휑 ↔ ⊤, and hence ⊢ 휑. We say that ⊨ 휑 iff 퐹(∥ 휑 ∥) = 푃퐹(ℌ). By duality, completeness follows. Theorem (Soundness and Completeness). ⊢ 휑 iff ⊨ 퐹(∥ 휑 ∥)

1.5 Intuitionistic completeness The proofs of Stone and Heyting duality were carried out in classical set theory, and make use of the Law of the Excluded Third, or Tertium Non Datur (TND) in several places. Moreover, they rely on the use of the prime ideal theorem and the ultrafilter lemma, which are valid theorems in ZF set theory with choice. From the point of view of categorical logic, these theorems are valid in the internal logic of a boolean, well-pointed topos, where epis split. They will not hold in any general topos, for example, a presheaf topos where choice fails.

8 It is not possible to prove a strong completeness theorem for Intuitionistic propositional logic using an intuitionistic metatheory, as shown in McCarty et al. [1991].

9 References

Laura van Schijndel and NP Landsman. Stone and heyting duality for classical and intuitionistic propositional logic. 2017.

David Charles McCarty et al. Incompleteness in intuitionistic metamathematics. Notre Dame journal of formal logic, 32(3):323–358, 1991.

10