Heyting algebra

A Heyting algebra ${\displaystyle H}$ is a bounded lattice that has all exponential objects.

The lattice ${\displaystyle H}$ is regarded as a category where meet, ${\displaystyle \wedge }$, is the product. The exponential condition means that for any objects ${\displaystyle Y}$ and ${\displaystyle Z}$ in ${\displaystyle H}$ an exponential ${\displaystyle Z^{Y}}$ uniquely exists as an object in ${\displaystyle H}$.

A Heyting implication (often written using ${\displaystyle \Rightarrow }$ or ${\displaystyle \multimap }$ to avoid confusions such as the use of ${\displaystyle \to }$ to indicate a functor) is just an exponential: ${\displaystyle Y\Rightarrow Z}$ is an alternative notation for ${\displaystyle Z^{Y}}$. From the definition of exponentials we have that implication (${\displaystyle \Rightarrow :H\times H\to H}$) is right adjoint to meet (${\displaystyle \wedge :H\times H\to H}$). This adjunction can be written as ${\displaystyle (-\wedge Y)\dashv (Y\Rightarrow -)}$ or more fully as:

$${\displaystyle (-\wedge Y):H{\stackrel {\longrightarrow }{\underset {\longleftarrow }{\top }}}H:(Y\Rightarrow -)}$$

An equivalent definition of Heyting algebras can be given by considering the mappings:

${\displaystyle {\begin{cases}f_{a}\colon H\to H\\f_{a}(x)=a\wedge x\end{cases}}}$

for some fixed a in H. A bounded lattice H is a Heyting algebra if and only if every mapping f_a is the lower adjoint of a monotone Galois connection. In this case the respective upper adjoint g_a is given by g_a(x) = a→x, where → is defined as above.

Yet another definition is as a residuated lattice whose monoid operation is ∧. The monoid unit must then be the top element 1. Commutativity of this monoid implies that the two residuals coincide as a→b.

Given a bounded lattice A with largest and smallest elements 1 and 0, and a binary operation →, these together form a Heyting algebra if and only if the following hold:

1. ${\displaystyle a\to a=1}$
2. ${\displaystyle a\wedge (a\to b)=a\wedge b}$
3. ${\displaystyle b\wedge (a\to b)=b}$
4. ${\displaystyle a\to (b\wedge c)=(a\to b)\wedge (a\to c)}$

where equation 4 is the distributive law for →.

This characterization of Heyting algebras makes the proof of the basic facts concerning the relationship between intuitionist propositional calculus and Heyting algebras immediate. (For these facts, see the sections "Provable identities" and "Universal constructions".) One should think of the element ${\displaystyle \top }$ as meaning, intuitively, "provably true." Compare with the axioms at Intuitionistic logic#Axiomatization ).

Given a set A with three binary operations →, ∧ and ∨, and two distinguished elements ${\displaystyle \bot }$ and ${\displaystyle \top }$, then A is a Heyting algebra for these operations (and the relation ≤ defined by the condition that ${\displaystyle a\leq b}$ when a→b = ${\displaystyle \top }$) if and only if the following conditions hold for any elements x, y and z of A:

1. ${\displaystyle {\mbox{If }}x\leq y{\mbox{ and }}y\leq x{\mbox{ then }}x=y,}$
2. ${\displaystyle {\mbox{If }}\top \leq y,{\mbox{ then }}y=\top ,}$
3. ${\displaystyle x\leq y\to x,}$
4. ${\displaystyle x\to (y\to z)\leq (x\to y)\to (x\to z),}$
5. ${\displaystyle x\land y\leq x,}$
6. ${\displaystyle x\land y\leq y,}$
7. ${\displaystyle x\leq y\to (x\land y),}$
8. ${\displaystyle x\leq x\lor y,}$
9. ${\displaystyle y\leq x\lor y,}$
10. ${\displaystyle x\to z\leq (y\to z)\to (x\lor y\to z),}$
11. ${\displaystyle \bot \leq x.}$

Finally, we define ¬x to be x→ ${\displaystyle \bot }$.

Condition 1 says that equivalent formulas should be identified. Condition 2 says that provably true formulas are closed under modus ponens. Conditions 3 and 4 are then conditions. Conditions 5, 6 and 7 are and conditions. Conditions 8, 9 and 10 are or conditions. Condition 11 is a false condition.

Of course, if a different set of axioms were chosen for logic, we could modify ours accordingly.

The ordering ${\displaystyle \leq }$ on a Heyting algebra H can be recovered from the operation → as follows: for any elements a, b of H, ${\displaystyle a\leq b}$ if and only if a→b = 1.

In contrast to some many-valued logics, Heyting algebras share the following property with Boolean algebras: if negation has a fixed point (i.e. ¬a = a for some a), then the Heyting algebra is the trivial one-element Heyting algebra.

Given a formula ${\displaystyle F(A_{1},A_{2},\ldots ,A_{n})}$ of propositional calculus (using, in addition to the variables, the connectives ${\displaystyle \land ,\lor ,\lnot ,\to }$, and the constants 0 and 1), it is a fact, proved early on in any study of Heyting algebras, that the following two conditions are equivalent:

1. The formula F is provably true in intuitionist propositional calculus.
2. The identity ${\displaystyle F(a_{1},a_{2},\ldots ,a_{n})=1}$ is true for any Heyting algebra H and any elements ${\displaystyle a_{1},a_{2},\ldots ,a_{n}\in H}$.

The metaimplication 1 ⇒ 2 is extremely useful and is the principal practical method for proving identities in Heyting algebras. In practice, one frequently uses the deduction theorem in such proofs.

Since for any a and b in a Heyting algebra H we have ${\displaystyle a\leq b}$ if and only if a→b = 1, it follows from 1 ⇒ 2 that whenever a formula F→G is provably true, we have ${\displaystyle F(a_{1},a_{2},\ldots ,a_{n})\leq G(a_{1},a_{2},\ldots ,a_{n})}$ for any Heyting algebra H, and any elements ${\displaystyle a_{1},a_{2},\ldots ,a_{n}\in H}$. (It follows from the deduction theorem that F→G is provable (unconditionally) if and only if G is provable from F, that is, if G is a provable consequence of F.) In particular, if F and G are provably equivalent, then ${\displaystyle F(a_{1},a_{2},\ldots ,a_{n})=G(a_{1},a_{2},\ldots ,a_{n})}$, since ≤ is an order relation.

1 ⇒ 2 can be proved by examining the logical axioms of the system of proof and verifying that their value is 1 in any Heyting algebra, and then verifying that the application of the rules of inference to expressions with value 1 in a Heyting algebra results in expressions with value 1. For example, let us choose the system of proof having modus ponens as its sole rule of inference, and whose axioms are the Hilbert-style ones given at Intuitionistic logic#Axiomatization. Then the facts to be verified follow immediately from the axiom-like definition of Heyting algebras given above.

1 ⇒ 2 also provides a method for proving that certain propositional formulas, though tautologies in classical logic, cannot be proved in intuitionist propositional logic. In order to prove that some formula ${\displaystyle F(A_{1},A_{2},\ldots ,A_{n})}$ is not provable, it is enough to exhibit a Heyting algebra H and elements ${\displaystyle a_{1},a_{2},\ldots ,a_{n}\in H}$ such that ${\displaystyle F(a_{1},a_{2},\ldots ,a_{n})\neq 1}$.

If one wishes to avoid mention of logic, then in practice it becomes necessary to prove as a lemma a version of the deduction theorem valid for Heyting algebras: for any elements a, b and c of a Heyting algebra H, we have ${\displaystyle (a\land b)\to c=a\to (b\to c)}$.

For more on the metaimplication 2 ⇒ 1, see the section "Universal constructions" below.

Heyting algebras are always distributive. Specifically, we always have the identities

1. ${\displaystyle a\wedge (b\vee c)=(a\wedge b)\vee (a\wedge c)}$
2. ${\displaystyle a\vee (b\wedge c)=(a\vee b)\wedge (a\vee c)}$

The distributive law is sometimes stated as an axiom, but in fact it follows from the existence of relative pseudo-complements. The reason is that, being the lower adjoint of a Galois connection, ${\displaystyle \wedge }$ preserves all existing suprema. Distributivity in turn is just the preservation of binary suprema by ${\displaystyle \wedge }$.

By a similar argument, the following infinite distributive law holds in any complete Heyting algebra:

${\displaystyle x\wedge \bigvee Y=\bigvee \{x\wedge y\mid y\in Y\}}$

for any element x in H and any subset Y of H. Conversely, any complete lattice satisfying the above infinite distributive law is a complete Heyting algebra, with

${\displaystyle a\to b=\bigvee \{c\mid a\land c\leq b\}}$

being its relative pseudo-complement operation.

An element x of a Heyting algebra H is called regular if either of the following equivalent conditions hold:

1. x = ¬¬x.
2. x = ¬y for some y in H.

The equivalence of these conditions can be restated simply as the identity ¬¬¬x = ¬x, valid for all x in H.

Elements x and y of a Heyting algebra H are called complements to each other if x∧y = 0 and x∨y = 1. If it exists, any such y is unique and must in fact be equal to ¬x. We call an element x complemented if it admits a complement. It is true that if x is complemented, then so is ¬x, and then x and ¬x are complements to each other. However, confusingly, even if x is not complemented, ¬x may nonetheless have a complement (not equal to x). In any Heyting algebra, the elements 0 and 1 are complements to each other. For instance, it is possible that ¬x is 0 for every x different from 0, and 1 if x = 0, in which case 0 and 1 are the only regular elements.

Any complemented element of a Heyting algebra is regular, though the converse is not true in general. In particular, 0 and 1 are always regular.

For any Heyting algebra H, the following conditions are equivalent:

1. H is a Boolean algebra;
2. every x in H is regular;[9]
3. every x in H is complemented.[10]

In this case, the element a→b is equal to ¬a ∨ b.

The regular (respectively complemented) elements of any Heyting algebra H constitute a Boolean algebra H_reg (respectively H_comp), in which the operations ∧, ¬ and →, as well as the constants 0 and 1, coincide with those of H. In the case of H_comp, the operation ∨ is also the same, hence H_comp is a subalgebra of H. In general however, H_reg will not be a subalgebra of H, because its join operation ∨_reg may be differ from ∨. For x, y ∈ H_reg, we have x ∨_reg y = ¬(¬x ∧ ¬y). See below for necessary and sufficient conditions in order for ∨_reg to coincide with ∨.

One of the two De Morgan laws is satisfied in every Heyting algebra, namely

${\displaystyle \forall x,y\in H:\qquad \lnot (x\vee y)=\lnot x\wedge \lnot y.}$

However, the other De Morgan law does not always hold. We have instead a weak de Morgan law:

${\displaystyle \forall x,y\in H:\qquad \lnot (x\wedge y)=\lnot \lnot (\lnot x\vee \lnot y).}$

The following statements are equivalent for all Heyting algebras H:

1. H satisfies both De Morgan laws,
2. ${\displaystyle \lnot (x\wedge y)=\lnot x\vee \lnot y{\mbox{ for all }}x,y\in H,}$
3. ${\displaystyle \lnot (x\wedge y)=\lnot x\vee \lnot y{\mbox{ for all regular }}x,y\in H,}$
4. ${\displaystyle \lnot \lnot (x\vee y)=\lnot \lnot x\vee \lnot \lnot y{\mbox{ for all }}x,y\in H,}$
5. ${\displaystyle \lnot \lnot (x\vee y)=x\vee y{\mbox{ for all regular }}x,y\in H,}$
6. ${\displaystyle \lnot (\lnot x\wedge \lnot y)=x\vee y{\mbox{ for all regular }}x,y\in H,}$
7. ${\displaystyle \lnot x\vee \lnot \lnot x=1{\mbox{ for all }}x\in H.}$

Condition 2 is the other De Morgan law. Condition 6 says that the join operation ∨_reg on the Boolean algebra H_reg of regular elements of H coincides with the operation ∨ of H. Condition 7 states that every regular element is complemented, i.e., H_reg = H_comp.

We prove the equivalence. Clearly the metaimplications 1 ⇒ 2, 2 ⇒ 3 and 4 ⇒ 5 are trivial. Furthermore, 3 ⇔ 4 and 5 ⇔ 6 result simply from the first De Morgan law and the definition of regular elements. We show that 6 ⇒ 7 by taking ¬x and ¬¬x in place of x and y in 6 and using the identity a ∧ ¬a = 0. Notice that 2 ⇒ 1 follows from the first De Morgan law, and 7 ⇒ 6 results from the fact that the join operation ∨ on the subalgebra H_comp is just the restriction of ∨ to H_comp, taking into account the characterizations we have given of conditions 6 and 7. The metaimplication 5 ⇒ 2 is a trivial consequence of the weak De Morgan law, taking ¬x and ¬y in place of x and y in 5.

Heyting algebras satisfying the above properties are related to De Morgan logic in the same way Heyting algebras in general are related to intuitionist logic.

Given two Heyting algebras H₁ and H₂ and a mapping f : H₁ → H₂, we say that ƒ is a morphism of Heyting algebras if, for any elements x and y in H₁, we have:

1. ${\displaystyle f(0)=0,}$
2. ${\displaystyle f(x\land y)=f(x)\land f(y),}$
3. ${\displaystyle f(x\lor y)=f(x)\lor f(y),}$
4. ${\displaystyle f(x\to y)=f(x)\to f(y),}$

It follows from any of the last three conditions (2, 3, or 4) that f is an increasing function, that is, that f(x) ≤ f(y) whenever x ≤ y.

Assume H₁ and H₂ are structures with operations →, ∧, ∨ (and possibly ¬) and constants 0 and 1, and f is a surjective mapping from H₁ to H₂ with properties 1 through 4 above. Then if H₁ is a Heyting algebra, so too is H₂. This follows from the characterization of Heyting algebras as bounded lattices (thought of as algebraic structures rather than partially ordered sets) with an operation → satisfying certain identities.

The identity map f(x) = x from any Heyting algebra to itself is a morphism, and the composite g ∘ f of any two morphisms f and g is a morphism. Hence Heyting algebras form a category.

Given a Heyting algebra H and any subalgebra H₁, the inclusion mapping i : H₁ → H is a morphism.

For any Heyting algebra H, the map x ↦ ¬¬x defines a morphism from H onto the Boolean algebra of its regular elements H_reg. This is not in general a morphism from H to itself, since the join operation of H_reg may be different from that of H.

The metaimplication 2 ⇒ 1 in the section "Provable identities" is proved by showing that the result of the following construction is itself a Heyting algebra:

1. Consider the set L of propositional formulas in the variables A₁, A₂,..., A_n.
2. Endow L with a preorder ≼ by defining F≼G if G is an (intuitionist) logical consequence of F, that is, if G is provable from F. It is immediate that ≼ is a preorder.
3. Consider the equivalence relation F∼G induced by the preorder F≼G. (It is defined by F∼G if and only if F≼G and G≼F. In fact, ∼ is the relation of (intuitionist) logical equivalence.)
4. Let H₀ be the quotient set L/∼. This will be the desired Heyting algebra.
5. We write [F] for the equivalence class of a formula F. Operations →, ∧, ∨ and ¬ are defined in an obvious way on L. Verify that given formulas F and G, the equivalence classes [F→G], [F∧G], [F∨G] and [¬F] depend only on [F] and [G]. This defines operations →, ∧, ∨ and ¬ on the quotient set H₀=L/∼. Further define 1 to be the class of provably true statements, and set 0=[⊥].
6. Verify that H₀, together with these operations, is a Heyting algebra. We do this using the axiom-like definition of Heyting algebras. H₀ satisfies conditions THEN-1 through FALSE because all formulas of the given forms are axioms of intuitionist logic. MODUS-PONENS follows from the fact that if a formula ⊤→F is provably true, where ⊤ is provably true, then F is provably true (by application of the rule of inference modus ponens). Finally, EQUIV results from the fact that if F→G and G→F are both provably true, then F and G are provable from each other (by application of the rule of inference modus ponens), hence [F]=[G].

As always under the axiom-like definition of Heyting algebras, we define ≤ on H₀ by the condition that x≤y if and only if x→y=1. Since, by the deduction theorem, a formula F→G is provably true if and only if G is provable from F, it follows that [F]≤[G] if and only if F≼G. In other words, ≤ is the order relation on L/∼ induced by the preorder ≼ on L.

In fact, the preceding construction can be carried out for any set of variables {A_i : i∈I} (possibly infinite). One obtains in this way the free Heyting algebra on the variables {A_i}, which we will again denote by H₀. It is free in the sense that given any Heyting algebra H given together with a family of its elements 〈a_i: i∈I 〉, there is a unique morphism f:H₀→H satisfying f([A_i])=a_i. The uniqueness of f is not difficult to see, and its existence results essentially from the metaimplication 1 ⇒ 2 of the section "Provable identities" above, in the form of its corollary that whenever F and G are provably equivalent formulas, F(〈a_i〉)=G(〈a_i〉) for any family of elements 〈a_i〉in H.

Given a set of formulas T in the variables {A_i}, viewed as axioms, the same construction could have been carried out with respect to a relation F≼G defined on L to mean that G is a provable consequence of F and the set of axioms T. Let us denote by H_T the Heyting algebra so obtained. Then H_T satisfies the same universal property as H₀ above, but with respect to Heyting algebras H and families of elements 〈a_i〉 satisfying the property that J(〈a_i〉)=1 for any axiom J(〈A_i〉) in T. (Let us note that H_T, taken with the family of its elements 〈[A_i]〉, itself satisfies this property.) The existence and uniqueness of the morphism is proved the same way as for H₀, except that one must modify the metaimplication 1 ⇒ 2 in "Provable identities" so that 1 reads "provably true from T," and 2 reads "any elements a₁, a₂,..., a_n in H satisfying the formulas of T."

The Heyting algebra H_T that we have just defined can be viewed as a quotient of the free Heyting algebra H₀ on the same set of variables, by applying the universal property of H₀ with respect to H_T, and the family of its elements 〈[A_i]〉.

Every Heyting algebra is isomorphic to one of the form H_T. To see this, let H be any Heyting algebra, and let 〈a_i: i∈I〉 be a family of elements generating H (for example, any surjective family). Now consider the set T of formulas J(〈A_i〉) in the variables 〈A_i: i∈I〉 such that J(〈a_i〉)=1. Then we obtain a morphism f:H_T→H by the universal property of H_T, which is clearly surjective. It is not difficult to show that f is injective.

The constructions we have just given play an entirely analogous role with respect to Heyting algebras to that of Lindenbaum algebras with respect to Boolean algebras. In fact, The Lindenbaum algebra B_T in the variables {A_i} with respect to the axioms T is just our H_T∪T1, where T₁ is the set of all formulas of the form ¬¬F→F, since the additional axioms of T₁ are the only ones that need to be added in order to make all classical tautologies provable.