# Heyting algebra: Wikis

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# Encyclopedia

In mathematics, Heyting algebras are special partially ordered sets that constitute a generalization of Boolean algebras, named after Arend Heyting. Heyting algebras arise as models of intuitionistic logic, a logic in which the law of excluded middle does not in general hold. Complete Heyting algebras are a central object of study in pointless topology.

## Formal definition

A Heyting algebra H is a bounded lattice such that for all a and b in H there is a greatest element x of H such that

$a \wedge x \le b.$

This element is the relative pseudo-complement of a with respect to b, and is denoted ab. We write 1 and 0 for the largest and the smallest element of H, respectively.

In any Heyting algebra, one defines the pseudo-complement ¬x of any element x by setting ¬x = (x → 0). By definition, a ∧ ¬a = 0, and ¬a is the largest element having this property. However, it is not in general true that a ∨ ¬a = 1, thus ¬ is only a pseudo-complement, not a true complement, as would be the case in a Boolean algebra.

A complete Heyting algebra is a Heyting algebra that is a complete lattice.

A subalgebra of a Heyting algebra H is a subset H1 of H containing 0 and 1 and closed under the operations ∧, ∨ and →. It follows that it is also closed under ¬. A subalgebra is made into a Heyting algebra by the induced operations.

## Alternative definitions

### Lattice-theoretic definitions

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

$f_a: H \to H \mbox{ defined by } f_a(x)=a\wedge x,$

for some fixed a in H. A bounded lattice H is a Heyting algebra if and only if all mappings ƒa are the lower adjoint of a monotone Galois connection. In this case the respective upper adjoints ga are given by ga(x) = ax, 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 ab.

### Bounded lattice with an implication operation

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. $a\rightarrow a = 1$
2. $a\wedge(a\rightarrow b)=a\wedge b$
3. $b\wedge(a\rightarrow b)= b$
4. $a\rightarrow (b\wedge c)= (a\rightarrow b)\wedge (a\rightarrow c)$

where 4 is the distributive law for →.

### Characterization using the axioms of intuitionistic logic

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 1 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 0 and 1, then A is a Heyting algebra for these operations (and the relation ≤ defined by the condition that ab when ab = 1) if and only if the following conditions hold for any elements x, y and z of A:

1. $\mbox{If } x \to y = 1 \mbox{ and } y \to x = 1 \mbox{ then } x = y ,$
2. $\mbox{If } 1 \to y = 1, \mbox{ then } y = 1 ,$
3. $x \to (y \to x) = 1 ,$
4. $(x \to (y \to z)) \to ((x \to y) \to (x \to z)) = 1 ,$
5. $x \and y \to x = 1 ,$
6. $x \and y \to y = 1 ,$
7. $x \to (y \to (x \and y)) = 1 ,$
8. $x \to x \or y = 1 ,$
9. $y \to x \or y = 1 ,$
10. $(x \to z) \to ((y \to z) \to (x \or y \to z)) = 1 ,$
11. $0 \to x = 1 .$

Finally, we define ¬x to be x → 0.

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.

## Examples

The free Heyting algebra over one generator (aka Rieger–Nishimura lattice)
• Every Boolean algebra is a Heyting algebra, with pq given by ¬pq.
• Every totally ordered set that is a bounded lattice is also a Heyting algebra, where pq is equal to q when p>q, and 1 otherwise.
• The simplest Heyting algebra that is not already a Boolean algebra is the totally ordered set {0, ½, 1} with → defined as above. Notice that ½ ∨ ¬½ = ½ ∨ (½ → 0) = ½ ∨ 0 = ½ falsifies the law of excluded middle.
• Every topology provides a complete Heyting algebra in the form of its open set lattice. In this case, the element AB is the interior of the union of Ac and B, where Ac denotes the complement of the open set A. Not all complete Heyting algebras are of this form. These issues are studied in pointless topology, where complete Heyting algebras are also called frames or locales.

## Properties

### General properties

The ordering ≤ on a Heyting algebra H can be recovered from the operation → as follows: for any elements a, b of H, ab if and only if ab = 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.

### Provable identities

Given a formula F(A1, A2,…, An) of propositional calculus (using, in addition to the variables, the connectives ∧, ∨, ¬, →, 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 F(a1, a2,…, an) = 1 is true for any Heyting algebra H and any elements a1, a2,…, anH.

The implication 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 ab if and only if ab = 1, it follows from 1 → 2 that whenever a formula a formula FG is provably true, we have F(a1, a2,…, an) ≤ G(a1, a2,…, an) for any Heyting algebra H, and any elements a1, a2,…, anH. (It follows from the deduction theorem that FG is provable [from nothing] if and only if G is a provable from F, that is, if G is a provable consequence of F.) In particular, if F and G are provably equivalent, then F(a1, a2,…, an) = G(a1, a2,…, an), 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 F(A1, A2,…, An) is not provable, it is enough to exhibit a Heyting algebra H and elements a1, a2,..., anH such that F(a1, a2,…, an) ≠ 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 (ab) → c = a → (bc).

For more on the implication 2 → 1, see the section "Universal constructions" below.

### Distributivity

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

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

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, $\wedge$ preserves all existing suprema. Distributivity in turn is just the preservation of binary suprema by ∧.

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

$x\wedge\bigvee Y = \bigvee \{x\wedge y : y \in Y\}$

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

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

being its relative pseudo-complement operation.

### Regular and complemented elements

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

1. $x=\lnot\lnot x.$
2. $x=\lnot y \mbox{ for some } y \in H.$

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

Elements x and y of a Heyting algebra H are called complements to each other if xy = 0 and xy = 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;[1]
3. every x in H is complemented.[2]

In this case, the element ab is equal to ¬ab.

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

### The De Morgan laws in a Heyting algebra

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

$\lnot(x \vee y)=\lnot x \wedge \lnot y$, for all $x,y\in H$.

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

$\lnot(x \wedge y)= \lnot \lnot (\lnot x \vee \lnot y)$, for all $x,y\in H$.

The following statements are equivalent for all Heyting algebras H:

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

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

We prove the equivalence. Clearly 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 Hcomp is just the restriction of ∨ to Hcomp, taking into account the characterizations we have given of conditions 6 and 7. The implication 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.

## Heyting algebra morphisms

### Definition

Given two Heyting algebras H1 and H2 and a mapping ƒ : H1H2, we say that ƒ is a morphism of Heyting algebras if, for any elements x and y in H1, we have:

1. $f(0) = 0, \,$
2. $f(1) = 1, \,$
3. $f(x \and y) = f(x) \and f(y),$
4. $f(x \or y) = f(x) \or f(y),$
5. $f(x \to y) = f(x) \to f(y),$
6. $f(\lnot x) = \lnot f(x).$

We put condition 6 in brackets because it follows from the others, as ¬x is just x0, and one may or may not wish to consider ¬ to be a basic operation.

It follows from conditions 3 and 5 (or 1 alone, or 2 alone) that f is an increasing function, that is, that f(x) ≤ f(y) whenever xy.

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 5 (or 1 through 6) 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.

### Properties

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

### Examples

Given a Heyting algebra H and any subalgebra H1, the inclusion mapping i : H1H 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 Hreg. This is not in general a morphism from H to itself, since the join operation of Hreg may be different from that of H.

## Quotients

Let H be a Heyting algebra, and let FH. We call F a filter on H if it satisfies the following properties:

1. $1 \in F,$
2. $\mbox{If } x,y \in F \mbox{ then } x \and y \in F,$
3. $\mbox{If } x \in F, \ y \in H, \ \mbox{and } x \le y \mbox{ then } y \in F.$

The intersection of any set of filters on H is again a filter. Therefore, given any subset S of H there is a smallest filter containing S. We call it the filter generated by S. If S is empty, F = {1}. Otherwise, F is equal to the set of x in H such that there exist y1, y2, …, ynS with y1y2 ∧ … ∧ ynx.

If H is a Heyting algebra and F is a filter on H, we define a relation ∼ on H as follows: we write xy whenever xy and yx both belong to F. Then ∼ is an equivalence relation; we write H/F for the quotient set. There is a unique Heyting algebra structure on H/F such that the canonical surjection pF : HH/F becomes a Heyting algebra morphism. We call the Heyting algebra H/F the quotient of H by F.

Let S be a subset of a Heyting algebra H and let F be the filter generated by S. Then H/F satisfies the following universal property:

• Given any morphism of Heyting algebras ƒ : HH′ satisfying ƒ(y) = 1 for every yS, ƒ factors uniquely through the canonical surjection pF : HH/F. That is, there is a unique morphism ƒ′ : H/FH′ satisfying ƒ′pF = ƒ. The morphism ƒ′ is said to be induced by ƒ.

Let f : H1H2 be a morphism of Heyting algebras. The kernel of ƒ, written ker ƒ, is the set ƒ−1[{1}]. It is a filter on H1. (Care should be taken because this definition, if applied to a morphism of Boolean algebras, is dual to what would be called the kernel of the morphism viewed as a morphism of rings.) By the foregoing, ƒ induces a morphism ƒ′ : H1/(ker ƒ) → H2. It is an isomorphism of H1/(ker ƒ) onto the subalgebra ƒ[H1] of H2.

## Universal constructions

### Heyting algebra of propositional formulas in n variables up to intuitionist equivalence

The implication 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,..., An.
2. Endow L with a preorder ≼ by defining FG 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 FG induced by the preorder F≼G. (It is defined by FG if and only if FG and GF. 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 [FG], [FG], [FG] 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 FG and GF 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 xy if and only if xy=1. Since, by the deduction theorem, a formula FG 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.

### Free Heyting algebra on an arbitrary set of generators

In fact, the preceding construction can be carried out for any set of variables {Ai: iI} (possibly infinite). One obtains in this way the free Heyting algebra on the variables {Ai}, 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 〈ai: iI 〉, there is a unique morphism f:H0H satisfying f([Ai])=ai. The uniqueness of f is not difficult to see, and its existence results essentially from the implication 1→2 of the section "Provable identities" above, in the form of its corollary that whenever F and G are provably equivalent formulas, F(〈ai〉)=G(〈a i〉) for any family of elements 〈ai〉in H.

### Heyting algebra of formulas equivalent with respect to a theory T

Given a set of formulas T in the variables {Ai}, viewed as axioms, the same construction could have been carried out with respect to a relation FG defined on L to mean that G is a provable consequence of F and the set of axioms T. Let us denote by HT the Heyting algebra so obtained. Then HT satisfies the same universal property as H0 above, but with respect to Heyting algebras H and families of elements 〈ai〉 satisfying the property that J(〈ai〉)=1 for any axiom J(〈Ai〉) in T. (Let us note that HT, taken with the family of its elements 〈[Ai]〉, 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 implication 1→2 in "Provable identities" so that 1 reads "provably true from T," and 2 reads "any elements a1, a2,..., an in H satisfying the formulas of T."

The Heyting algebra HT 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 HT, and the family of its elements 〈[Ai]〉.

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

### Comparison to Lindenbaum algebras

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 BT in the variables {Ai} with respect to the axioms T is just our HTT1, where T1 is the set of all formulas of the form ¬¬FF, since the additional axioms of T1 are the only ones that need to be added in order to make all classical tautologies provable.

## Heyting algebras as applied to intuitionistic logic

If one interprets the axioms of the intuitionistic propositional logic as terms of a Heyting algebra, then they will evaluate to the largest element, 1, in any Heyting algebra under any assignment of values to the formula's variables. For instance, $(P \land Q) \rightarrow P$ is, by definition of the pseudo-complement, the largest element x such that $P \land Q \land x \le P$. This inequation is satisfied for any x, so the largest such x is 1.

Furthermore the rule of modus ponens allows us to derive the formula Q from the formulas P and P → Q. But in any Heyting algebra, if P has the value 1, and P → Q has the value 1, then it means that $P \land 1 \le Q$, and so $1 \land 1 \le Q$; it can only be that Q has the value 1.

This means that if a formula is deducible from the laws of intuitionistic logic, being derived from its axioms by way of the rule of modus ponens, then it will always have the value 1 in all Heyting algebras under any assignment of values to the formula's variables. However one can construct a Heyting algebra in which the value of Peirce's law is not always 1. Consider the 3-element algebra {0,½,1} as given above. If we assign ½ to P and 0 to Q, then the value of Peirce's law ((P → Q) → P) → P is ½. It follows that Peirce's law cannot be intuitionistically derived. See Curry-Howard isomorphism for the general context of what this implies in type theory.

The converse can be proven as well: if a formula always has the value 1, then it is deducible from the laws of intuitionistic logic, so the intuitionistically valid formulas are exactly those that always have a value of 1. This is similar to the notion that classically valid formulas are those formulas that have a value of 1 in the two-element Boolean algebra under any possible assignment of true and false to the formula's variables — that is, they are formulas which are tautologies in the usual truth-table sense. A Heyting algebra, from the logical standpoint, is then a generalization of the usual system of truth values, and its largest element 1 is analogous to 'true'. The usual two-valued logic system is a special case of a Heyting algebra, and the smallest non-trivial one, in which the only elements of the algebra are 1 (true) and 0 (false).

## Word problem

The word problem on free Heyting algebras is difficult.[3] The only known results are that the free Heyting algebra on one generator is infinite, and that the free complete Heyting algebra on one generator exists (and has one more element than the free Heyting algebra).

## Notes

1. ^ Rutherford (1965), Th.26.2 p.78.
2. ^ Rutherford (1965), Th.26.1 p.78.
3. ^ Peter T. Johnstone, Stone Spaces, (1982) Cambridge University Press, Cambridge, ISBN 0-521-23893-5. (See paragraph 4.11)

## References

• Rutherford, Daniel Edwin (1965). Introduction to Lattice Theory. Oliver and Boyd.
• F. Borceux, Handbook of Categorical Algebra 3, In Encyclopedia of Mathematics and its Applications, Vol. 53, Cambridge University Press, 1994.
• G. Gierz, K.H. Hoffmann, K. Keimel, J. D. Lawson, M. Mislove and D. S. Scott, Continuous Lattices and Domains, In Encyclopedia of Mathematics and its Applications, Vol. 93, Cambridge University Press, 2003.
• S. Ghilardi. Free Heyting algebras as bi-Heyting algebras, Math. Rep. Acad. Sci. Canada XVI., 6:240–244, 1992.