Heyting algebra

Formal definition

Alternative definitions

Category-theoretic definition

Lattice-theoretic definitions

Bounded lattice with an implication operation

Characterization using the axioms of intuitionistic logic

Examples

• Every Boolean algebra is a Heyting algebra, with p→q given by ¬p∨q.
  
• Every totally ordered set that has a least element 0 and a greatest element 1 is a Heyting algebra. In this case p→q equals to 1 when p≤q, and q otherwise.
  
• The simplest Heyting algebra that is not already a Boolean algebra is the totally ordered set , yielding the operations:
  

  In this example, that 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 A→B 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. Every interior algebra provides a Heyting algebra in the form of its lattice of open elements. Every Heyting algebra is of this form as a Heyting algebra can be completed to a Boolean algebra by taking its free Boolean extension as a bounded distributive lattice and then treating it as a generalized topology in this Boolean algebra. The Lindenbaum algebra of propositional intuitionistic logic is a Heyting algebra. The global elements of the subobject classifier Ω of an elementary topos form a Heyting algebra; it is the Heyting algebra of truth values of the intuitionistic higher-order logic induced by the topos. Łukasiewicz–Moisil algebras are also Heyting algebras for any n. 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, 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, then the Heyting algebra is the trivial one-element Heyting algebra. Provable identities Given a formula of propositional calculus, it is a fact, proved early on in any study of Heyting algebras, that the following two conditions are equivalent: The formula F is provably true in intuitionist propositional calculus. The identity is true for any Heyting algebra H and any elements. The metaimplication 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 if and only if a→b = 1, it follows from that whenever a formula F→G is provably true, we have for any Heyting algebra H, and any elements. In particular, if F and G are provably equivalent, then, 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 is not provable, it is enough to exhibit a Heyting algebra H and elements such that. 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. For more on the metaimplication 2 ⇒ 1, see the section "Universal constructions" below. Distributivity Heyting algebras are always distributive. Specifically, we always have the identities 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, 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: 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 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: x = ¬¬x. 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. 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: H is a Boolean algebra; every x in H is regular; every x in H is complemented. In this case, the element is equal to The regular elements of any Heyting algebra H constitute a Boolean algebra Hreg, 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 we have 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 However, the other De Morgan law does not always hold. We have instead a weak de Morgan law: The following statements are equivalent for all Heyting algebras H: H satisfies both De Morgan laws, 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 the metaimplications and are trivial. Furthermore, and result simply from the first De Morgan law and the definition of regular elements. We show that by taking ¬x and ¬¬x in place of x and y in 6 and using the identity Notice that follows from the first De Morgan law, and 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 metaimplication 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 we say that ƒ is a morphism of Heyting algebras if, for any elements x and y in H1, we have: It follows from any of the last three conditions that f is an increasing function, that is, that whenever. Assume H1 and H2 are structures with operations →, ∧, ∨ 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 with an operation → satisfying certain identities. Properties The identity map from any Heyting algebra to itself is a morphism, and the composite of any two morphisms f and g is a morphism. Hence Heyting algebras form a category. Examples Given a Heyting algebra H and any subalgebra H1, the inclusion mapping is a morphism. For any Heyting algebra H, the map 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 We call F a filter on H if it satisfies the following properties: 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, Otherwise, F is equal to the set of x in H such that there exist with If H is a Heyting algebra and F is a filter on H, we define a relation ∼ on H as follows: we write whenever and both belong to F. Then ∼ is an equivalence relation; we write for the quotient set. There is a unique Heyting algebra structure on such that the canonical surjection becomes a Heyting algebra morphism. We call the Heyting algebra 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: Let be a morphism of Heyting algebras. The kernel of f, written ker f, is the set It is a filter on H1. By the foregoing, f induces a morphism It is an isomorphism of onto the subalgebra f of H2. Universal constructions Heyting algebra of propositional formulas in ''n'' variables up to intuitionist equivalence The metaimplication in the section "Provable identities" is proved by showing that the result of the following construction is itself a Heyting algebra: Consider the set L of propositional formulas in the variables A1, A2,..., An. Endow L with a preorder ≼ by defining F≼G if G is an logical consequence of F, that is, if G is provable from F. It is immediate that ≼ is a preorder. Consider the equivalence relation F∼G induced by the preorder F≼G. Let H0 be the quotient set L/∼. This will be the desired Heyting algebra. We write 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 , , and depend only on and . This defines operations →, ∧, ∨ and ¬ on the quotient set H0=L/∼. Further define 1 to be the class of provably true statements, and set 0=. 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. 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, hence =. 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 ≤ 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 . One obtains in this way the free Heyting algebra on the variables, 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: i∈I 〉, there is a unique morphism f:H0→H satisfying f=ai. The uniqueness of f is not difficult to see, and its existence results essentially from the metaimplication of the section "Provable identities" above, in the form of its corollary that whenever F and G are provably equivalent formulas, F=G 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, 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 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=1 for any axiom J in T. The existence and uniqueness of the morphism is proved the same way as for H0, except that one must modify the metaimplication 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 〈〉. 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. Now consider the set T of formulas J in the variables 〈Ai: i∈I〉 such that J=1. Then we obtain a morphism f:HT→H 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 with respect to the axioms T is just our HT∪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. 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 is, by definition of the pseudo-complement, the largest element x such that. 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, and so ; 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 as given above. If we assign ½ to P and 0 to Q, then the value of Peirce's law →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 and 0. Decision problems The problem of whether a given equation holds in every Heyting algebra was shown to be decidable by S. Kripke in 1965. The precise computational complexity of the problem was established by R. Statman in 1979, who showed it was PSPACE-complete and hence at least as hard as deciding equations of Boolean algebra and conjectured to be considerably harder. The elementary or first-order theory of Heyting algebras is undecidable. It remains open whether the universal Horn theory of Heyting algebras, or word problem, is decidable. À propos of the word problem it is known that Heyting algebras are not locally finite, in contrast to Boolean algebras which are locally finite and whose word problem is decidable. It is unknown whether free complete Heyting algebras exist except in the case of a single generator where the free Heyting algebra on one generator is trivially completable by adjoining a new top.