- Heyting algebra
In
mathematics , Heyting algebras are specialpartially ordered set s that constitute a generalization of Boolean algebras, named afterArend Heyting . Heyting algebras arise as models ofintuitionistic logic , a logic in which thelaw of excluded middle does not in general hold.Complete Heyting algebra s are a central object of study inpointless 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
:
This element is the relative pseudo-complement of "a" with respect to "b", and is denoted . We write 1 and 0 for the largest and the smallest element of "H", respectively.
In any Heyting algebra, one defines the pseudo-complement of any element "x" by setting . By definition, , and is the largest element having this property. However, it is not in general true that , 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 acomplete lattice .A subalgebra of a Heyting algebra "H" is a subset "H"1 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
: defined by
for some fixed "a" in "H". A bounded lattice "H" is a Heyting algebra
if and only if all mappings are the lower adjoint of a monotoneGalois connection . In this case the respective upper adjoints are given by , 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 .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:
#
#
#
# (distributive law for )Characterization using the axioms of intuitionist 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 "a" ≤ "b" when "a"→"b" = 1) if and only if the following conditions hold for any elements "x", "y" and "z" of "A":
: EQUIV: if "x" → "y" = 1 and "y" → "x" = 1, then "x" = "y" ("equivalent formulas should be identified");: MODUS-PONENS: if 1 → "y" = 1, then "y" = 1 ("provably true formulas are closed under modus ponens");: THEN-1: "x" → ("y" → "x") = 1: THEN-2: ("x" → ("y" → "z")) → (("x" → "y") → ("x" → "z")) = 1: AND-1: "x" ∧ "y" → "x" = 1: AND-2: "x" ∧ "y" → "y" = 1: AND-3: "x" → ("y" → ("x" ∧ "y")) = 1: OR-1: "x" → "x" ∨ "y" = 1: OR-2: "y" → "x" ∨ "y" = 1: OR-3: ("x" → "z") → (("y" → "z") → ("x" ∨ "y" → "z")) = 1: FALSE: 0 → "x" = 1
Finally, we define ¬"x" to be "x"→0.
Of course, if a different set of axioms were chosen for logic, we could modify ours accordingly.
Examples
* Every totally ordered set that is a bounded lattice is also a Heyting algebra, where and for all "a" other than 0.
* Every Boolean algebra is a Heyting algebra, with "p" → "q" given by ¬"p" ∨ "q".
* the simplest Heyting algebra that is not already a Boolean algebra is the linearly ordered set {0, ½, 1} equipped with the following operations:
width="30" | |:Notice that ½½ = ½ (½0) = ½0 = ½ falsifies the law of excluded middle.
* Every
* The
* The
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", "a"≤"b" if and only if "a"→"b"=1.
In contrast to some
Provable identities
Given a formula "F"("A"1, "A"2,..., "A""n") 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:
# the formula "F" is provably true in intuitionist propositional calculus;
# the identity "F"("a"1, "a"2,..., "a""n") = 1 is true for any Heyting algebra "H" and any elements "a"1, "a"2,..., "a""n" in "H".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
Since for any "a" and "b" in a Heyting algebra "H" we have "a"≤"b" if and only if "a"→"b"=1, it follows from 1→2 that whenever a formula a formula "F"→"G" is provably true, we have "F"("a"1, "a"2,..., "a""n")≤"G"("a"1, "a"2,..., "a""n") for any Heyting algebra "H", and any elements "a"1, "a"2,..., "a""n" of "H". (It follows from the deduction theorem that "F"→"G" 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"("a"1, "a"2,..., "a""n") = "G"("a"1, "a"2,..., "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
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"("A"1, "A"2,..., "A""n") is not provable, it is enough to exhibit a Heyting algebra "H" and elements "a"1, "a"2,..., "a""n" of "H" such that "F"("a"1, "a"2,..., "a""n")≠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 ("a"∧"b")→"c"="a"→("b"→"c").
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
#
#
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
By a similar argument, the following
:
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:
# .
# for some "y" in "H".The equivalence of these conditions can be restated simply as the identity , valid for all "x" in "H".
Elements "x" and "y" of a Heyting algebra "H" are called complements to each other if and . If it exists, any such "y" is unique and must in fact be equal to . We call an element "x" complemented if it admits a complement. It is true that "if" "x" is complemented, then so is , and then "x" and are complements to each other. However, confusingly, even if "x" is not complemented, may nonetheless have a complement (not equal to "x"). In any Heyting algebra, the elements 0 and 1 are complements to each other.
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; [Rutherford (1965), Th.26.2 p.78.]
# every "x" in "H" is complemented. [Rutherford (1965), Th.26.1 p.78.] In this case, the element is equal to .
The regular (resp. complemented) elements of any Heyting algebra "H" constitute a Boolean algebra "H"reg (resp. "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 ∨.
The De Morgan laws in a Heyting algebra
One of the two
The following statements are equivalent for all Heyting algebras "H":
# "H" satisfies both De Morgan laws;
# , for all ;
# , for all regular ;
# , for all "x", "y" in "H";
# , for all regular "x", "y" in "H";
# , for all regular "x", "y" in "H";
# , for all .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. 1→2, 2→3 and 4→5 are trivial. 3↔4 and 5↔6 result simply from the first De Morgan law and the definition of regular elements. 7 results from 6 taking and for "x" and "y" and using the identity . 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 implication 5→2 is a trivial consequence of the weak De Morgan law, taking and for "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 "H"1 and "H"2 and a mapping "f":"H"1→"H"2, we say that "f" is a morphism of Heyting algebras if, for any elements "x" and "y" in "H"1, we have:
#"f"("x"∧"y")="f"("x")∧"f"("y");
#"f"("x"∨"y")="f"("x")∨"f"("y");
#"f"(1)=1;
#"f"(0)=0;
#"f"("x"→"y")="f"("x")→"f"("y"); and
# ["f"(¬"x")=¬"f"("x")] .
We put condition 6 in brackets because it follows from the others, as ¬"x" is just "x"→"0", 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 "x"≤"y".
Assume "H"1 and "H"2 are structures with operations →, ∧, ∨ (and possibly ¬) and constants 0 and 1, and "f" is a surjective mapping from "H"1 to "H"2 with properties 1 through 5 (or 1 through 6) above. Then if "H"1 is a Heyting algebra, so too is "H"2. 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 "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.
Examples
Given a Heyting algebra "H" and any subalgebra "H"1, the inclusion mapping "i":"H"1→"H" is a morphism.
For any Heyting algebra "H", the map 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".
Quotients
Let "H" be a Heyting algebra, and let "F"⊆"H". We call "F" a filter on "H" if it satisfies the following properties:
#1∈"F";
#if "x","y"∈"F", then "x"∧"y"∈"F";
#if "x"∈"F", "y"∈"H", and "x"≤"y", then "y"∈"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 "y"1, "y"2, ..., "y""n" in "S" with "y"1∧"y"2∧...∧"y""n"≤"x".
If "H" is a Heyting algebra and "F" is a filter on "H", we define a relation ∼ on "H" as follows: we write "x"∼"y" whenever "x"→"y" and "y"→"x" 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 "p""F":"H"→"H"/"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 "f":"H"→"H′" satisfying "f"("y") = 1 for every "y" in "S", "f" factors uniquely through the canonical surjection "p""F":"H"→"H"/"F". That is, there is a unique morphism "f′":"H"/"F"→"H′" satisfying "f′p""F"="f". The morphism "f′" is said to be "induced" by "f".
Let "f":"H"1→"H"2 be a morphism of Heyting algebras. The kernel of "f", written ker "f", is the set "f"-1 [{1}] . It is a filter on "H"1. (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, "f" induces a morphism "f′":"H"1/(ker "f")→"H"2. It is an isomorphism of "H"1/(ker "f") onto the subalgebra "f" ["H"1] of "H"2.
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:
#Consider the set "L" of propositional formulas in the variables "A"1, "A"2,..., "A""n".
# 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.
# 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.)
# Let "H"0 be the quotient set "L"/∼. This will be the desired Heyting algebra.
# 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 "H"0="L"/∼. Further define 1 to be the class of provably true statements, and set 0= [⊥] .
# Verify that "H"0, together with these operations, is a Heyting algebra. We do this using the axiom-like definition of Heyting algebras. "H"0 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 "H"0 by the condition that "x"≤"y" if and only if "x"→"y"=1. Since, by the
Free Heyting algebra on an arbitrary set of generators
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 "H"0. 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":"H"0→"H" satisfying "f"( ["A""i"] )="a""i". 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"(〈"a""i"〉)="G"(〈"a""i"〉) for any family of elements 〈"a""i"〉in "H".
Heyting algebra of formulas equivalent with respect to a theory "T"
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 "H"0 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 "H"0, 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 "a"1, "a"2,..., "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 "H"0 on the same set of variables, by applying the universal property of "H"0 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.
Comparison to Lindenbaum algebras
The constructions we have just given play an entirely analogous role with respect to Heyting algebras to that of
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, 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
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
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
Word problem
The word problem on free Heyting algebras is difficult. [Peter T. Johnstone, "Stone Spaces", (1982) Cambridge University Press, Cambridge, ISBN 0-521-23893-5. "(See paragraph 4.11)"] 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
References
*
* 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.
ee also
*
External links
* (
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