properties of Heyting algebras

Proposition 1.

Let H be a Brouwerian lattice. The following properties hold:

  1. 1.


  2. 2.


  3. 3.


  4. 4.



The first three equations are proved in this entry ( We prove the last equation here. For any xH, xa(bc) iff xabc iff xab and xac iff xab and xac iff x(ab)(ac). Hence the equation holds. ∎

Proposition 2.

Conversely, a latticeMathworldPlanetmath with a binary operationMathworldPlanetmath satisfying the four conditions above is a Brouwerian lattice.


Let H be a lattice with a binary operation on it satisfying the identitiesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath above. We want to show that xab iff xab for any xH. First, suppose xab. Then xaa(ab)=abb. Conversely, suppose xab. Then a(xa)ab by the property 6 in this entry ( As a result, x=x(ax)(aa)(ax)=a(ax)ab. ∎

Corollary 1.

The class of Brouwerian lattices is equational. The class of Heyting algebrasMathworldPlanetmath is equational.


The first fact is the result of the two propositionsPlanetmathPlanetmath above. The second comes from the fact that 0 is not used in the proofs of the propositions. ∎

Proposition 3.

Let H be a Heyting algebra. Then aa*=1 iff a**=a for all aH.


Suppose aa*=1. Since aa** in any Heyting algebra, we only need to show that a**a. Since H is distributive, we have a**=a**(aa*)=(a**a)(a**a*)=a**a. The last equation comes from the fact that a**a*=0. As a result, a**a. Conversely, suppose a**=a. Now, (aa*)*a*a**=0, and therefore aa*=(aa*)**=0*=1. ∎

Note, the last inequality in the proof above comes from the inequality (ab)*a*b*, which is a direct consequence of the fact that pseudocomplementation is order-reversing: xy implies that y*x*.

Corollary 2.

A Heyting algebra where psuedocomplentation * satisfies the equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath conditions above is a Boolean algebraMathworldPlanetmath. Conversely, a Boolean algebra with ab:=a*b is a Heyting algebra.


Since aa*=0 and aa*=1, the pseudocomplementation operationMathworldPlanetmath * is the complementation operation. And because any Heyting algebra is distributive, it is Boolean as a result. Conversely, assume B is Boolean. Then cab=a*b, so that caa(a*b)=abb. On the other hand, if cab, then cca*=(ca)a*a*b=ab. ∎

Proposition 4.

A subset F of a Heyting algebra H is an ultrafilterMathworldPlanetmath iff there is a Heyting algebra homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath f:H{0,1} with F=f-1(1).


First, assume f:H{0,1} is a Heyting algebra homomorphism, and F=f-1(1). Clearly, F is a filter. Suppose 0aF, then f(a)=0. Now, f(a*)=f(a)*=0*=1, so a*F. If F is not maximal, let G be a proper filter containing F and a, then a*G, so that 0aa*G, and hence G=H, contradicting the fact that G is proper. So F is maximal.

Conversely, suppose F is an ultrafilter of H. Define f:H{0,1} by f(x)=1 iff xF. Let a,bH. We first show that f is a lattice homomorphismMathworldPlanetmath:

  • First, f(ab)=1 iff abF iff a,bF (since F is a filter) iff f(a)=f(b)=1. So f respects .

  • Next, if f(ab)=0, then abF, which means neither a nor b is in F, or that f(a)=f(b)=0. On the other hand, if f(a)=f(b)=0, then neither a nor b is in F, since F is an ultrafilter. As a result, neither is abF, which means f(ab)=0. So f respects .

So f is a lattice homomorphism. Next, we show that f is a Heyting algebra homomorphism, which means showing that f respects : f(ab)=f(a)f(b). It suffices to show f(ab)=0 iff f(a)=1 and f(b)=0.

  • First, if f(a)=1 and f(b)=0 then aF and bF. If abF, then (ab)aF. Since (ab)ab, bF, a contradictionMathworldPlanetmathPlanetmath. So abF.

  • On the other hand, suppose f(ab)=0. So abF. Now, since bab, bF, or f(b)=0. If f(a)=0, then aF, so there is some cF with 0=ac. But this means ca*, or a*F. Since a*ab, we would have abF, a contradiction. Hence f(a)=1.

Therefore f is a Heyting algebra homomorphism. ∎

In the proof above, we use the fact that, for any ultrafilter F in a bounded latticeMathworldPlanetmath L, if xF, then there is yF such that 0=xy (for otherwise, the filter generated by x and F would be proper and properly contains F, contradicting the maximality of F). If in addition L were distributive, then abF implies that either aF or bF. To see this, suppose aF. Then there is cF such that 0=ac. Similarly, if bF, there is dF such that 0=bd. Let e=cdF. So e0, and ae=0=be. Furthermore, 0=(ae)(be)=(ab)e. If abF, so would 0F, a contradiction. Hence abF.

Title properties of Heyting algebras
Canonical name PropertiesOfHeytingAlgebras
Date of creation 2013-03-22 19:31:45
Last modified on 2013-03-22 19:31:45
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 19
Author CWoo (3771)
Entry type Definition
Classification msc 03G10
Classification msc 06D20