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'Heyting algebra'
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| Title of object: |
Heyting algebra |
| Canonical Name: |
HeytingAlgebra |
| Type: |
Definition |
| Created on: |
2007-01-09 15:20:58 |
| Modified on: |
2007-01-23 18:45:08 |
| Classification: |
msc:06D20 |
Preamble:
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Content:
A \emph{Heyting algebra} $L$ is a Brouwerian lattice with $0$. Equivalently, it is a relatively pseudocomplemented and pseudocomplemented lattice.
Let $a^*$ denote the pseudocomplement of $a$ and $a\to b$ the pseudocomplement of $a$ relative to $b$. Then we have the following properties:
\begin{enumerate}
\item $a^*=a\to 0$ (equivalence of definitions)
\item $a^*=0$ iff $a=1$ (if $a^*=0$, then $0=a\to 0$. But this means $c\wedge a=0$ iff $c=0$, or that $c\wedge a=c$ for all $c$. So $a=1$.)
\item $a^*=1$ iff $a=0$ (dual of the above statement)
\item $a\le a^{**}$ and $a^*=a^{***}$ (already true in any pseudocomplemented lattice)
\item $a^*\le a\to b$ (since $a^*\wedge a=0\le b$)
\item $(a\to b)\wedge (a\to b^*)=a^*$
\begin{proof}
If $c\wedge a=0$, then $c\wedge a\le b$ so $c\le (a\to b)$, and $c\le (a\to b^*)$ likewise, so $c\le (a\to b)\wedge (a\to b^*)$. This means precisely that $a^*=(a\to b)\wedge (a\to b^*)$.
\end{proof}
\item $a\to b\le b^*\to a^*$ (since $(a\to b)\wedge b^*\le (a\to b)\wedge (a\to b^*)=a^*)$
\item $a^*\vee b\le a\to b$ (since $b\wedge a\le b$ and $a^* \wedge a=0\le b$)
\end{enumerate}
Note that in property 4, $a\le a^{**}$, whereas $a^{**}\le a$ is in general not true, contrasting with the equality $a=a^{\prime\prime}$ in a Boolean lattice, where $^{\prime}$ is the complement operator. It can be shown that if $a^{**}\le a$ for all $a$ in a Heyting algebra $L$, then $L$ is a Boolean lattice. In this case, the pseudocomplement coincides with the complement of an element $a^*=a^{\prime}$, and we have the equality in property 7: $a^*\vee b=a\to b$, meaning that the concept of \PMlinkname{relative pseudocomplementation}{RelativelyPseudocomplemented} coincides with the material implication in classical propositional logic.
\textbf{Remark.} In the literature, the assumption that a Heyting algebra contains $0$ is sometimes dropped. Here, we call it a Brouwerian lattice instead. |
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