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In a co-Heyting algebra it is possible to define a negation operator $\sim$ which validates the law of excluded middle but invalidates the law of non-contradiction dual to the negation in a Heyting algebra.
Let $a$ be an element of a co-Heyting algebra $L$ with subtraction $\backslash$. The co-Heyting negation of $a$ , called non a (Lawvere 1991) or the supplement of $a$, is defined as $\sim a:=1\backslash a$.
A bi-Heyting algebra is naturally equipped with two negation operators: the Heyting negation $\neg$ and the co-Heyting supplement $\sim$. Both coincide in a Boolean algebra considered as a bi-Heyting algebra.
In applications, such co-occuring pairs of negation operators $\neg$ and $\sim$ are the most interesting cases as their combination give rise to adjoint modalities e.g. in mereology they yield thickened boundaries (Stell&Worboys 1997). Beside mereology they have found applications in linguistics, intuitionistic logic and physics.
$\sim a$ minimally supplements $a$ to truth in the sense that $\sim a$ is the least $x$ with $a\vee x=1$. This follows from the adjointness of $\backslash$ which unwinds as $\sim a\leq x$ iff $1=a\vee x$.
$\sim 1=0$ and $\sim 0= 1$.
$\sim$ can be used to define the co-Heyting boundary operator $\partial :L\to L$ by $\partial a:=a\wedge\sim a$. That $\partial a$ is not necessary trivial is dual to the non-validity of the tertium non datur for general Heyting algebras and already points to the utility of the co-Heyting negation for paraconsistent logic.
The co-Heyting supplement satisfies the dual de Morgan rule $\sim (a\wedge b)=(\sim a)\vee(\sim b)$, but not $\sim (a\vee b) = (\sim a)\wedge (\sim b)$ in general.
For $a\in L$ define its core as $\sim\sim a$. Then $a=\partial a\vee(\sim\sim a)$. Call $a$ with $a=\sim\sim a$ regular. Lawvere (1986) proposes in the vein of classical mereology e.g. Tarski 1927 on regions as regular open sets, to consider only regular subbodies as bodies in the full sense.
Suppose the element $a$ of a co-Heyting algebra has a complement $x$ i.e. $a\vee x = 1$ and $a\wedge x = 0$, then the complement $x$ coincides with $\sim a$. Because from $a\vee x=1$ follows $\sim a\leq x$ since $\sim a$ is the least element with this property; conversely, $\sim a=\sim a\vee (a\wedge x)=(\sim a\vee a)\wedge (\sim a\vee x)=\sim a\vee x$ whence $x\leq \sim a$.
A complemented element is obviously regular. Conversely, a regular element $a$ is complemented since $0=\sim 1=\sim (a\vee\sim a)=(\sim a )\wedge (\sim\sim a)=\sim a\wedge a$.
In a bi-Heyting algebra: $\neg a=\neg a\wedge 1=\neg a\wedge (a\vee\sim a)=(\neg a\wedge a)\vee (\neg a\wedge \sim a)= 0\vee (\neg a\wedge\sim a)=\neg a\wedge\sim a$ hence $\neg a\leq\sim a$ and we see that $\neg$ is more strongly negative than $\sim$.
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Last revised on July 29, 2016 at 17:33:11. See the history of this page for a list of all contributions to it.