*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **ties of tuples. So, R _ S is the \either-R-or-S " property. Now since R+ and
S + are the sets of tuples for which the properties R and S , respectively, are believed
to hold, the set of tuples for which the property \either-R-or-S " is believed to hold is
clearly R+ S +. Moreover, since R? and S ? are the sets of tuples for which properties
R and S , respectively, are believed to not hold, the set of tuples for which the property
\either-R-or-S " is believed to not hold is similarly R? \ S ?.
7 In an informal symbolic notation, R+ = ft j R(t)g, and R? = ft j not R(t)g.
Now, (R _ S )+ = ft j R(t) or S (t)g, which is R+ S +. Similarly, (R _ S )? =
ft j not (R(t) or S (t))g = ft j not R(t) and not S (t)g, which is R? \ S ?.
The de nition of complement and of all the other operators on paraconsistent relations
de ned later can (and should) be understood in the same way.
_
Proposition 3 The operators _ and unary ? on paraconsistent relations are strong generalisations of the usual operators and unary ? on ordinary relations. Proof Let R and S be consistent relations on scheme . Then comps (R _ S ) is the
set fQ j R+ S + Q ( ) ? (R? \ S ?)g. This set is the same as the set fr s j R+
r ( ) ? R?; S + s ( ) ? S ?g, which is S ( )(comps (R); comps (S )). Such a
_
result for unary ? can also be shown similarly. 2
For sake of completeness, we de ne the following two related set-theoretic operators
on paraconsistent relations: De nition 9 Let R and S be paraconsistent relations on scheme . Then,
_
(a) the intersection of R and S , denoted R \ S , is a paraconsistent relation on scheme
, given by _
(R \ S )+ = R+ \ S +; _
(R \ S )? = R? S ?; _
(b) the di erence of R and S , denoted R ? S , is a paraconsistent relation on scheme ,
given by _
(R ? S )+ = R+ \ S ?; _
(R ? S )? = R? S +: 2 Again, to obtain an intuitive grasp of such de nitions, let us consider the di erence
_
operator. We should interpret R ? S as the \R-but-not-S " property. Now the tuples that
_
de nitely have this property are exactly those in R+ \ S ?. Thus, (R ? S )+ = R+ \ S ?.
Moreover, the tuples that de nitely do not have this property are the ones in R? S +.
_
Hence, (R ? S )? = R? S +.
_
In our informal symbolic notation, (R ? S )+ = ft j R(t) and not S (t)g, which is
+ \ S ?. Similarly, (R ? S )? = ft j not (R(t) and not S (t))g = ft j not R(t) or S (t)g,
_
R
? S +.
which is R
Although we have given independent de nitions of intersection and di erence, these
two operators can be derived from the fundamental operators union and complement as
intuitively expected.
8 Proposition 4 For any paraconsistent relations R and S on a common scheme, we have
__
_
_
R \ S = ?(?R _ ?S ); and
_
__
R ? S = ?(?R _ S ):
__
_
_
_
_
_
_
Proof (?(?R _ ?S ))+ = (?R _ ?S )? = (?R)? \ (?S )? = R+ \ S + = (R \ S )+.
? = (R \ S )? . The second part of the result can be shown
_ (?R _ ?S ))
_
_
_
Similarly, (?
similarly. 2
Table 1 gives some more algebraic laws involving the set-theoretic opera...

View
Full
Document