Jon Awbrey
Registered: Feb 2004
Posts: 558 
Language Of Cacti
LOC. Note 36
1.3.10.13. Stretching Exercises (concl.)
Taking up the preceding arrays of particular connections, namely,
the boolean functions on two or less variables, it is possible to
illustrate the use of the stretch operation in a variety of
concrete cases.
For example, suppose that F is a connection of the form F : %B%^2 > %B%,
that is, any one of the sixteen possibilities in Table 16, while p and q
are propositions of the form p, q : X > %B%, that is, propositions about
things in the universe X, or else the indicators of sets contained in X.
Then one has the imagination #f# = <f_1, f_2> = <p, q> : (X > %B%)^2,
and the stretch of the connection F to #f# on X amounts to a proposition
F^$ <p, q> : X > %B%, usually written as "F^$ (p, q)" and vocalized as
the "stretch of F to p and q". If one is concerned with many different
propositions about things in X, or if one is abstractly indifferent to
the particular choices for p and q, then one can detach the operator
F^$ : (X > %B%)^2 > (X > %B%), called the "stretch of F over X",
and consider it in isolation from any concrete application.
When the "cactus notation" is used to represent boolean functions,
a single "$" sign at the end of the expression is enough to remind
a reader that the connections are meant to be stretched to several
propositions on a universe X.
For instance, take the connection F : %B%^2 > %B% such that:
 F(x, y) = F^2_06 (x, y) = (x, y)
This connection is the boolean function on a couple of variables x, y
that yields a value of %1% if and only if just one of x, y is not %1%,
that is, if and only if just one of x, y is %1%. There is clearly an
isomorphism between this connection, viewed as an operation on the
boolean domain %B% = {%0%, %1%}, and the dyadic operation on binary
values x, y in !B! = GF(2) that is otherwise known as "x + y".
The same connection F : %B%^2 > %B% can also be read as a proposition
about things in the universe X = %B%^2. If S is a sentence that denotes
the proposition F, then the corresponding assertion says exactly what one
otherwise states by uttering "x is not equal to y". In such a case, one
has [S] = F, and all of the following expressions are ordinarily taken
as equivalent descriptions of the same set:
 [ [S] ]
= [ F ]
= F^(1)(%1%)
= {<x, y> in %B%^2 : S}
= {<x, y> in %B%^2 : F(x, y) = %1%}
= {<x, y> in %B%^2 : F(x, y)}
= {<x, y> in %B%^2 : (x, y) = %1%}
= {<x, y> in %B%^2 : (x, y) }
= {<x, y> in %B%^2 : x exclusiveor y}
= {<x, y> in %B%^2 : just one true of x, y}
= {<x, y> in %B%^2 : x not equal to y}
= {<x, y> in %B%^2 : x <=/=> y}
= {<x, y> in %B%^2 : x =/= y}
= {<x, y> in %B%^2 : x + y}.
Notice the slight distinction, that I continue to maintain at this point,
between the logical values {false, true} and the algebraic values {0, 1}.
This makes it legitimate to write a sentence directly into the right side
of the setbuilder expression, for instance, weaving the sentence S or the
sentence "x is not equal to y" into the context "{<x, y> in %B%^2 : ... }",
thereby obtaining the corresponding expressions listed above, while the
proposition F(x, y) can also be asserted more directly without equating
it to %1%, since it already has a value in {false, true}, and thus can
be taken as tantamount to an actual sentence.
If the appropriate safeguards can be kept in mind, avoiding all danger of
confusing propositions with sentences and sentences with assertions, then
the marks of these distinctions need not be forced to clutter the account
of the more substantive indications, that is, the ones that really matter.
If this level of understanding can be achieved, then it may be possible
to relax these restrictions, along with the absolute dichotomy between
algebraic and logical values, which tends to inhibit the flexibility
of interpretation.
This covers the properties of the connection F(x, y) = (x, y),
treated as a proposition about things in the universe X = %B%^2.
Staying with this same connection, it is time to demonstrate how
it can be "stretched" into an operator on arbitrary propositions.
To continue the exercise, let p and q be arbitrary propositions about
things in the universe X, that is, maps of the form p, q : X > %B%,
and suppose that p, q are indicator functions of the sets P, Q c X,
respectively. In other words, one has the following set of data:
 p = {P} : X > %B%
 q = {Q} : X > %B%
 <p, q> = < {P} , {Q} > : (X > %B%)^2
Then one has an operator F^$, the stretch of the connection F over X,
and a proposition F^$ (p, q), the stretch of F to <p, q> on X, with
the following properties:
 F^$ = ( , )^$ : (X > %B%)^2 > (X > %B%)
 F^$ (p, q) = (p, q)^$ : X > %B%
As a result, the application of the proposition F^$ (p, q) to each x in X
yields a logical value in %B%, all in accord with the following equations:
oo
 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` 
 ` F^$ (p, q)(x) ` = ` (p, q)^$ (x) `in `%B% ` 
 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` 
 ` ` ` ` ^ ` ` ` ` ` ` ` ` ` ^ ` ` ` ` ` ` ` ` ` 
 ` ` ` `  ` ` ` ` ` ` ` ` `  ` ` ` ` ` ` ` ` ` 
 ` ` ` ` = ` ` ` ` ` ` ` ` ` = ` ` ` ` ` ` ` ` ` 
 ` ` ` `  ` ` ` ` ` ` ` ` `  ` ` ` ` ` ` ` ` ` 
 ` ` ` ` v ` ` ` ` ` ` ` ` ` v ` ` ` ` ` ` ` ` ` 
 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` 
 ` F(p(x), q(x)) ` = ` (p(x), q(x))` in` %B% ` 
 ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` ` 
oo
For each choice of propositions p and q about things in X, the stretch of
F to p and q on X is just another proposition about things in X, a simple
proposition in its own right, no matter how complex its current expression
or its present construction as F^$ (p, q) = (p, q)^$ makes it appear in
relation to p and q. Like any other proposition about things in X, it
indicates a subset of X, namely, the fiber that is variously described
in the following ways:
 [ F^$ (p, q) ]
= [ (p, q)^$ ]
= (F^$ (p, q))^(1)(%1%)
= {x in X : F^$ (p, q)(x)}
= {x in X : (p, q)^$ (x)}
= {x in X : (p(x), q(x)) }
= {x in X : p(x) ± q(x)}
= {x in X : p(x) =/= q(x)}
= {x in X : {P} (x) =/= {Q} (x)}
= {x in X : x in P <=/=> x in Q}
= {x in X : x in PQ or x in QP}
= {x in X : x in PQ _ QP}
= {x in X : x in P ± Q}
= P ± Q c X
= [p] ± [q] c X.
Which was to be shown.
Jon Awbrey

Last edited by Jon Awbrey on 12202004 at 02:09 AM
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