ONT Re: Propositional Equation Reasoning Systems
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PERS. Note 13
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Analysis of Contingent Expressions (cont.)
We are in the process of examining various proofs of the
propositional equation "(p (q))(p (r)) = (p (q r))", and
viewing them in the light of their character as semiotic
processes, in essence, as sign-theoretic transformations.
Here is a reminder of the equation in question:
o-----------------------------------------------------------o
| Equation E1 |
o-----------------------------------------------------------o
| |
| q r |
| q o o r o |
| | | | |
| p o o p p o |
| \ / | |
| @ = @ |
| |
| (p (q)) (p (r)) = (p (q r)) |
| |
| [p=>q] & [p=>r] = [p=>[q&r]] |
| |
o-----------------------------------------------------------o
The second way of establishing the truth of this equation is
one that I see, rather loosely, as "model-theoretic", for no
better reason than the sense of its ending with a pattern of
expression, a variant of the "disjunctive normal form" (DNF),
that is commonly recognized to be the form that one extracts
from a truth table by pulling out the rows of the table that
evaluate to true and constructing the disjunctive expression
that sums up the senses of the corresponding interpretations.
In order to apply this "model-theoretic" method to an equation
between a couple of contingent expressions, one must transform
each expression into its associated DNF and then compare those
to see if they are equal. In the current setting, these DNF's
may indeed end up as identical expressions, but it is possible,
also, for them to turn out slightly off-kilter from each other,
and so the ultimate comparison may not be absolutely immediate.
The explanation of this is that, for the sake of computational
efficiency, it is useful to tailor the DNF that gets developed
as the output of a DNF algorithm to the particular form of the
propositional expression that is given as input.
o-----------------------------------------------------------o
| Equation E1. Proof 2. (=>=) |
o-----------------------------------------------------------o
| |
| q o o r |
| | | |
| p o o p |
| \ / |
| @ |
| |
o=============================< CAST "p" >==================o
| |
| q r |
| q o o r o o o o |
| | | \| |/ |
| o o o o |
| \ / \ / |
| p o-----------o---o p |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< Domination >================o
| |
| q o o r o o |
| | | \ / |
| o o o o |
| \ / \ / |
| p o-----------o---o p |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< Cancellation >==============o
| |
| q o o r |
| | | |
| o o |
| \ / |
| p o-----------o---o p |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< CAST "q" >==================o
| |
| o |
| | |
| o o r o o r |
| | | | | |
| o o o o |
| \ / \ / |
| q o-----------o---o q |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< Cancellation >==============o
| |
| o r o r |
| | | |
| o o o |
| / \ / |
| q o-----------o---o q |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< Domination >================o
| |
| o r |
| | |
| o o |
| / \ |
| q o-----------o---o q |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< CAST "r" >==================o
| |
| o |
| | |
| o o |
| | | |
| o o |
| / / |
| r o-----------o---o r |
| \ / |
| \ / |
| \ / |
| \ / o |
| \ / | |
| q o-------o---o q |
| \ / |
| \ / |
| \ / |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< Cancellation >==============o
| |
| o |
| | |
| r o-------o---o r |
| \ / |
| \ / o |
| \ / | |
| q o-------o---o q |
| \ / |
| \ / |
| \ / |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o===========================================================o
What we have harvested is the succulent equivalent of
a "disjunctive normal form" (DNF) for the proposition
with which we started. Remembering that a blank node
is the graphical equivalent of a logical value "true",
we can read this brand of DNF in the following manner:
o-----------------------------------------------------------o
| DNF of "(p (q))(p (r))" |
o-----------------------------------------------------------o
| |
| o |
| | |
| r o-------o---o r |
| \ / |
| \ / o |
| \ / | |
| q o-------o---o q |
| \ / |
| \ / |
| \ / |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o-----------------------------------------------------------o
| |
| Either not 'p' and thus 'true' |
| Or 'p' and thus |
| Either not 'q' and thus 'false' |
| Or 'q' and thus |
| Either not 'r' and thus 'false' |
| Or 'r' and thus 'true'. |
| |
o-----------------------------------------------------------o
I think that I will leave the other half of the equation
as an "exercise for the reader", since it is much easier
to do this work with a few scribbles on paper than it is
for me to lay it out in Ascii-glyphics. At any rate, no
doubt the reader will not be too surprised to learn that
it all works out, and that if one cases the variables in
the usual "leftmost shallowest" order that one will even
derive identical DNF's for the two sides of the equation.
Jon Awbrey
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