ONT Re: Propositional Equation Reasoning Systems
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PERS. Note 14
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Analysis of Contingent Expressions (cont.)
We are still in the middle of contemplating a particular example
of a propositional equation, namely, "(p (q))(p (r)) = (p (q r))",
and we are still considering the second of three formal methods that
I intend to illustrate in the process of thrice-over establishing it.
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
I know that it must seem tedious, but I probably ought to go ahead and
carry out the second half of this analogically "model-theoretic" strategy,
just so that we will have the security of this concrete and shared experience
on which to fall back at every later point in what may quickly become a rather
abstruse discussion. Here then is the rest of the necessary chain of equations:
o-----------------------------------------------------------o
| Equation E1. Proof 2. (=<=) |
o-----------------------------------------------------------o
| |
| q r |
| o |
| | |
| p o |
| | |
| @ |
| |
o=============================< CAST "p" >==================o
| |
| q r q r |
| o o o |
| | \| |
| o o |
| | | |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< Domination >================o
| |
| q r |
| o o |
| | \ |
| o o |
| | | |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< Cancellation >==============o
| |
| q r |
| o |
| | |
| o |
| | |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< CAST "q" >==================o
| |
| o |
| | |
| o r o r |
| | | |
| o o |
| | | |
| q o-------o---o q |
| \ / |
| \ / |
| \ / |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< Domination >================o
| |
| o |
| | |
| o r o |
| | | |
| o o |
| | | |
| q o-------o---o q |
| \ / |
| \ / |
| \ / |
| p o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
o=============================< Cancellation >==============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=============================< QED >=======================o
This is not only a logically equivalent DNF,
but exactly the same DNF expression that we
obtained before, so we have established the
given equation "(p (q))(p (r)) = (p (q r))".
Incidentally, one may wish to note that this DNF
expression quickly folds into the following form:
o-----------------------------------------------------------o
| |
| pqr o-------o---o p |
| \ / |
| \ / |
| \ / |
| @ |
| |
| (p q r, (p)) |
| |
o-----------------------------------------------------------o
This can be read to say "either p q r, or not p",
which gives us yet another expression equivalent
to the sentences "(p (q))(p (r))" and "(p (q r))".
Jon Awbrey
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