Difference between revisions of "Talk:Logical graph"

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==Logical Equivalence Problem==
 
==Logical Equivalence Problem==
 
===Problem===
 
  
 
* [http://mathforum.org/kb/message.jspa?messageID=6513648 Problem posted by Mike1234 on the Discrete Math List at the Math Forum].
 
* [http://mathforum.org/kb/message.jspa?messageID=6513648 Problem posted by Mike1234 on the Discrete Math List at the Math Forum].
  
===Solution===
+
* [http://mathforum.org/kb/plaintext.jspa?messageID=6514666 Solution posted by Jon Awbrey, working by way of logical graphs].
 
 
* [http://mathforum.org/kb/plaintext.jspa?messageID=6514666 Solution posted by Jon Awbrey, working in the medium of logical graphs].
 
  
 
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===Discussion===
 
 
o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o
 
 
Back to the initial problem:
 
 
* Show that ~(p <=> q) is equivalent to (~q) <=> p
 
 
We can translate this into logical graphs by supposing that we
 
have to express everything in terms of negation and conjunction,
 
using parentheses for negation -- that is, "(x)" for "not x" --
 
and simple concatenation for conjunction -- "xyz" or "x y z"
 
for "x and y and z".
 
 
In this form of representation, for historical reasons called
 
the "existential interpretation" of logical graphs, we have
 
the following expressions for basic logical operations:
 
 
The disjunction "x or y" is written "((x)(y))".
 
 
This corresponds to the logical graph:
 
 
        x  y
 
        o  o
 
        \ /
 
          o
 
          |
 
          O
 
 
The disjunction "x or y or z" is written "((x)(y)(z))".
 
 
This corresponds to the logical graph:
 
 
        x y z
 
        o o o
 
        \|/
 
          o
 
          |
 
          O
 
 
Etc.
 
 
The implication "x => y" is written "(x (y)),
 
which can be read "not x without y" if that
 
helps to remember the form of expression.
 
 
This corresponds to the logical graph:
 
 
        y o
 
          |
 
        x o
 
          |
 
          O
 
 
Thus, the equivalence "x <=> y" has to be written somewhat
 
inefficiently as a conjunction of to and fro implications:
 
"(x (y))(y (x))".
 
 
This corresponds to the logical graph:
 
 
      y o  o x
 
        |  |
 
      x o  o y
 
        \ /
 
          O
 
 
Putting all the pieces together, the problem given
 
amounts to proving the following equation, expressed
 
in parse string and logical graph forms, respectively:
 
 
* Show that ~(p <=> q) is equivalent to (~q) <=> p
 
 
      q o  o p          q o
 
        |  |              |
 
      p o  o q            o  o p
 
        \ /                |  |
 
          o              p o  o--o q
 
          |                  \ /
 
          O        =        O
 
 
( (p (q)) (q (p)) ) = (p ( (q) )) ((p)(q))
 
 
No kidding ...
 
 
o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o12:30, 2 December 2008 (PST)[[User:Jon Awbrey|Jon Awbrey]] 12:30, 2 December 2008 (PST)o
 
 
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Revision as of 20:34, 2 December 2008

Notes & Queries

Place for Discussion


\(\ldots\)

Logical Equivalence Problem

Date: 30 Nov 2008, 2:00 AM
Author: Jon Awbrey
Subject: Re: logical equivalence problem

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required to show:  ~(p <=> q) is equivalent to (~q) <=> p

in logical graphs, the required equivalence looks like this:

      q o   o p           q o
        |   |               |
      p o   o q             o   o p
         \ /                |   |
          o               p o   o--o q
          |                  \ / 
          @         =         @

we have a theorem that says:

        y o                xy o
          |                   |
        x @        =        x @

see: http://www.mywikibiz.com/Logical_graph#C2.__Generation_theorem

applying this twice to the left hand side of the required equation:

      q o   o p          pq o   o pq
        |   |               |   |
      p o   o q           p o   o q
         \ /                 \ /
          o                   o
          |                   |
          @         =         @

by collection, the reverse of distribution, we get:

          p   q
          o   o
       pq  \ / 
        o   o
         \ /
          @

but this is the same result that we get from one application of
double negation to the right hand side of the required equation.

QED

Jon Awbrey

PS.  I will copy this to the Inquiry List:
     http://stderr.org/pipermail/inquiry/
     since I know it preserves the trees.

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