Difference between revisions of "Talk:Logical graph"

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(logical equivalence problem from the drexel math forum)
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==Place for Discussion==
 
==Place for Discussion==
  
<math>\ldots\!</math>
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<br><math>\ldots</math><br>
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 +
==Logical Equivalence Problem==
 +
 
 +
* [http://mathforum.org/kb/message.jspa?messageID=6513648&tstart=0 Problem posted by Mike1234 on the Discrete Math list at the Math Forum].
 +
 
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* [http://mathforum.org/kb/plaintext.jspa?messageID=6514666 Solution posted by Jon Awbrey, working by way of logical graphs].
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 +
<pre>
 +
Date: 30 Nov 2008, 2:00 AM
 +
Author: Jon Awbrey
 +
Subject: Re: logical equivalence problem
 +
 
 +
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
 +
 
 +
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.
 +
 
 +
o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o
 +
</pre>

Revision as of 03:44, 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

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o

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.

o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o~~~~~~~~~o