Re: logical equivalence problem

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Mike et al.,

This is a very good example, as it illustrates many important
issues and themes that go beyond the present task of proving
the equation at hand, so I think I'll expatiate on it a bit.

I'm in the middle of beaming up to a new computer,
and it's taking me a while to get all my settings
set in the manner to which I've become e-customed,
but I think I have enough things tied down now to
try and run through that equivalence problem in a
more leisurely fashion.

It looks like the <Plain Text> button in the upper right corner of the
Forum webpage lets you see the trees as they are meant to be seen, but
I'll post extra copies of this stuff to the Inquiry List and MyWikiBiz
so that I can work with it easier later on.

http://stderr.org/pipermail/inquiry/2008-November/thread.html#3522
http://www.mywikibiz.com/Talk:Logical_graph

Here is where I left off --

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

Back in a flash ...

Jon Awbrey

CC: Conceptual Graphs, Discrete Math, Inquiry

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