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* [http://mathforum.org/kb/plaintext.jspa?messageID=6514666 Solution posted by Jon Awbrey, working in the medium of logical graphs].
* [http://mathforum.org/kb/plaintext.jspa?messageID=6514666 Solution posted by Jon Awbrey, working in the medium of logical graphs].
Required to show: ~(p <=> q) is equivalent to (~q) <=> p.
In logical graphs, the required equivalence looks like this:
<pre>
q o o p q o
q o o p q o
| | |
| | |
| \ /
| \ /
@ = @
@ = @
</pre>
We have a theorem that says:
<pre>
y o xy o
y o xy o
| |
| |
x @ = x @
x @ = x @
</pre>
See [http://www.mywikibiz.com/Logical_graph#C2.__Generation_theorem Logical Graph : C<sub>2</sub>. Generation Theorem].
Applying this twice to the left hand side of the required equation, we get:
<pre>
q o o p pq o o pq
q o o p pq o o pq
| | | |
| | | |
| |
| |
@ = @
@ = @
</pre>
By collection, the reverse of distribution, we get:
<pre>
p q
p q
o o
o o
\ /
\ /
@
@
</pre>
But this is the same result that we get from one application of double negation to the right hand side of the required equation.
double negation to the right hand side of the required equation.
QED
QED
Jon Awbrey
Jon Awbrey
PS. I will copy this to the Inquiry List:
PS. I will copy this to the [http://stderr.org/pipermail/inquiry/ Inquiry List], since I know it preserves the trees.
===Discussion===
===Discussion===