Difference between revisions of "Talk:Logical graph"
MyWikiBiz, Author Your Legacy — Tuesday November 26, 2024
Jump to navigationJump to searchJon Awbrey (talk | contribs) (logical equivalence problem from the drexel math forum) |
Jon Awbrey (talk | contribs) |
||
Line 5: | Line 5: | ||
==Place for Discussion== | ==Place for Discussion== | ||
− | <br><math>\ldots</math><br> | + | <br> |
+ | <math>\ldots</math> | ||
+ | <br> | ||
==Logical Equivalence Problem== | ==Logical Equivalence Problem== |
Revision as of 03:46, 2 December 2008
Notes & Queries
- Jon Awbrey 20:20, 1 February 2008 (PST)
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