Difference between revisions of "Talk:Logical graph"
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==Place for Discussion== | ==Place for Discussion== | ||
− | <math>\ldots\ | + | <br><math>\ldots</math><br> |
+ | |||
+ | ==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]. | ||
+ | |||
+ | * [http://mathforum.org/kb/plaintext.jspa?messageID=6514666 Solution posted by Jon Awbrey, working by way of logical graphs]. | ||
+ | |||
+ | <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
- 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