Can someone please give me a hint how to solve my task ? Show that every Boolean algebra is an ortholattice, but not conversely. 1^{st} and 2^{nd} are done by definition.What about 3 and 4 ?
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Hey xmegx. The 3rd condition is known as De-Morgans Laws. You might want to read up on them for more information. The 4th one could be done in a few ways - including the use of De-Morgans Laws (Part 3) where you find (x' OR 0)' and simplify.
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