How can I show a set B with 8 elements and two operations, such that the axioms of huntington for boolean algebra holds?
I found it with set of 2 elemtnts. but can't understand how to start with 8 elements. I got an hint about use just the two elements and make it like 2^3.
One possibility is to "vectorize" the ordinary Booleans. So your set will consist of all triples of Booleans, and the NOT and OR operations will apply in parallel to the three elements; i.e., $$ \neg(b_1,b_2,b_3)=(\neg b_1,\neg b_2, \neg b_3) $$ and $$ (b_1,b_2,b_3) \vee (b_1', b_2', b_3')=(b_1\vee b_1',b_2\vee b_2', b_3\vee b_3'). $$ Clearly this set satisfies associativity and commutativity of OR and the Huntington axiom for NOT and OR.