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In boolean algebra the following is true (From Wolfram):

  1. $ \emptyset \wedge a = \emptyset $
  2. $ \emptyset \vee a = a $
  3. $ I \wedge a = a $
  4. $ I \vee a = I $

Where $I$ = the universal set

Why are 3 and 4 true?

If $I = \{0,1\}^n$ and $a=\{0,1\}^n$ wouldn't $I \wedge a = \emptyset$, as not all the bits in $I$ or $a$ are $1$?

BanksySan
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    If $I = {0,1}^n$ and $a = {0,1}^n$, then $I = a$ and the idempotent laws would imply that $I \wedge a = I \wedge I = a \wedge a = I = a$. I don't know enough about Boolean algebra to address in any further detail, hence the comment as opposed to a full answer. It just looks like $I \wedge a = \emptyset$ violates two laws in that case (idempotency and rule 3 above). –  Mar 29 '17 at 16:09

1 Answers1

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If, as you suggest, the Boolean Algebra is the one of the power-set of an $n$-element set, then $I$, the greatest element, is not $\{0,1\}^n$, as you say, but rather $I = (1, \ldots, 1)$ ($n$ times).
So if you take an arbitrary $a \in \{0,1\}^n$, that is $a = (a_1, \ldots, a_n)$, in which $a_i \in \{0,1\}$, then $I \wedge a = a$ and $I \vee a = I$, because $$(a_1, \ldots, a_n) \vee (b_1, \ldots, b_n) = (\max\{a_1,b_1\}, \ldots, \max\{a_n,b_n\}),$$ and similarly for $a \wedge b$ (with min in place of max).
Then just consider $b=I = (1, \ldots, 1)$.

amrsa
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