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Section 6.4 The Duality Principle

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Subsection 6.4.1

permalinkIn Section 4.2, we observed that each of the Table 6.2.1 labeled 1 through 9 had an analogue 1 through 9. We notice that each of the laws in one column can be obtained from the corresponding law in the other column by replacing by , by , by U, U by , and leaving the complement unchanged.

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Definition 6.4.1. Duality Principle for Sets.

Let S be any identity involving sets and the operations complement, intersection and union. If S is obtained from S by making the substitutions , , U , and U, then the statement S is also true and it is called the dual of the statement S.

The dual of \((A \cap B) \cup \left(A \cap \overline{B} \right) = A\) is \((A\cup B)\cap \left(A\cup \overline{B}\right)=A\text{.}\)

permalinkOne should not underestimate the importance of this concept. It gives us a whole second set of identities, theorems, and concepts. For example, we can consider the dual of minsets and minset normal form to obtain what is called maxsets and maxset normal form.

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Exercises 6.4.2 Exercises for Section 6.4

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1.

State the dual of of each of the following:

  1. A(BA)=A.

  2. A¯((¯BA)B)=U

  3. ¯(A¯B)B=¯AB

Answer
  1. \(A\cap (B\cup A)=A\)

  2. \(A \cap \overline{\left(\left(\overline{B}\cap A\right)\cup B\right)}=\emptyset\)

  3. \(\overline{\left(A\cap \overline{B}\right)} \cup B=\overline{A}\cup B\)

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3.

Write the dual of of each of the following:

  1. p¬((¬qp)q)1

  2. (¬(p(¬q)))q(¬pq).

Answer
  1. \((p \land \neg (\neg q \land p)\lor g)) \Leftrightarrow 0\)

  2. \((\neg (p \lor (\neg q)) \land q)\Leftrightarrow ((\neg p) \land q)\)

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4.

Use the principle of duality and the definition of minset to write the definition of maxset.

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5.

Let A={1,2,3,4,5,6} and let B1={1,3,5} and B2={1,2,3}.

  1. Find the maxsets generated by B1 and B2. Note the set of maxsets does not constitute a partition of A. Can you explain why?

  2. Write out the definition of maxset normal form.

  3. Repeat Exercise 6.3.3.4 for maxsets.

Answer

The maxsets are:

  • \(B_1\cup B_2=\{1,2,3,5\}\)

  • \(B_1\cup \overline{B_2}=\{1,3,4,5,6\}\)

  • \(\overline{B_1}\cup B_2=\{1,2,3,4,6\}\)

  • \(\overline{B_1}\cup \overline{B_2}=\{2,4,5,6\}\)

They do not form a partition of \(A\) since it is not true that the intersection of any two of them is empty. A set is said to be in maxset normal form when it is expressed as the intersection of distinct nonempty maxsets or it is the universal set \(U\text{.}\)