The algebra of sets describes basic laws on set operations: union, intersection, complement and other relations, which are mentioned as below.
Let \(A, B,C\) be subsets of a universal set \(U\) then
Laws | Identities | |
Over union | Over Intersection | |
Idempotent Laws | \(A \cup A = A\) | \(A \cap A = A\) |
Identities Laws | \(A \cup U = U , A \cup \phi = A\) | \(A \cap U = A , A \cap \phi = \phi \) |
Complement Laws | \(A\cup \overline{A} = U , \overline{(\overline{A} )}=A\) | \(A \cap \overline{A} = \phi , \overline{U}=\phi \) |
Commutative Laws | \( A \cup B = B \cup A\) | \(A \cap B = B \cap A\) |
Associative Laws | \(( A \cup B )\cup C = A\cup (B\cup A)\) | \((A\cap B)\cap C=A\cap (B\cap A)\) |
Distributive Laws | \(A\cup (B\cap C)=( A \cup B )\cap (A\cup C)\) | \(A\cap (B\cup C)=(A\cap B)\cup (A\cap C)\) |
De-Morgan’s Laws | \(\overline{(A\cup B)}=\overline{A} \cap \overline{B}\) | \(\overline{(A\cup B)}=\overline{A} \cap \overline{B}\) |
Prove that \((A \cap B)'=A' \cup B' \)
Proof- Inductive method (example)
Let \(U = \{1, 2, 3, 4, 5, 6\}\) be the universal set with \(A = \{1, 2, 3, 4\}, B = \{3, 4, 5, 6\}\)
, then- Calculate \( A \cap B = \{3, 4\}\)
- Calculate \((A \cap B)' = U \setminus (A \cap B) = \{1, 2, 5, 6\}\) (1)
- Calculate \( A' = U \setminus A = \{5, 6\}\)
- Calculate \(B' = U \setminus B = \{1, 2\}\)
- Calculate \( A' \cup B' = \{1, 2\} \cup \{5, 6\} = \{1, 2, 5, 6\}\) (2)
- Conclusion: \((A \cap B)' = \{1, 2, 5, 6\} = A' \cup B'\)
- Venn-diagram method (figure)
- Tabular method (logic)
- Deductive method (set-builder)
Let \(x\in (A\cap B)'\) then
\(x \notin (A\cap B)\)
or\(x \notin A \text{ or } x \notin B\)
or\(x \in A' \text{ or } x \in B'\)
or\(x \in (A' \cup B')\)
Hence, \((A\cap B)'=A'\cup B'\)
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