CSE 20 Lecture 9 Boolean Algebra: Theorems and Transformations

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CSE 20 Lecture 9Boolean Algebra Theorems and
Transformations

• CK Cheng

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Theorems Proofs4 Postulates

• P1 ab ba, abba (commutative)
• P2 abc (ab)(ac) (distributive)
• a(bc) ab ac
• P3 a0a, a1 a (identity)
• P4 aa1, aa 0 (complement)

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• Theorem 6 (Involution Laws)
• For every element a in B, (a')' a
• Proof a is one complement of a'.
• The complement of a' is unique.
• Thus a (a').
• Theorem 7 (Absorption Law) For every pair a,b in
  B, a(ab) a a ab a.
• Proof a(ab)
• (a0)(ab) (P3)
• a0b (P2)
• a 0 (P3)
• a (P3)

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Theorems and Proofs

• Theorem 8 For every pair a, b in B
• a ab a b a(a b) ab
• Proof a ab
• (a a)(a b) (P2)
• (1)(a b) (P4)
• a b (P3)

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Theorem 9 De Morgans Law

• Theorem For every pair a, b in set B
• (ab) ab, and (ab) ab.
• Proof We show that ab and ab are
  complementary.
• In other words, we show that both of the
  following are true (P4)
• (ab)(ab) 1, (ab)(ab) 0.

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Theorem 9 De Morgans Law (cont.)
Proof (Continue) (ab)(ab) (aba)(abb)
(P2) (1b)(a1) (P4) 1 (Theorem 3)
(ab)(ab) (ab)(ab) (P1) abaabb
(P2) 0ba0 (P4) 00 (Theorem 3) 0 (P3)
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3. Theorems Switching Algebra vs. Multiple
Valued Boolean Algebra

• Boolean Algebra is termed Switching Algebra when
  B 0, 1
• When B gt 2, the system is multiple valued.
• Example M (0, 1, 2, 3), ,

0 1 2 3
0 0 0 0 0
1 0 1 0 1
2 0 0 2 2
3 0 1 2 3
0 1 2 3
0 0 1 2 3
1 1 1 3 3
2 2 3 2 3
3 3 3 3 3
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• iClicker M (0, 1, 2, 3), ,
• Boolean algebra can have only two elements 0,
  1.
• The identity elements are 0 and 3
• a 0 a
• a 3 a
• C. The complement of 1 is 2
• Two of the above
• None of the above.

0 1 2 3
0 0 1 2 3
1 1 1 3 3
2 2 3 2 3
3 3 3 3 3
0 1 2 3
0 0 0 0 0
1 0 1 0 1
2 0 0 2 2
3 0 1 2 3
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• Example M (0, 1, 2, 3), ,
• P1 Commutative Laws
• a b b a
• a b b a
• P2 Distributive Laws
• a (b c) (a b) ( a c)
• a (b c) (a b) (a c)
• P3 Identity Elements
• a 0 a
• a 3 a
• P4 Complement Laws
• a a 3
• a a 0

0 1 2 3
0 0 1 2 3
1 1 1 3 3
2 2 3 2 3
3 3 3 3 3
0 1 2 3
0 0 0 0 0
1 0 1 0 1
2 0 0 2 2
3 0 1 2 3
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Boolean Transform

• Given a Boolean expression, we reduce the
  expression (literals, terms) using laws and
  theorems of Boolean algebra.
• When B0,1, we can use tables to visualize the
  operation.
• The approach follows Shannons expansion.
• The tables are organized in two dimension space
  and called Karnaugh maps.

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4. Boolean Transformations

• Show that ababab ab
• Proof 1 ababab ab(aa)b P2
• ab b
  P4
• a b
  Theorem 8
• Proof 2 ababab
• abababab Theorem 5
• ab ab abab P1
• a(bb) (aa)b P2
• a1 1b
  P4
• a b
  P3

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Boolean Transformation

• (abc)(ab)(bac)
• (abc)(ab)(b(ac)) (DeMorgans)
• (abc)(ab)b(ac) (DeMorgans)
• (abc)b(ac) (Absorption)
• (abbbc)(ac) (P2)
• (0bc)(ac) (P4)
• bc(ac) (P3)
• abcbcc (P2)
• abc0 (P4)
• abc (P3)