MELJUN CORTES - Mathematics of Computing
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MELJUN CORTES - Mathematics of Computing

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MELJUN CORTES - Mathematics of Computing

MELJUN CORTES - Mathematics of Computing

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MELJUN CORTES - Mathematics of Computing MELJUN CORTES - Mathematics of Computing Document Transcript

  • Year 1 LESSON 6 DE MORGAN’S LAWIntroduction♦ In addition to our ‘ordinary’ Boolean algebra, De Morgan’s Law is one tool used for simplifying complex expressions in Boolean algebra or complex switching circuits CS113/0401/v1 Lesson 6 - 1
  • Year 1 ALGEBRAIC LAWS (1) A ARULES: i. A.O = O ii. A+O = A iii. A.1 = A iv. A+1 = 1 v. A.A = A vi. A+A = A vii. A.A = O viii. A+A = 11 Universal Set0 Empty Set CS113/0401/v1 Lesson 6 - 2
  • Year 1 ALGEBRAIC LAWS (2) INTERSECTING SETS ε A BRULES: ix. A . (A + B) = A x. A+A.B = A xi. A . (A + B) = A . B xii. A+A.B = A+B CS113/0401/v1 Lesson 6 - 3
  • Year 1 ALGEBRAIC LAWS (2)♦ Communicative Laws A.B=B.A A+B=B+A♦ Associative Laws A . (B . C) = (A . B) . C = A . B . C A + (B + C) = (A + B) + C = A + B + C (A + B) + (C + D) = A + B + C + D CS113/0401/v1 Lesson 6 - 4
  • Year 1 ALGEBRAIC LAWS (2)♦ Distributive Laws A . (B + A) = A . B + A . C (A + B) . (A + C) = A.A+A.C+B.A+B.C = A . (A + B + C) + B . C = A + BC♦ Double negatives (A) = A CS113/0401/v1 Lesson 6 - 5
  • Year 1 DE MORGAN’ S LAWSThe following two rules constitute DeMorgan’s laws: (A + B) = A . B (A . B) = A + BUsing these rules, two steps are used toform a complement:Step 1. The + symbols are replaced with .symbols and . symbols with + symbols.Step 2. Each of the terms in the expressionis complemented. CS113/0401/v1 Lesson 6 - 6
  • Year 1DE MORGAN’S LAWS (2) ExamplesExample One Example Two A+B A.B1 A.B 1 A+B2 A.B 2 A+B3 A.B 3 A+B =A.B =A+B CS113/0401/v1 Lesson 6 - 7
  • Year 1 DE MORGAN’S LAWS (3) A+B = A.B ♦ Circuit for A + B: A A+B A+B B ♦ Circuit for A.B : AA A.BB B Same Output A B A+B A+B A B A.B 0 0 0 1 1 1 1 0 1 1 0 1 0 0 1 0 1 0 0 1 0 1 1 1 0 0 0 0 Truth Table CS113/0401/v1 Lesson 6 - 8
  • Year 1DE MORGAN’S LAWS (4)♦ Simplify (A+B).CSTEP 1: A+B = A.BSTEP 2: A.B.C=A+B+CALSO A.B.C.D = A+B+C+ D CS113/0401/v1 Lesson 6 - 9
  • Year 1 DE MORGAN’S LAWSThe use of De Morgan’s laws may bedemonstrated by finding the complement ofthe expression A + BC. First, note that amultiplication sign has been omitted, and theexpression could be written A + (B.C). Tocomplement this, the addition symbols, andthe two terms are complemented, givingA .(B.C); then the remaining term iscomplemented : A (B + C). The followingequivalence has been found:( A + BC) + A (B + C)The complement of AB + CD maybe formedby two steps:Step 1. The addition symbol is changed.Step 2. The complementof each term is formed: (A.B)(C.D)This become (A + B) (C + D) CS113/0401/v1 Lesson 6 - 10
  • Year 1 SIMPLIFICATION OF EXPRESSIONConsider the expression (A + B) (A + B) (A + C)The first two terms consist of A + B andA + B; these maybe multiplied and sinceA + AB + AB = A and BB = 0, reduced A.The expression has been reduced now toA(A + C), which maybe expressed asAA + AC. And since AA is equal to 0. The entire expression (A + B) (A + B) (A + C)maybe reduced to AC CS113/0401/v1 Lesson 6 - 11