Workshop on Boolean Algebra

The Workshop on Boolean Algebra is designed for you to practice Boolean equations and its laws.

Minimising Boolean Expressions

We can use Commutative, Associative, and Distributive Laws to manipulate Boolean expressions.


BOOLEAN LAWS

  1. Identity Laws:

    • A + 0 = A
    • A . 1 = A
  2. Complement Laws:

    • A + \(\overline{A}\) = 1
    • A . \(\overline{A}\) = 0
  3. Idempotent Laws:

    • A + A = A
    • A . A = A
  4. Double Negation Law:

    • \( '( \overline{A}) \) = A
  5. Annulment Law:

    • A . 0 = 0
    • A + 1 = 1
  6. Absorption Laws:

    • A + (A . B) = A
    • A . (A + B) = A
  7. Commutative Laws:

    • A + B = B + A
    • A . B = B . A
  8. Associative Laws:

    • (A + B) + C = A + (B + C)
    • (A . B) . C = A . (B . C)
  9. Distributive Laws:

    • A . (B + C) = (A . B) + (A . C)
    • A + (B . C) = (A + B) . (A + C)
  10. De Morgan's Laws:

    • \( \overline{(}A + B) = \overline{A} . \overline{B} \)
    • \( \overline{(}A . B) = \overline{A} + \overline{B} \)

Axioms

The following rules (axioms) can also be used to minimise Boolean Expressions:

  1. A literal by itself cancels out any term that contains it (Absorption):

    • \(A + A.B + A.B.C \Longrightarrow A + \)A.B + A.B.C \(\Longrightarrow A\)
  1. A literal by itself knocks out its NOT'ed opposite that appears in any 'minterm' (Absorption):

    • \(\overline{N} + N.\overline{B} + N.C \Longrightarrow \overline{N} +\) N \(.\overline{B} +\) N \(.C \Longrightarrow \overline{N} + \overline{B} + C\)
  1. A(A + B) = A --> if A is true, then A will always be true regardless of the value of B.
  1. A + A.B = ** --> if A is true, then A will always be true regardless of the value of B.
  1. A(A+B) = A.B --> It states that when a variable (A) is multiplied by the negation of itself ORed with another variable (¬A+B), the result is the variable (A) ANDed with the other variable (B)
  1. A + \(\overline{A}.B = A+B \) --> It states that the logical OR of a variable (A) and the logical NOT of that variable ANDed with another variable ( \(\overline{A}.B) \) equals the logical OR of the two variables (A+B).

Exercises:

Minimise the following Boolean Expressions:

  1. \(\hspace{1.75em}S = \overline{R} + R.C + \overline{B}.C\)


    Click for Solution

    \(S = \overline{R} + \)R\(.C + \overline{B}.C\) \ Axioms-2 above

    \(S = \overline{R} + C +\) B.C

    \(S = \overline{R} + C\)


  1. \(\hspace{1.75em}S = A.C + A.B.C + A.B.\overline{C}\)


    Click for Solution

    \(S = A.C + A.B.(C + \overline{C})\)

    \(S = A.C + A.B.(1)\)

    \(S = A.C + A.B\)


  1. \(\hspace{1.75em}S = \overline{B} + A.B + A.\overline{B}.C\)


    Click for Solution

    \( \overline{B} + A(B+\overline{B}.C) \)

    \( \overline{B} + A(B+.C) \)

    \( \overline{B} + AB+AC) \)

    \( \overline{B} + A +AC) \)

    \( \overline{B} + A(1+C) \)

    \( \overline{B} + A \)


  1. \(\hspace{1.75em}S = \overline{A}.\overline{B}.\overline{C} + A.\overline{B}.\overline{C} + A.C\)


    Click for Solution

    \(S = \overline{B}.\overline{C}.(A + \overline{A}) + A.C\)

    \(S = \overline{B}.\overline{C}.(1) + A.C\)

    \(S = \overline{B}.\overline{C} + A.C\)


  1. \(\hspace{1.75em}S = C + A.\overline{C} + B.C + A.B.C\)


    Click for Solution

    \(S = C + A.\)C \(+\) B.C \(+\) A.B.C

    \(S = C + A\)


  1. \(\hspace{1.75em}S = A.B.\overline{C} + A.B.C + A.\overline{B}.\overline{C} + A.\overline{B}.C + \overline{A}.B.C + \overline{A}.B.\overline{C}\)


    Click for Solution

    \(S = A.B.(\overline{C} + C) + A.\overline{B}.(\overline{C} + C) + \overline{A}.B.(C + \overline{C})\)

    \(S = A.B.(1) + A.\overline{B}.(1) + \overline{A}.B.(1)\)

    \(S = A.B + A.\overline{B} + \overline{A}.B\)

    \(S = A.(B + \overline{B}) + \overline{A}.B\)

    \(S = A.(1) + \overline{A}.B\)

    \(S = A +\)A\(.B\)

    \(S = A + B\)


  1. \(\hspace{1.75em}S = A + B.C + \overline{A}.B.C\)


    Click for Solution

    \(S = A + B.C +\) A \(.B.C\)

    \(S = A + B.C +\) B.C

    \(S = A + B.C\)


  1. \(\hspace{1.75em}S = B + \overline{A}.B.\overline{C} + A.B.\overline{C} + \overline{A}.C\)


    Click for Solution

    \(S = B +\) A.B.C \(+\) A.B.C \(+ \overline{A}.C\)

    \(S = B + \overline{A}.C\)