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
Commutative Laws
-
\(A + B \Longleftrightarrow B + A\)
-
\(A.B \Longleftrightarrow B.A\)
Associative Laws
-
\((A + B) + C \Longleftrightarrow A + (B + C)\)
-
\((A.B).C \Longleftrightarrow A.(B.C)\)
Distributive Laws
-
\(A + (B + C) \Longleftrightarrow (A + B) + (A + C)\)
-
\(A.(B.C) \Longleftrightarrow (A.B) + (A.C)\)
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\(\Longrightarrow A\)
- \(A + A.B + A.B.C \Longrightarrow A + \)
2.
-
A literal by itself knocks out its NOT'ed opposite that appears in any 'minterm' (Absorption):
- \(\overline{A} + A.\overline{B} + A.C \Longrightarrow \overline{A} +\)
A\(.\overline{B} +\)A\(.C \Longrightarrow \overline{A} + \overline{B} + C\)
- \(\overline{A} + A.\overline{B} + A.C \Longrightarrow \overline{A} +\)
Exercise: One
Minimise the following Boolean Expressions:
1. \(\hspace{1.75em}S = \overline{A} + A.C + \overline{B}.C\)
Click for Solution
\(S = \overline{A} + \)A\(.C + \overline{B}.C\)
\(S = \overline{A} + C +\) B.C
\(S = \overline{A} + C\)
2. \(\hspace{1.75em}S = A.C + A.B.C + A.B.\overline{C}\)
Click for Solution
\(S = A.C + A.B.(C + \overline{C})\)
\(S = A.C + A.B.(1)\)
\(S = A.C + A.B\)
3. \(\hspace{1.75em}S = \overline{B} + A.B + A.\overline{B}.C\)
Click for Solution
\(S = \overline{B} + A.\) B \(+\) A.B.C
\(S = \overline{B} + A\)
4. \(\hspace{1.75em}S = \overline{A}.\overline{B}.\overline{C} + A.\overline{B}.\overline{C} + A.C\)
Click for Solution
\(S = \overline{B}.\overline{C}.(A + \overline{A}) + A.C\)
\(S = \overline{B}.\overline{C}.(1) + A.C\)
\(S = \overline{B}.\overline{C} + A.C\)
5. \(\hspace{1.75em}S = C + A.\overline{C} + B.C + A.B.C\)
Click for Solution
\(S = C + A.\)C \(+\) B.C \(+\) A.B.C
\(S = C + A\)
6. \(\hspace{1.75em}S = A.B.\overline{C} + A.B.C + A.\overline{B}.\overline{C} + A.\overline{B}.C + \overline{A}.B.C + \overline{A}.B.\overline{C}\)
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\(.B\)
\(S = A + B\)
7. \(\hspace{1.75em}S = A + B.C + \overline{A}.B.C\)
Click for Solution
\(S = A + B.C +\) A \(.B.C\)
\(S = A + B.C +\) B.C
\(S = A + B.C\)
8. \(\hspace{1.75em}S = B + \overline{A}.B.\overline{C} + A.B.\overline{C} + \overline{A}.C\)
Click for Solution
\(S = B +\) A.B.C \(+\) A.B.C \(+ \overline{A}.C\)
\(S = B + \overline{A}.C\)
Exercise: Two
For each of the following truth tables produce the standard Sum of the Product Terms for the output \(S\). The reduce the Boolean expression to a simpler expression using Boolean algebra.
1.
| A | B | S |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Click for Solution
\(S = \overline{A}.\overline{B} + \overline{A}.B + A.\overline{B}\)
\(S = \overline{A}.(\overline{B} + B) + A.\overline{B}\)
\(S = \overline{A}.(1) + A.\overline{B}\)
\(S = \overline{A} + A.\overline{B}\)
\(S = \overline{A} +\) A}\(.\overline{B}\)
\(S = \overline{A} + \overline{B}\)
2.
| A | B | S |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Click for Solution
\(S = \overline{A}B + A\overline{B}\)
3.
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 |
Click for Solution
\(S = \overline{A}.\overline{B}.\overline{C} + \overline{A}.\overline{B}.C\)
\(S = \overline{A}.\overline{B}.(\overline{C} + C)\)
\(S = \overline{A}.\overline{B}.(1)\)
\(S = \overline{A}.\overline{B}\)
4.
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Click for Solution
\(S = \overline{A}.\overline{B}.C + \overline{A}.B.C + A.\overline{B}.C + A.B.C\)
\(S = \overline{B}.C.(\overline{A} + A) + B.C.(\overline{A} + A)\)
\(S = \overline{B}.C.(1) + B.C.(1)\)
\(S = \overline{B}.C + B.C\)
\(S = C.(\overline{B} + B)\)
\(S = C.(1)\)
\(S = C\)
5.
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Click for Solution
\(S = \overline{A}.\overline{B}.\overline{C} + \overline{A}.\overline{B}.C + \overline{A}.B.C + A.B.C\)
\(S = \overline{A}.\overline{B}.(\overline{C} + C) + B.C.(\overline{A} + A)\)
\(S = \overline{A}.\overline{B}.(1) + B.C.(1)\)
\(S = \overline{A}.\overline{B} + B.C\)
6.
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Click for Solution
\(S = \overline{A}.\overline{B}.\overline{C} + \overline{A}.\overline{B}.C + \overline{A}.B.C + A.B.C\)
\(S = \overline{A}.\overline{B}.(\overline{C} + C) + B.C.(\overline{A} + A)\)
\(S = \overline{A}.\overline{B}.(1) + B.C.(1)\)
\(S = \overline{A}.\overline{B} + B.C\)
7.
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Click for Solution
\(S = \overline{A}.B.\overline{C} + \overline{A}.B.C + A.\overline{B}.C + A.B.\overline{C} + A.B.C\)
\(S = \overline{A}.B.(\overline{C} + C) + A.\overline{B}.C + A.\overline{B}.(\overline{C} + C)\)
\(S = \overline{A}.B.(1) + A.\overline{B}.C + A.\overline{B}.(1)\)
\(S = \overline{A}.B + A.\overline{B}.C + A.\overline{B}\)
\(S = B.(\overline{A} + A) + A.\overline{B}.C\)
\(S = B.(1) + A.\overline{B}.C\)
\(S = B + A.\overline{B}.C\)
\(S = B + A.\) B \(.C\)
\(S = B + A.C\)
8.
| A | B | C | S |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Click for Solution
\(S = \overline{A}.B.C + A.\overline{B}.\overline{C} + A.\overline{B}.C + A.B.\overline{C} + A.B.C\)
\(S = \overline{A}.B.C + A.\overline{B}.(\overline{C} + C) + A.B.(\overline{C} + C)\)
\(S = \overline{A}.B.C + A.\overline{B}.(1) + A.B.(1)\)
\(S = \overline{A}.B.C + A.\overline{B} + A.B\)
\(S = \overline{A}.B.C + A.(\overline{B} + B)\)
\(S = \overline{A}.B.C + A.(1)\)
\(S = \overline{A}.B.C + A\)
\(S =\) A \(B.C + A\)
\(S = B.C + A\)