Boolean Algebra Questions with Solutions

Definition: The algebra of logic is a Boolean algebra. It works with variables with two different values, such as 0 (False) and 1 (True), as well as logically significant operations. George Boole invented the first way of manipulating symbolic logic, which later became known as Boolean Algebra. Because of its vast applications in switching theory, developing basic electronic circuits, and designing digital computers, Boolean algebra has become a vital tool in computer science. The following are the basic operations of Boolean algebra: AND or conjunction operation OR operation or disjunction Not or Negative operation

Rules: Suppose X and Y are two Boolean variables, then the three operations are defined as follows; X AND Y, satisfies X ∧ Y = True, if X = Y = True or else X ∧ Y = False. X OR Y, satisfies X ∨ Y = False, if X = Y = False, else X ∨ Y = True. ¬X satisfies ¬X = False, if X = True and ¬X = True if X = False

Boolean Algebra Questions with Solutions

1. Define Boolean expression.

Answer:

The result of a Boolean expression is always a Boolean value. A Boolean expression is made up of the Boolean constants, logical connectives and Boolean variables and logical connectives. A Boolean function is represented by each Boolean expression. The Boolean expression XY′Z is an example.

2. What is meant by isomorphic Boolean algebra?

Answer:

Isomorphic Boolean algebras B and B1 are those that have a one-to-one correspondence f: B ⟶B1 that retains the three operations +,*, and ‘ for all elements x, y in B,

• f (x+y)=f(x)+f(y)
• f (x*y)=f(x)*f(y) and
• f(x’)=f(x)’

3. Define Boolean function.

Answer:

A Boolean function is a type of mathematical function with the formula f:Xn→X of degree n, where ‘n’ is a non-negative integer, and X = {0,1} is a Boolean domain. It explains how Boolean input generates Boolean output.

Let F(X, Y)=X′Y′ as an example. This is a degree 2 function from the collection of ordered pairs (two numbers) of Boolean variables to set 0,1 with F(0, 1)=0, F(0,0) = 1, F(1,1)=0, and F(1,0) = 0.

Boolean Identities Double Complement Law ∼(∼X) = X Complement Law X+∼X = 1 (OR Form) X.∼X = 0 (AND Form) Idempotent Law X+X = X (OR Form) X.X = X (AND Form) Identity Law X+0 = X (OR Form) X.1 = X (AND Form) Dominance Law X+1 = 1 (OR Form) X.0 = 0 (AND Form) Commutative Law X+Y = Y+X (OR Form) X.Y = Y.X (AND Form) Associative Law X+(Y+Z) = (X+Y)+Z (OR Form) X.(Y.Z) = (X.Y).Z (AND Form) Absorption Law X.(X+Y) = X X+(X.Y) = X Simplification Law X.(∼X+Y) = X.Y X+(∼X.Y) = X+Y Distributive Law X+(Y.Z) = (X+Y).(X+Z) X.(Y+Z) = (X.Y)+(X.Z) De-Morgan’s Law ∼(X.Y)=∼X+∼Y ∼(X+Y)=∼X.∼Y

4. Using Boolean identities, reduce the given Boolean expression:

F(X, Y, Z) = X′Y + YZ′ + YZ + XY′Z′

Solution:

Given,F(X, Y, Z) = X′Y + YZ′ + YZ + XY′Z′

Using the idempotent law, we can write YZ’ = YZ’ + YZ’

⇒ F(X, Y, Z) = X′Y+(YZ′+YZ′)+YZ + XY′Z′

Now, interchange the second and third term, we get

⇒ F(X, Y, Z) = X′Y+(YZ′+YZ)+(YZ′+XY′Z′)

By using distributive law,

⇒ F(X, Y, Z) = X′Y+Y(Z′+Z)+Z′(Y+XY′)

Using Z’ + Z = 1 and absorption law (Y + XY’)= (Y + X),

⇒ F(X, Y, Z) = X′Y+Y.1+Z′(Y+X)

⇒ F(X, Y, Z) = X′Y+Y+Z′(Y+X) [Since Y.1 = Y ]

⇒ F(X, Y, Z) = Y(X′+1)+Z′(Y+X)

⇒ F(X, Y, Z) = Y.1+Z′(Y+X) [ As (X’ + 1) = 1 ]

⇒ F(X, Y, Z) = Y +Z′(Y+X) [ As, Y.1 = Y ]

⇒ F(X, Y, Z) = Y+YZ’+XZ’

⇒ F(X, Y, Z) = Y(1+Z′)+XZ′

⇒ F(X, Y, Z) = Y.1+XZ′ [Since (1 + Z’) = 1]

⇒ F(X, Y, Z) = Y+XZ′ [Since Y.1 = Y]

Hence, the simplified form of the given Boolean expression is F(X, Y, Z) = Y+XZ′.

5. Reduce the following Boolean expression: F(P ,Q, R)=(P+Q)(P+R)

Solution:

Given, F(P ,Q, R)=(P+Q)(P+R)

Using distributive law,

⇒ F(P, Q, R) = P.P + P.R +Q.P + Q.R

Using Idempotent law,

⇒ F(P, Q, R) = P + P.R +Q.P + Q.R

Again using distributive law, we get

⇒ F(P, Q, R) = P(1+R) + Q.P + Q.R

Using dominance law, we can write

⇒ F(P, Q, R) = P + Q.P + Q.R

Again using distributive law, we get

⇒ F(P, Q, R) = (P+1).P+ Q.R

Therefore, using dominance law, we can get the reduced form as follows:

⇒ F(P, Q, R) = 1.P+Q.R

⇒ F(P, Q, R) = P+Q.R

Hence, the reduced form of F(P, Q, R) = (P+Q)(P+R) is F(P, Q, R) = P+Q.R.

6. What is the equivalent expression for the Boolean expression x’y’z +yz+ xz?

Solution:

Answer: z

Given Boolean expression: x’y’z +yz+ xz

x’y’z +yz+ xz = z(x’y’+y+x)

Now, apply distributive law for the first two terms inside the bracket.

x’y’z +yz+ xz = z[(x’+y) (y+y’)+ x]

x’y’z +yz+ xz = z [(x’ + y) . 1 + x] [Since A+A’ = 1]

x’y’z +yz+ xz = z [x’ + y + x]

Further x+x’ = 1

So, x’y’z +yz+ xz = z(1+y)

Now, using null law, 1+y = 1

x’y’z +yz+ xz = z.1

Now, using identity law, A.1 = A

Therefore, x’y’z +yz+ xz = z.

Hence, the Boolean expression equivalent to x’y’z +yz+ xz is z.

7. What is the simplified sum of product form for the Boolean expression:

(A + B’ + C’)(A + B’ + C)(A + B + C’)

Solution:

Given Boolean expression: (A + B’ + C’)(A + B’ + C)(A + B + C’)

Now, using the distributive law, we can write

(A + B’ + C’)(A + B’ + C)(A + B + C’) = [(A+B’) + C’C](A + B + C’)

Using AA’ = 0, it becomes

(A + B’ + C’)(A + B’ + C)(A + B + C’) = [(A+B’) + 0](A + B + C’)

Since A + 0 = A,

(A + B’ + C’)(A + B’ + C)(A + B + C’) = (A+B’)(A + B + C’)

(A + B’ + C’)(A + B’ + C)(A + B + C’) = A + B’.(B+C’)

Again, using distributive law

(A + B’ + C’)(A + B’ + C)(A + B + C’) = A + B’B +B’C’

(A + B’ + C’)(A + B’ + C)(A + B + C’) = A + 0 + B’C’

(A + B’ + C’)(A + B’ + C)(A + B + C’) = A + B’C’.

8. Write the reduced form for the Boolean expression (AB’(C+BD) + A’B’)C.

Solution:

Given expression: (AB’(C+BD) + A’B’)C

Using distributive law, we can write;

(AB’(C+BD) + A’B’)C = (AB’C + AB’BD +A’B’) C

Since, BB’ = 0

(AB’(C+BD) + A’B’)C = (AB’C + 0 + A’B’)C

(AB’(C+BD) + A’B’)C = (AB’C + A’B’)C

(AB’(C+BD) + A’B’)C = AB’C + A’B’C

Now, take B’C as common,

(AB’(C+BD) + A’B’)C = B’C (A + A’)

(AB’(C+BD) + A’B’)C = B’C (1)

(AB’(C+BD) + A’B’)C = B’C.

Hence, the reduced form of (AB’(C+BD) + A’B’)C is B’C.

9. Reduce the Boolean expression: A = XY + X(Y+Z) + Y(Y+Z)

Solution:

Given: A = XY + X(Y+Z) + Y(Y+Z)

The given Boolean expression can be written as follows:

A = XY + XY + XZ + YY + YZ

Using Idempotent law, Y.Y = Y

Thus, A = XY + XY + XZ + Y + YZ

Again, XY + XY = XY, we get

A = XY + XZ + Y + YZ

A = XY + XZ + Y(1+Z)

Using null law, (i.e., 1+A = 1), we can write

A = XY + XZ + Y.1

Using Identity law, 1.Y = Y.

So, A = XY + XZ + Y

Now, the above form can be written as:

A = Y(1+X) + XZ

A = Y.1 + XZ

A = Y + XZ

Hence, the reduced form of the Boolean expression A = XY + X(Y+Z) + Y(Y+Z) is A = Y+XZ.

10. Check whether AC + ABC = AC is true or false?

Solution:

Given Boolean expression: AC + ABC = AC

AC (1+B) = AC

Now, using the null law 1+B = 1, the above expression can be written as:

AC.1 = AC

Thus, using the identity law, 1.A = A, the above expression becomes:

AC = AC

Hence, AC + ABC = AC is true.