CBSE Class 12 Computer Science Boolean Logic Concepts

Boolean algebra is an algebra that deals with Boolean values((TRUE and FALSE) . Everyday we have to make logic decisions: “Should I carry the book or not?” , “Should I watch TV or not?” etc. Each question will have two answers yes or no, true or false. In Boolean Algebra we use 1 for true and 0 for false which are known as truth values.

Truth table:

A truth table is composed of one column for each input variable (for example, A and B), and onefinal column for all of the possible results of the logical operation that the table is meant to represent (forexample, A XOR B). Each row of the truth table therefore contains one possible configuration of the input variables (for instance, A = true B = false), and the result of the operation for those values. Logical Operators: In Algebraic function e use +,-,*,/ operator but in case of Logical Function or Compound statement we use AND,OR & NOT operator.

Example: He prefers Computer Science NOT IP.

There are three Basic Logical Operator:

1. NOT

2. OR

3. AND

Low Order Thinking Questions: (Boolean Algebra)

Question a. State and verify absorption law in Boolean algebra.
Answer: Absorption law states that combining operations can absorb a variable. The two forms are:
1. \(X + XY = X\)
2. \(X(X + Y) = X\)
Verification for \(X + XY = X\):
\(= X(1 + Y)\)
\(= X(1)\) (since \(1 + Y = 1\))
\(= X\)
Verification for \(X(X + Y) = X\):
\(= XX + XY\)
\(= X + XY\) (since \(XX = X\))
\(= X(1 + Y)\)
\(= X(1) = X\)
In simple words: The absorption law allows you to drop redundant variables from an expression to simplify it.

Exam Tip: Always look for terms common to both parts of an expression to apply the absorption law quickly.

Question b. Verify \(X' \cdot Y + X \cdot Y' = (X' + Y') \cdot (X + Y)\) algebraically.
Answer: LHS = \(X'Y + XY'\)
\(= (X'Y + X) \cdot (X'Y + Y')\) (using distributive law)
\(= (X' + X)(Y + X) \cdot (X' + Y')(Y + Y')\)
\(= 1 \cdot (X + Y) \cdot (X' + Y') \cdot 1\) (since \(X' + X = 1\) and \(Y + Y' = 1\))
\(= (X' + Y')(X + Y)\)
= RHS, hence proved.
In simple words: Algebraic verification involves expanding and applying basic Boolean postulates step-by-step until both sides match.

Exam Tip: This is a standard expansion identity often used in converting between SOP and POS forms.

Question c. Write the equivalent Boolean Expression F for the following circuit diagram:
Answer: Examining the given circuit diagram containing AND, OR, and NOT gates, the output expression derived from the logic gates is:
\(F = A'B + AB + B'C\)
In simple words: Break down the circuit gate by gate from inputs to the final output to write the expression.

Exam Tip: Trace each sub-path carefully, noting where variables pass through NOT gates (inverters).

Question d. If \(F(P, Q, R, S) = \prod (3, 4, 5, 6, 7, 13, 15)\), obtain the simplified form using K-Map.
Answer: Plotting the maxterms on a 4-variable Karnaugh Map and grouping adjacent zeros into quads and pairs:
Quad 1 = \(M_4 \cdot M_5 \cdot M_6 \cdot M_7 = P + Q'\)
Quad 2 = \(M_5 \cdot M_7 \cdot M_{13} \cdot M_{15} = Q' + S'\)
Pair = \(M_3 \cdot M_7 = P + R' + S'\)
Therefore, the simplified POS expression is:
\(F(P, Q, R, S) = (P + Q')(Q' + S')(P + R' + S')\)
In simple words: For POS simplification using K-maps, group the zeros into largest possible powers of 2.

Exam Tip: Remember that in POS forms, a variable with a value of 1 is represented in its complemented form.

Question e. Simplify \(F(a, b, c, d) = \sum (0, 2, 4, 5, 7, 8, 10, 12, 13, 15)\).
Answer: Grouping the minterms into blocks:
\(B_1 = m_0 + m_4 + m_{12} + m_8 = c'd'\)
\(B_2 = m_5 + m_7 + m_{13} + m_{15} = bd\)
\(B_3 = m_0 + m_2 + m_8 + m_{10} = b'd'\)
Combining all groups, the simplified SOP expression is:
\(F(a, b, c, d) = c'd' + bd + b'd'\)
In simple words: Combine adjacent 1s in the K-map to eliminate changing variables.

Exam Tip: Always check if any minterm is covered by multiple groups to ensure maximum simplification.

Question f. Write the equivalent Boolean expression for the given truth table:

Answer: Extracting minterms where output F = 1:
\(F(X, Y, Z) = \sum (1, 5, 7)\)
In canonical SOP form: \(F(X, Y, Z) = X'YZ' + XYZ' + XYZ\)
In simple words: Identify the rows where the output is 1 and write them as minterms.

Exam Tip: Double-check binary to decimal conversion of input combinations to avoid index errors.

1 / 2 Marks Practice Questions:

Question 1. State and Prove DeMorgan Law using Truth Table.
Answer: DeMorgan's First Law: \((X + Y)' = X' \cdot Y'\)
DeMorgan's Second Law: \((X \cdot Y)' = X' + Y'\)
Truth Table Proof:
X | Y | X+Y | (X+Y)' | X' | Y' | X'Y'
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
Since the columns for \((X+Y)'\) and \(X'Y'\) are identical, the law is verified.
In simple words: DeMorgan's laws show how complementation distributes over OR and AND operations.

Exam Tip: Memorize both forms of DeMorgan's theorem as they are essential for gate conversions.

Question 2. State and prove Absorption Law algebraically.
Answer: Statement: \(X + XY = X\) and \(X(X + Y) = X\).
Proof for \(X + XY = X\):
\(= X(1 + Y)\)
\(= X(1)\) (since \(1 + Y = 1\))
\(= X\)
In simple words: Factoring out common terms helps prove absorption instantly.

Exam Tip: State both parts of the law unless only one is specified.

Question 3. State and Prove Distributive Law algebraically.
Answer: Statement:
1. \(X(Y + Z) = XY + XZ\)
2. \(X + YZ = (X + Y)(X + Z)\)
Proof for 1 using truth table or boolean postulates follows standard expansion where multiplication distributes over addition.
In simple words: Distributive law works similarly to regular algebra for the first rule, with a dual rule in Boolean algebra.

Exam Tip: The second distributive law is unique to Boolean algebra and frequently tested.

Question 4. Write the equivalent Boolean Expression for the following Logic Circuit.
Answer: Tracing the inputs P and Q through the given combination of NOT, OR, and AND gates:
\(F = (P' + Q) \cdot (P + Q')\)
In simple words: Combine intermediate gate outputs step by step from inputs to output.

Question 5. Write the equivalent Boolean Expression F for the following circuit diagram.
Answer: Analyzing the multi-stage NAND/AND gate configuration with inputs A and B:
\(F = A'B + AB' + AB\)
In simple words: Combine the product terms generated by each parallel branch.

Exam Tip: Simplify expressions derived from circuits if required by the question.

Question 6. Write the equivalent Boolean Expression F for the following circuit diagram.
Answer: Tracing inputs X, Y, and Z through NOT, OR, and AND gates:
\(F = (X' + Y + Z) \cdot Z'\)
In simple words: Follow each input line through its respective gate operation.

Exam Tip: Pay close attention to parentheses when combining OR gate outputs into an AND gate.

Question 7. Convert the following Boolean expression into its equivalent Canonical Sum of Product Form (SOP): \((X' + Y + Z') \cdot (X' + Y + Z) \cdot (X' + Y' + Z) \cdot (X' + Y' + Z')\).
Answer: Converting the given maxterm expression to minterms (SOP form):
The missing minterms correspond to the given maxterms, yielding:
\(F = \sum (0, 1, 2, 3)\) or expanded in canonical SOP form.
In simple words: Convert maxterm indices to equivalent minterm indices by taking the unlisted binary combinations.

Exam Tip: Canonical SOP uses minterms (\(\sum\)), while POS uses maxterms (\(\prod\)).

Question 8. Convert the following Boolean expression into its equivalent Canonical Product of Sum form (POS): \(A \cdot B' \cdot C + A' \cdot B \cdot C + A' \cdot B \cdot C'\).
Answer: Converting the given SOP expression into canonical POS form by identifying absent minterms / maxterms:
Corresponding maxterms: \(\prod (0, 3, 4, 6, 7)\)
In simple words: Identify the decimal equivalents of the terms and select the remaining index values for POS form.

Exam Tip: Ensure all variables appear in every term for canonical representation.

Question 9. Draw a Logical Circuit Diagram for the following Boolean expression: \(A \cdot (B + C')\).
Answer: Construction Steps:
1. Take input C and pass it through a NOT gate to get \(C'\).
2. Pass inputs B and \(C'\) into an OR gate to produce \((B + C')\).
3. Pass input A and the output of the OR gate into an AND gate to get the final output \(A \cdot (B + C')\).
In simple words: Build the circuit from inside the parentheses outward to the final operation.

Exam Tip: Draw clear connection dots and label all gate inputs and outputs.

Question 10. Write the equivalent Boolean Expression F for the following circuit diagram.
Answer: Tracing the inputs through the OR, NOT, and AND gates in the diagram:
\(F = (A + C) \cdot (A' + B)\)
In simple words: Combine the gate blocks sequentially from inputs to output.

Exam Tip: Use brackets properly to maintain operator precedence between OR and AND operations.

Question 11. Prove that \(XY + YZ + YZ' = Y\) algebraically.
Answer: LHS = \(XY + YZ + YZ'\)
\(= XY + Y(Z + Z')\) (factoring out Y)
\(= XY + Y(1)\) (since \(Z + Z' = 1\))
\(= XY + Y\)
\(= Y(X + 1)\)
\(= Y(1)\) (since \(X + 1 = 1\))
\(= Y\)
= RHS, hence proved.
In simple words: Factor out common variables and apply complementarity laws to simplify the expression.

Exam Tip: Look for complementary pairs like \(Z\) and \(Z'\) to eliminate variables quickly.

Question 12. Design \((A + B) \cdot (C + D)\) using NOR Gate.
Answer: Using DeMorgan's law and universal gate properties:
\((A + B) \cdot (C + D) = ((A + B)' + (C + D)')'\)
This can be implemented entirely using NOR gates by passing A and B into a NOR gate, C and D into another NOR gate, and feeding their outputs into a final NOR gate.
In simple words: Universal gates like NOR can replicate any Boolean function through double complementation.

Exam Tip: Always apply double negation when converting expressions into pure NAND or NOR logic.

3 Marks Practice Questions:

Question 13. If \(F(a, b, c, d) = \sum (0, 2, 4, 5, 7, 8, 10, 12, 13, 15)\), obtain the simplified form using K-Map.
Answer: Plotting the given minterms on a 4-variable Karnaugh Map:
- Group 1 (\(m_0, m_4, m_8, m_{12}\)) gives \(c'd'\)
- Group 2 (\(m_5, m_7, m_{13}, m_{15}\)) gives \(bd\)
- Group 3 (\(m_0, m_2, m_8, m_{10}\)) gives \(b'd'\)
Simplified SOP Expression:
\(F(a, b, c, d) = c'd' + bd + b'd'\)
In simple words: Map the minterms, form quads of size 4, and write the reduced product terms.

Exam Tip: Verify that all given minterms are covered by at least one group.

Question 14. If \(F(a, b, c, d) = \sum (0, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15)\), obtain the simplified form using K-Map.
Answer: Plotting the minterms on a 4-variable K-map and combining adjacent cells into octets and quads:
- Octet for lower/middle columns yields simplified terms.
- Remaining isolated minterms are covered by quads and pairs.
Simplified SOP Expression:
\(F(a, b, c, d) = bd + a'c'd' + b'c + ad'\)
In simple words: Look for octets (8 cells) first to eliminate three variables at once.

Exam Tip: Prioritize grouping larger blocks (octets over quads, quads over pairs) for maximum reduction.

Question 15. Obtain a simplified form for a boolean expression \(F(U, V, W, Z) = \prod (0, 1, 3, 5, 6, 7, 10, 14, 15)\).
Answer: Plotting the maxterms on a 4-variable K-map and grouping zeros:
- Grouping zeros into quads and pairs.
Simplified POS Expression:
\(F(U, V, W, Z) = (U + V + W)(V' + W')(\dots)\)
In simple words: Group zeros in the K-map for POS expressions just as you group ones for SOP expressions.

Exam Tip: Be careful with variable mapping conventions when switching between SOP and POS K-maps.

Question 16. Reduce the following boolean expression using K-Map: \(F(A, B, C, D) = \sum (5, 6, 7, 8, 9, 12, 13, 14, 15)\).
Answer: Plotting the minterms on a 4-variable K-map:
- Form quads and pairs covering all specified minterms.
Simplified SOP Expression:
\(F(A, B, C, D) = AC + BD + A'BC + AB'D'\)
In simple words: Combine wrapped-around edges and adjacent cells to simplify the expression completely.

Exam Tip: Remember that K-maps wrap around top-to-bottom and left-to-right.