Boolean Algebra for ISC Class 12: Laws, Canonical SOP and POS, and Simplification

Boolean algebra makes up Section A of the ISC Computer Science theory paper: proving laws, simplifying expressions, converting to canonical form and drawing circuits. This guide covers the laws and the algebraic methods. Pair it with our K-map guide for map-based reduction.

Basic laws (each with its dual)

To get the dual of a law, swap + with ·, and swap 0 with 1.

• Identity: A + 0 = A, and A · 1 = A
• Null (domination): A + 1 = 1, and A · 0 = 0
• Idempotent: A + A = A, and A · A = A
• Complement: A + A' = 1, and A · A' = 0
• Involution: (A')' = A
• Commutative: A + B = B + A, and A · B = B · A
• Associative: A + (B + C) = (A + B) + C
• Distributive: A · (B + C) = AB + AC, and A + BC = (A + B)(A + C)
• Absorption: A + AB = A, and A(A + B) = A
• Redundancy: A + A'B = A + B, and A(A' + B) = AB
• De Morgan's: (A + B)' = A' · B', and (A · B)' = A' + B'

Consensus theorem

AB + A'C + BC = AB + A'C. The term BC is redundant.

Proof:

AB + A'C + BC
= AB + A'C + BC(A + A')          [A + A' = 1]
= AB + A'C + ABC + A'BC
= AB(1 + C) + A'C(1 + B)
= AB + A'C                        [1 + X = 1]

Algebraic simplification: solved examples

Example 1: AB + AB'

= A(B + B') = A · 1 = A

Example 2: (A + B)(A + B')

= A + BB'    [distributive: (A+B)(A+C) = A + BC]
= A + 0 = A

Example 3: A'B'C + A'BC + AB'C + ABC

= C(A'B' + A'B + AB' + AB)
= C(A'(B' + B) + A(B' + B))
= C(A' + A) = C

Example 4: (A + B)' + (A' + B')'

= A'B' + AB        [De Morgan's law on each term]
= A ⊙ B (XNOR)

Minterms and maxterms

For n variables, there are 2ⁿ input combinations.

• A minterm is a product (AND) term containing every variable once. A variable that is 0 is complemented. For example, ABC = 011 gives the minterm A'BC, which is m3.
• A maxterm is a sum (OR) term containing every variable once. A variable that is 1 is complemented. For example, ABC = 011 gives the maxterm (A + B' + C'), which is M3.

Canonical SOP is the OR of the minterms where F = 1, written Σ(...). Canonical POS is the AND of the maxterms where F = 0, written π(...). The two lists always complement each other.

Converting to canonical form

Express F(A, B, C) = A + B'C in canonical SOP and POS.

Multiply each term by (X + X') for every missing variable X:

A    = A(B + B')(C + C') = ABC + ABC' + AB'C + AB'C'
B'C  = (A + A')B'C       = AB'C + A'B'C

Removing the duplicate AB'C:

F = A'B'C + AB'C' + AB'C + ABC' + ABC = Σ(1, 4, 5, 6, 7)

The remaining combinations give the canonical POS:

F = π(0, 2, 3) = (A + B + C)(A + B' + C)(A + B' + C')

Check with the truth table: at m1 (001), A = 0 and B'C = 1, so F = 1. At m3 (011), A = 0 and B'C = 0, so F = 0.

Proving with a truth table

When a question says "verify using a truth table", list every combination, show a column for each intermediate expression, and point out that the final two columns are identical. Example: verifying De Morgan's law (A + B)' = A'B' needs the columns A, B, A + B, (A + B)', A', B' and A'B'.

Exam tips

• State the law you use at each step. ISC awards marks for justified steps, not just the final answer.
• Check a simplified answer by substituting two or three input combinations into both the original and the result.
• For "draw the logic circuit", use NOT gates for the complemented inputs, AND gates for each product term and one OR gate for the sum. Label every input.

Practice

1. Simplify: A + A'B + AB'
2. Write the dual of: A · (B + 0) = AB
3. Express F(X, Y) = X' + Y in canonical SOP.

Answers:

1. A + A'B = A + B, then + AB' is absorbed by A, giving A + B.
2. A + (B · 1) = A + B.
3. X' + Y = X'Y' + X'Y + XY = Σ(0, 1, 3).

Continue with Karnaugh Maps (K-Maps) Step by Step and test yourself in the solved ISC Class 12 Computer Science Sample Paper 2027.