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Boolean Theorems, SOP, POS and K-Map ISC Class 12

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Computer ScienceISC Class 12 · Boolean Algebra

Boolean Theorems, SOP and POS Forms and Karnaugh Maps (ISC Class 12 Computer Science)

Updated 17 October 2026 · Draft, not yet fact-checked

Reduce a Boolean function by theorems (absorption, De Morgan, duality) or by a Karnaugh map: mark the minterms (or maxterms) on a "Gray"-coded grid, circle groups of 1, 2, 4 or 8 adjacent cells, and write one product (or sum) term per group. Canonical SOP lists minterms; canonical POS lists maxterms.

Circle neighbouring 1s in pairs (blank cells are 0); each loop keeps only the variables that do not change, so F = A'B + AC.

Illustration drawn for QuestionBuddy.

Understand Boolean Theorems, SOP and POS Forms and Karnaugh Maps

Basic theorems. For variables A, B, C: idempotence A + A = A, A·A = A; operations with 0 and 1: A + 0 = A, A·1 = A, A + 1 = 1, A·0 = 0; complement A + A' = 1, A·A' = 0; involution (A')' = A; commutative, associative and distributive laws (A + BC = (A + B)(A + C)); absorption A + AB = A and A(A + B) = A; also A + A'B = A + B. Duality: swap + with ·, and 0 with 1; the dual of a true identity is true. De Morgan: (A + B)' = A'·B' and (AB)' = A' + B'. Laws are verified with a truth table by showing both sides have identical columns.

Minterms and maxterms. For n variables, a minterm is a product containing every variable once (complemented if its bit is 0); a maxterm is a sum containing every variable once (complemented if its bit is 1). For A, B, C: m5 = 101 = AB'C and M5 = A' + B + C'. Minterm m_i = (M_i)'.

Canonical forms. Canonical SOP is a sum of minterms of the rows where F = 1: F = Σm(...). Canonical POS is a product of maxterms of the rows where F = 0: F = ΠM(...). The shorthand with the index numbers is the cardinal form. A function with 8 rows, ones at 2, 3, 5, 7, has zeros at 0, 1, 4, 6.

Karnaugh map (up to four variables). Rows and columns are labelled in "Gray" code (00, 01, 11, 10) so neighbouring cells differ in one variable. Rules:
1. Put 1s (SOP) or 0s (POS) on the map.
2. Group cells in rectangles of 1, 2, 4, 8 or 16; groups may wrap round the edges and the four corners are adjacent.
3. Use the fewest, largest groups; overlap is allowed; every 1 must be covered.
4. A group yields the variables that stay constant. For SOP write the product (constant 1 means plain, 0 means complemented). For POS write the sum (constant 0 means plain, 1 means complemented).

Map layout for four variables ABCD: rows AB = 00, 01, 11, 10; columns CD = 00, 01, 11, 10; cell index = 8A + 4B + 2C + D.

Gates, adders and NAND/NOR realisation are in the Computer Hardware chapter; propositional logic is a separate topic.

Boolean Theorems, SOP and POS Forms and Karnaugh Maps: key facts

Absorption and its useful form
  • A + AB = A
  • A(A + B) = A
  • A + A'B = A + B
Most reductions finish with one of these.
De Morgan's theorems
  • (A + B)' = A'·B'
  • (A·B)' = A' + B'
  • Break the bar, change the sign
Valid for any number of variables.
Minterm and maxterm
  • Minterm: bit 1 plain, bit 0 complemented
  • Maxterm: bit 0 plain, bit 1 complemented
  • m_i is the complement of M_i
Index = binary value of ABC.
K-map group sizes
  • Pair eliminates 1 variable
  • Quad eliminates 2
  • Octet eliminates 3
  • Group size is a power of 2
For 4 variables: octet leaves 1 variable.

How to answer Boolean Theorems, SOP and POS Forms and Karnaugh Maps questions

Use this routine for a K-map reduction question.

  1. 1Write the function in minterm (or maxterm) number form; convert a product expression by expanding missing variables.
  2. 2Draw the map with "Gray"-code labels and place a 1 (SOP) or 0 (POS) in each listed cell.
  3. 3Circle octets first, then quads, then pairs, then singles, using the fewest groups.
  4. 4Check the wrap-around: top-bottom edges, left-right edges and the four corners.
  5. 5For each group, read off the variables that do not change and write its term.
  6. 6Join the terms with + (SOP) or · (POS), then verify two or three cells.

Boolean Theorems, SOP and POS Forms and Karnaugh Maps in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Majority vote of three sensors

Suppose a smart-room controller switches the fan on when at least two of three sensors A, B, C report heat.

Write the reduced Boolean expression for the fan.

Try it first, then show the solutionSolution
  1. Fan is on for minterms 3, 5, 6, 7.
  2. K-map pairs: m3 with m7 (BC), m5 with m7 (AC), m6 with m7 (AB).

Answer: Fan = AB + AC + BC

Memory hook: Any two of three: three pairs.

Locker rule with a forbidden pattern

Suppose a locker opens (F = 1) for all codes ABC except 000, 001, 011 and 111.

Write F as a POS expression.

Try it first, then show the solutionSolution
  1. Zeros at 0, 1, 3, 7.
  2. Pair 0 and 1 gives (A + B); pair 3 and 7 gives (B' + C').

Answer: F = (A + B)(B' + C')

Memory hook: POS follows the zeros.

Tricks to remember Boolean Theorems, SOP and POS Forms and Karnaugh Maps

Short memory aids. Say each one aloud once, then use it on a question the same day.

  • "Gray" code order

    Label the map 00, 01, 11, 10, never 00, 01, 10, 11.

    Use it when: Drawing the grid.

  • Min is one, max is zero

    Minterms are the rows where F is 1, maxterms are the rows where F is 0.

    Use it when: Choosing SOP or POS.

  • Bigger and fewer

    Make groups as big as possible and use as few as possible.

    Use it when: Avoiding a non-minimal answer.

  • Keep what stays

    Write only the variables that do not change inside the group.

    Use it when: Reading a term from a group.

Boolean Theorems, SOP and POS Forms and Karnaugh Maps: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Dual of an identity · 1 mark · MCQBoard level

Write the dual of A + 0 = A.
(a) A·1 = A (b) A·0 = A (c) A + 1 = A (d) A' + 0 = A'

Try it first, then show the solutionSolution
  1. Duality swaps + with · and 0 with 1.
  2. A + 0 = A becomes A·1 = A.

Answer: (a) A·1 = A

Example 2: Three-variable K-map · 3 marks · Short answerHarder board level

Reduce F(A, B, C) = Σm(0, 2, 4, 5, 6) using a Karnaugh map and show the groups.

Try it first, then show the solutionSolution
  1. Map with rows A = 0, 1 and columns BC = 00, 01, 11, 10. Ones at m0, m2 (A = 0, columns 00 and 10) and m4, m5, m6 (A = 1, columns 00, 01, 10). (1 mark)
  2. Quad m0, m2, m4, m6: columns 00 and 10 in both rows, where C = 0 always, so C'. Pair m4, m5: A = 1, B = 0, so AB'. (1 mark)
  3. F = C' + AB'. Check m5 = 101: AB' = 1, so F = 1; check m3 = 011: C' = 0 and AB' = 0, so F = 0. (1 mark)

Answer: F = C' + AB'

Example 3: Four-variable K-map with corners · 4 marks · Short answerHarder board level

Reduce F(P, Q, R, S) = Σm(0, 2, 5, 7, 8, 10, 13, 15). Name the groups.

Try it first, then show the solutionSolution
  1. Index = 8P + 4Q + 2R + S. Ones: m0 = 0000, m2 = 0010, m8 = 1000, m10 = 1010 (the four corners); m5 = 0101, m7 = 0111, m13 = 1101, m15 = 1111 (the middle two columns, rows 01 and 11). (1 mark)
  2. Corner quad: Q = 0 and S = 0 throughout, so Q'S'. (1.5 marks)
  3. Middle quad: Q = 1 and S = 1 throughout, so QS. (1 mark)
  4. F = Q'S' + QS, which is the XNOR of Q and S. (0.5 mark)

Answer: F = Q'S' + QS (two quads)

Example 4: POS reduction · 3 marks · Short answerHarder board level

Reduce F(A, B, C) = ΠM(0, 1, 3, 7) to a product of sums and verify it for minterm 2.

Try it first, then show the solutionSolution
  1. Zeros at 000, 001, 011, 111. Pair m0, m1: A = 0, B = 0, so the sum is (A + B). (1 mark)
  2. Pair m3, m7: B = 1, C = 1, so the sum is (B' + C'). (1 mark)
  3. F = (A + B)(B' + C'). For 010: A + B = 1 and B' + C' = 0 + 1 = 1, so F = 1, correct since 2 is not a maxterm number. (1 mark)

Answer: F = (A + B)(B' + C')

Example 5: Case: committee vote · 4 marks · Case-basedHOTS / competency

Suppose a motion in a club passes (F = 1) when at least two of the three officers A, B, C vote yes.
(i) List the voting patterns for which F = 1 and write the sum of those product terms. (1 mark)
(ii) Reduce the function using a Karnaugh map. (2 marks)
(iii) How many gates does the reduced circuit need, and of what kind? (1 mark)

Try it first, then show the solutionSolution
  1. (i) At least two yes: rows 011, 101, 110, 111, so F = Σm(3, 5, 6, 7) = A'BC + AB'C + ABC' + ABC.
  2. (ii) Pairs m3-m7 (BC), m5-m7 (AC), m6-m7 (AB): F = AB + AC + BC.
  3. (iii) Three 2-input AND gates and one 3-input OR gate.

Answer: (i) Σm(3, 5, 6, 7) (ii) F = AB + AC + BC (iii) 3 AND gates and 1 OR gate

Where students lose marks in Boolean Theorems, SOP and POS Forms and Karnaugh Maps

The slips examiners see most often, and what to write instead.

  • Mistake: Ordinary binary order on the map

    Neighbouring cells must differ in one bit; 10 next to 11 breaks this.

    Wrong working

    Columns CD labelled 00, 01, 10, 11

    Right working

    Columns CD labelled 00, 01, 11, 10

    Do this: Always label in "Gray" code.

    Usually costs: 1 mark

  • Mistake: Missing the wrap-around group

    Edge cells and corners are neighbours on a K-map.

    Wrong working

    Four corner ones written as four separate terms

    Right working

    One quad such as B'D' (with the correct constant variables)

    Do this: Check the opposite edges before finishing.

    Usually costs: 1 mark

  • Mistake: Using SOP rules for a POS group

    In POS a constant 0 is written plain and a constant 1 is complemented.

    Wrong working

    Zeros pair with A = 0, B = 0 written as A'B'

    Right working

    (A + B)

    Do this: For zeros, write a sum with the opposite choice.

    Usually costs: 1 mark

Exam tips

  • Draw the map neatly with variable labels and "Gray"-code headings; marks are given for the grid.
  • Show each group as a labelled loop (octet, quad, pair) as the question asks.
  • Write the term of every group beside the map before the final expression.
  • For canonical forms, write the cardinal numbers first and then the full expression.
  • Verify one cell inside each group and one cell outside all groups.

Practice questions on Boolean Theorems, SOP and POS Forms and Karnaugh Maps

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

  1. Q1 · 1 mark · MCQBoard level

    By De Morgan's theorem, (A + B)' equals
    (a) A' + B' (b) A'·B' (c) A·B (d) A + B'

    Hint

    Break the bar, change the sign.

    Answer

    (b) A'·B'

  2. Q2 · 2 marks · Short answerBoard level

    State the absorption law A + AB = A and verify it with a truth table.

    Hint

    Four rows for A and B.

    Answer

    Rows (A, B): 00 gives A + AB = 0; 01 gives 0; 10 gives 1 + 0 = 1; 11 gives 1 + 1 = 1. The column 0, 0, 1, 1 equals A, so the law holds.

  3. Q3 · 3 marks · Short answerHarder board level

    A function is given as F(A, B, C) = A'B + AC. Expand it into lists of minterm and maxterm numbers, then write out the full maxterm product.

    Hint

    Expand each term with the missing variable.

    Answer

    A'B = A'BC' + A'BC (m2, m3). AC = AB'C + ABC (m5, m7). F = Σm(2, 3, 5, 7) and F = ΠM(0, 1, 4, 6) = (A + B + C)(A + B + C')(A' + B + C)(A' + B' + C).

  4. Q4 · 4 marks · Short answerHarder board level

    Reduce F(A, B, C, D) = Σm(0, 1, 2, 3, 6, 7, 8, 9, 10, 11, 14, 15) using a K-map.

    Hint

    Look for an octet in the rows where B = 0.

    Answer

    Ones at 0-3 and 8-11 (B = 0, an octet) give B'. Ones at 6, 7, 14, 15 (B = 1, C = 1, a quad) give BC. F = B' + BC = B' + C. Zeros are 4, 5, 12, 13, where B = 1 and C = 0, which confirms it.

  5. Q5 · 3 marks · Case-basedHOTS / competency

    Suppose a hall light is on (F = 1) for switch settings A'B + AB' + AB. Reduce it (i) by Boolean algebra and (ii) by a K-map, and state the common result.

    Hint

    The last two terms share A.

    Answer

    (i) A'B + A(B' + B) = A'B + A = A + B. (ii) Ones at m1, m2, m3: the pair m2, m3 gives A and the pair m1, m3 gives B, so F = A + B. Both methods give A + B.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

  1. 1.State the dual of A·1 = A.Show answer

    A + 0 = A.

  2. 2.What is the complement of AB?Show answer

    A' + B' (De Morgan).

  3. 3.How many variables does a quad eliminate?Show answer

    Two.

  4. 4.Which rows give the canonical POS?Show answer

    The rows where the function is 0.

  5. 5.Write the maxterm for 101 in A, B, C.Show answer

    A' + B + C'.

  6. 6.Are the four corners of a 4-variable map adjacent?Show answer

    Yes, they form one quad.

Boolean Theorems, SOP and POS Forms and Karnaugh Maps: frequently asked questions

What is the difference between minterm and maxterm?

A minterm is a product of all variables that equals 1 for exactly one row; a maxterm is a sum of all variables that equals 0 for exactly one row.

What is the cardinal form of a Boolean function?

It is the short form that lists only the index numbers of the minterms (Σm) or maxterms (ΠM), for example F = Σm(2, 3, 5, 7).

Why does a K-map use "Gray" code?

Adjacent cells then differ in one variable, so a group of adjacent cells can drop the changing variable and give a simpler term.

Is it enough to use only the minterm form for the answer?

Not when the question asks for a reduced form; the canonical form is only the starting point and the K-map or algebra reduction gives the final expression.