Product of sums (POS) form: Worked examples — Hardware deep dive

Notation used on this page

Everything here is built from just three operations on the values $0$ and $1$ . We define them before using them anywhere.

Definition Minterm/maxterm index notation (

$\sum m$ and $\prod M$ ) Number every truth-table row by reading its input bits as a binary number (see the row-index formula below). Then:

A ==minterm $m_{i}$ == is the AND of all variables that is $1$ at exactly one row (row $i$ ) and $0$ everywhere else. Complement a variable if it is $0$ in that row.
A ==maxterm $M_{i}$ == is the OR of all variables that is $0$ at exactly one row (row $i$ ) and $1$ everywhere else. Complement a variable if it is $1$ in that row.
$\sum m (i, j, \dots)$ = "OR these minterms together" = the SOP listing all rows where $F = 1$ .
$\prod M (i, j, \dots)$ = "AND these maxterms together" = the POS listing all rows where $F = 0$ .

The $\sum$ ( $Σ$ ) means "sum/OR the listed minterms"; the $\prod$ ( $Π$ ) means "product/AND the listed maxterms". The numbers in brackets are row indices, not arithmetic.

The scenario matrix

Before any example, here is the full space of cases a POS problem can throw. Every cell below is covered by at least one worked example.

Cell	Case class	What is tricky	Example
C1	Standard 3-var, several $0$ rows	Get the complement rule right	Ex 1
C2	Degenerate: $F = 1$ everywhere (no $0$ rows)	What is an empty product?	Ex 2
C3	Degenerate: $F = 0$ everywhere (all $0$ rows)	Product of all maxterms	Ex 3
C4	2-variable baby case	Fewer literals per term	Ex 4
C5	Given SOP, convert to POS	Use complementary row set	Ex 5
C6	Simplify POS with a K-map (group $0$ s)	Non-canonical (reduced) POS	Ex 6
C7	Real-world word problem	Translate English → truth table → POS	Ex 7
C8	Exam twist: expression given, not a table	Build the table first, spot hidden $0$ s	Ex 8

Two quick reminders we lean on everywhere:

Ex 1 — Standard 3-variable (Cell C1)

Worked example Statement

$F (A, B, C)$ is $1$ everywhere except rows $1$ , $4$ , $6$ . Write the canonical POS.

Forecast: How many factors will the answer have? (Count the $0$ rows.) Which literals get complemented in row $6$ ?

Step 1 — list the $0$ rows. Rows $1, 4, 6$ . Why this step? POS is built from the rows where $F = 0$ ; each one contributes exactly one maxterm.

Step 2 — decode each row into bits using $index = 4 A + 2 B + C$ .

Row $1 = 001 \Rightarrow A = 0, B = 0, C = 1$ .
Row $4 = 100 \Rightarrow A = 1, B = 0, C = 0$ .
Row $6 = 110 \Rightarrow A = 1, B = 1, C = 0$ . Why this step? The maxterm rule reads off individual bits, so we must expand the index into $A, B, C$ first.

Step 3 — apply the complement rule (flip the $1$ s).

Row $1$ : $C = 1$ flips → $(A + B + \overset{ˉ}{C})$ .
Row $4$ : $A = 1$ flips → $(\overset{ˉ}{A} + B + C)$ .
Row $6$ : $A = 1, B = 1$ flip → $(\overset{ˉ}{A} + \overset{ˉ}{B} + C)$ . Why this step? An OR is $0$ only when every literal inside is $0$ ; complementing the $1$ -valued variables makes each literal $0$ at its own forbidden row.

Step 4 — AND them (write the factors side by side = juxtaposition = AND). $F = (A + B + \overset{ˉ}{C}) (\overset{ˉ}{A} + B + C) (\overset{ˉ}{A} + \overset{ˉ}{B} + C) = \prod M (1, 4, 6)$ Why AND? AND outputs $0$ if any factor is $0$ , so each maxterm can veto its one bad row while the others stay $1$ .

Verify: Plug in row $6$ $(1, 1, 0)$ : first factor $(1 + 1 + 0) = 1$ , second $(0 + 1 + 0) = 1$ , third $(0 + 0 + 0) = 0$ → product $= 0$ . ✓ Plug in row $0$ $(0, 0, 0)$ : $(0 + 0 + 1) = 1$ , $(1 + 0 + 0) = 1$ , $(1 + 1 + 0) = 1$ → product $= 1$ . ✓ Answer has 3 factors, matching 3 zero-rows.

Ex 2 — Degenerate: $F = 1$ everywhere (Cell C2)

Ex 3 — Degenerate: $F = 0$ everywhere (Cell C3)

Worked example Statement

$F (A, B) = 0$ for all four rows. Write its POS.

Forecast: Now every row is forbidden. How many factors? What must their product equal?

Step 1 — list $0$ rows using $index = 2 A + B$ (so $00 \to 0, 01 \to 1, 10 \to 2, 11 \to 3$ ). All of them: $0, 1, 2, 3$ . Why this step? One factor per $0$ row → four factors here.

Step 2 — write all four maxterms (flip the $1$ s in each row).

Row $0 = 00$ : $(A + B)$ .
Row $1 = 01$ : $(A + \overset{ˉ}{B})$ .
Row $2 = 10$ : $(\overset{ˉ}{A} + B)$ .
Row $3 = 11$ : $(\overset{ˉ}{A} + \overset{ˉ}{B})$ . Why this step? Each maxterm kills one row; killing all four rows means using all four.

Step 3 — AND them. $F = (A + B) (A + \overset{ˉ}{B}) (\overset{ˉ}{A} + B) (\overset{ˉ}{A} + \overset{ˉ}{B}) = \prod M (0, 1, 2, 3) = 0$ Why this equals $0$ : every row is somebody's forbidden row, so at every input at least one factor is $0$ , dragging the product to $0$ everywhere.

Verify: Try row $1$ $(A = 0, B = 1)$ : factor $(A + \overset{ˉ}{B}) = (0 + 0) = 0$ → product $0$ . ✓ Every other row similarly has one zero factor. So $F \equiv 0$ , the contradiction function. ✓

Ex 4 — Two-variable baby case (Cell C4)

Worked example Statement

$F (A, B) = 0$ only at rows $1$ and $2$ . Write the POS.

Forecast: With only two variables, each sum term has just two literals. Which two rows survive as $F = 1$ ?

Step 1 — decode the $0$ rows with $index = 2 A + B$ .

Row $1 = 01 \Rightarrow A = 0, B = 1$ .
Row $2 = 10 \Rightarrow A = 1, B = 0$ . Why this step? The two-bit index turns each row number back into $A, B$ values so we can apply the maxterm rule.

Step 2 — build the two maxterms.

Row $1$ : $B = 1$ flips → $(A + \overset{ˉ}{B})$ .
Row $2$ : $A = 1$ flips → $(\overset{ˉ}{A} + B)$ . Why this step? Same complement rule, just two literals per OR.

Step 3 — AND. $F = (A + \overset{ˉ}{B}) (\overset{ˉ}{A} + B) = \prod M (1, 2)$ Why this shape? Two forbidden rows → two factors.

Answering the Forecast — the surviving $F = 1$ rows. All four rows are ${0, 1, 2, 3}$ ; the forbidden rows are ${1, 2}$ , so the rows that survive as $F = 1$ are the leftovers ${0, 3}$ . Those are $A = 0, B = 0$ and $A = 1, B = 1$ — i.e. the two rows where $A$ and $B$ agree. Why this matters: it names the function ( $F = 1$ exactly when $A = B$ ), which we use in the check below.

Verify: This is the well-known XNOR. Check row $0$ $(0, 0)$ : $(0 + 1) (1 + 0) = 1 \cdot 1 = 1$ . Row $3$ $(1, 1)$ : $(1 + 0) (0 + 1) = 1$ . Rows $1, 2$ give $0$ . So $F = 1$ exactly on rows $0, 3$ (when $A = B$ ) — that's XNOR. ✓ Cross-check with SOP: $F = \sum m (0, 3)$ , the complementary rows. ✓

Ex 5 — Convert given SOP to POS (Cell C5)

Worked example Statement

You are handed $F (A, B, C) = \sum m (0, 3, 5, 7)$ . Write $F$ as canonical POS without re-deriving the whole table by hand.

Forecast: SOP lists the $F = 1$ rows. POS wants the $F = 0$ rows. What is the shortcut?

Step 1 — recognise the complementary-set trick. $\sum m (0, 3, 5, 7)$ means $F = 1$ at rows $0, 3, 5, 7$ . With 3 variables there are $2^{3} = 8$ rows: ${0, 1, 2, 3, 4, 5, 6, 7}$ . The $F = 0$ rows are the rest. Why this step? Every row is $1$ or $0$ — never both — so the two index lists partition all $8$ rows. This is exactly the " $\overline{\sum m_{i}} = \prod M_{i}$ " identity from the parent note in disguise.

Step 2 — subtract the sets. ${0, \dots, 7} ∖ {0, 3, 5, 7} = {1, 2, 4, 6}$ . Why this step? Those are the rows where $F = 0$ , so they become the maxterm indices.

Step 3 — write the POS symbol, then expand. $F = \prod M (1, 2, 4, 6)$

Row $1 = 001$ : $(A + B + \overset{ˉ}{C})$ .
Row $2 = 010$ : $(A + \overset{ˉ}{B} + C)$ .
Row $4 = 100$ : $(\overset{ˉ}{A} + B + C)$ .
Row $6 = 110$ : $(\overset{ˉ}{A} + \overset{ˉ}{B} + C)$ . $F = (A + B + \overset{ˉ}{C}) (A + \overset{ˉ}{B} + C) (\overset{ˉ}{A} + B + C) (\overset{ˉ}{A} + \overset{ˉ}{B} + C)$ Why this step? Same maxterm rule as always; the only new idea was finding the row list by set complement.

Verify: Row $3 = 011$ should be $F = 1$ (it's in the minterm list). Check: $(0 + 1 + 0) = 1$ , $(0 + 0 + 1) = 1$ , $(1 + 1 + 1) = 1$ , $(1 + 0 + 1) = 1$ → product $1$ . ✓ Row $2 = 010$ should be $0$ : second factor $(0 + 0 + 0) = 0$ → product $0$ . ✓

Ex 6 — Simplify POS with a K-map (Cell C6)

Worked example Statement

$F (A, B, C) = 0$ at rows $0, 1, 4, 5$ . Give the simplified (non-canonical) POS by grouping $0$ s on a Karnaugh map.

Forecast: Four zeros in a nice pattern. Can one small sum term cover all four? What single literal condition do rows $0, 1, 4, 5$ share?

A Karnaugh map is just the truth table drawn as a grid so that neighbouring cells differ in exactly one bit. The figure below plots each row's $F$ value; the four red cells are our $0$ rows and the green outline is the group we will read. Below the figure is the same grid written out as plain text so you can follow even without seeing the image.

Text version of the K-map (rows = $A$ , columns = $B C$ ; each cell shows $F$ and its row number $m_{i}$ ):

$A \ B C$	00	01	11	10
0	0 ( $m_{0}$ )	0 ( $m_{1}$ )	1 ( $m_{3}$ )	1 ( $m_{2}$ )
1	0 ( $m_{4}$ )	0 ( $m_{5}$ )	1 ( $m_{7}$ )	1 ( $m_{6}$ )

Step 1 — plot the $0$ s and find the largest rectangle of $0$ s. The zeros sit in the two columns $B C = 00$ and $B C = 01$ , across both rows $A = 0$ and $A = 1$ — a $2 \times 2$ block (green outline). Why this step? For POS we group the zeros (mirror image of SOP, where you group ones). Bigger groups → fewer, smaller factors.

Step 2 — see which variables change inside the group. Walk the four grouped cells: $A$ takes both $0$ and $1$ ; $C$ takes both $0$ and $1$ ; but $B$ is $0$ in every one of them (columns $B C = 00$ and $B C = 01$ both have $B = 0$ ). Why this step? A variable that changes across the group cannot help describe it — its two halves cancel. Only a variable that stays constant survives into the sum term. Here only $B$ is constant.

Step 3 — turn the surviving constant into a sum term via the maxterm rule. The group is a block of $0$ s, so it behaves like one big maxterm covering only the constant variable $B$ . Apply the maxterm rule to that constant: $B = 0$ throughout, and the rule says "complement a variable only if it is $1$ ." Since $B = 0$ , we keep it plain. A one-literal OR is just the literal itself, so the sum term is $(B)$ . $F = (B)$ Why this step? The literal $B$ is $0$ exactly on the $B = 0$ half of the table (rows $0, 1, 4, 5$ ) and $1$ on the other half — precisely the required $F$ . One factor replaces the four canonical maxterms $\prod M (0, 1, 4, 5)$ .

Verify: $F = B$ is $0$ whenever $B = 0$ (rows $0, 1, 4, 5$ ) and $1$ whenever $B = 1$ (rows $2, 3, 6, 7$ ). ✓ Compare with the canonical POS $\prod M (0, 1, 4, 5)$ — expanding and simplifying it collapses to $B$ , so the two agree while the K-map version needs far fewer gates. ✓

Ex 7 — Real-world word problem (Cell C7)

Worked example Statement

A greenhouse fan should run ( $F = 1$ ) unless it is both cold and dark. Inputs: $H$ = "warm" ( $1$ =warm, $0$ =cold), $L$ = "light" ( $1$ =lit, $0$ =dark). Write the POS controller for $F$ .

Forecast: "Both cold and dark" is one specific input combination. How many $0$ rows is that? So how many factors in the POS?

Step 1 — translate English into a truth table. "Cold and dark" means $H = 0$ (cold) and $L = 0$ (dark). That is exactly row $00$ . Everywhere else the fan runs.

$H$	$L$	$F$
0	0	0
0	1	1
1	0	1
1	1	1

Why this step? A verbal spec is ambiguous until pinned to rows; POS then reads straight off the $0$ rows.

Step 2 — one $0$ row → one maxterm. Row $0 = 00$ : both variables are $0$ , so nothing gets complemented → $(H + L)$ . Why this step? The complement rule: flip only the $1$ s, and here there are none.

Step 3 — the controller. $F = (H + L)$ Why one factor? Only one forbidden combination exists, so only one tripwire (sum term) is needed. Any single $0$ input would need both $H$ and $L$ at $0$ to drive $(H + L)$ to $0$ , which matches "cold and dark" perfectly.

Verify: $(H + L)$ is $0$ only when $H = 0$ and $L = 0$ (cold and dark) — exactly the "fan off" case; $1$ otherwise. ✓ Sanity: warm+dark $(1, 0)$ gives $(1 + 0) = 1$ , fan runs — correct, warmth alone keeps it on. ✓

Ex 8 — Exam twist: expression, not a table (Cell C8)

Worked example Statement

An exam gives $G (A, B, C) = \overset{ˉ}{A} \overset{ˉ}{B} + A C$ (an SOP-looking expression) and asks for the canonical POS. There is no table printed.

Forecast: You cannot read $0$ rows off an expression directly. What must you build first? Watch for rows the two product-terms both miss.

Step 1 — evaluate $G$ at all 8 rows to build the table. Recall $\overset{ˉ}{A} \overset{ˉ}{B}$ (juxtaposition = AND) is $1$ only when $A = 0$ and $B = 0$ ; $A C$ is $1$ only when $A = 1$ and $C = 1$ ; the $+$ ORs the two.

#	$A B C$	$\overset{ˉ}{A} \overset{ˉ}{B}$	$A C$	$G$
0	000	1	0	1
1	001	1	0	1
2	010	0	0	0
3	011	0	0	0
4	100	0	0	0
5	101	0	1	1
6	110	0	0	0
7	111	0	1	1

Why this step? POS is defined by $0$ rows, and the only reliable way to find them from an expression is to test every input — the "hidden $0$ s" are rows both terms fail (here rows $2, 3, 4, 6$ ).

Step 2 — collect the $0$ rows. Rows $2, 3, 4, 6$ . Why this step? These become the maxterm indices.

Step 3 — write each maxterm (flip the $1$ s).

Row $2 = 010$ : $B = 1$ → $(A + \overset{ˉ}{B} + C)$ .
Row $3 = 011$ : $B = 1, C = 1$ → $(A + \overset{ˉ}{B} + \overset{ˉ}{C})$ .
Row $4 = 100$ : $A = 1$ → $(\overset{ˉ}{A} + B + C)$ .
Row $6 = 110$ : $A = 1, B = 1$ → $(\overset{ˉ}{A} + \overset{ˉ}{B} + C)$ . $G = \prod M (2, 3, 4, 6) = (A + \overset{ˉ}{B} + C) (A + \overset{ˉ}{B} + \overset{ˉ}{C}) (\overset{ˉ}{A} + B + C) (\overset{ˉ}{A} + \overset{ˉ}{B} + C)$ Why this step? Standard rule once the table exists — the twist was purely getting to the table.

Verify: Row $5 = 101$ should be $1$ : $(1 + 0 + 1) = 1$ , $(1 + 0 + 0) = 1$ , $(0 + 0 + 1) = 1$ , $(0 + 0 + 1) = 1$ → product $1$ . ✓ Row $3 = 011$ should be $0$ : second factor $(0 + 0 + 0) = 0$ → product $0$ . ✓

Active recall

Empty product (no

F = 0

rows) equals?

1

(the AND identity — a tautology).

POS of

F = 0

everywhere over 2 vars?

\prod M (0, 1, 2, 3) = (A + B) (A + \overset{ˉ}{B}) (\overset{ˉ}{A} + B) (\overset{ˉ}{A} + \overset{ˉ}{B}) = 0

From

\sum m (0, 3, 5, 7)

, the POS indices are?

\prod M (1, 2, 4, 6)

(the complementary set).

For a POS via K-map you group which cells?

The

0

cells; a constant variable at

0

stays plain, at

1

is complemented.

Fan runs unless cold and dark; POS controller?

(H + L)

, with

H

=warm,

L

=light.

Connections

Parent: POS form — the rule these examples drill.
Sum of products (SOP) form — the dual we converted from in Ex 5.
Minterms and maxterms — the atoms of every example here.
De Morgan's theorems — hides inside the SOP→POS conversion (Ex 5, Ex 8).
Truth tables — every example starts or ends here.
Karnaugh maps — grouping $0$ s in Ex 6.
Logic gates — each POS is an AND-of-ORs circuit.

Worked examples — Product of sums (POS) form

How to read this page

Notation used on this page

The scenario matrix

Ex 1 — Standard 3-variable (Cell C1)

Ex 2 — Degenerate: $F = 1$ everywhere (Cell C2)

Ex 3 — Degenerate: $F = 0$ everywhere (Cell C3)

Ex 4 — Two-variable baby case (Cell C4)

Ex 5 — Convert given SOP to POS (Cell C5)

Ex 6 — Simplify POS with a K-map (Cell C6)

Ex 7 — Real-world word problem (Cell C7)

Ex 8 — Exam twist: expression, not a table (Cell C8)

Active recall

Connections