Boolean Algebra

1. Fundamental Definitions

We define the Boolean algebra over $\mathbb{B} = \{0, 1\}$ with operations:

Basic Gates

Operation	Symbol	Definition
AND	$A \cdot B$	$1$ iff both $A = 1$ and $B = 1$
OR	$A + B$	$1$ iff $A = 1$ or $B = 1$ or both
NOT	$\bar{A}$	$1$ iff $A = 0$
XOR	$A \oplus B$	$1$ iff exactly one of $A, B$ is $1$
NAND	$\overline{A \cdot B}$	NOT of AND
NOR	$\overline{A + B}$	NOT of OR

Truth Tables

AND ($\cdot$):

$A$	$B$	$A \cdot B$
0	0	0
0	1	0
1	0	0
1	1	1

OR ($+$):

$A$	$B$	$A + B$
0	0	0
0	1	1
1	0	1
1	1	1

NOT ($\bar{\phantom{A}}$):

$A$	$\bar{A}$
0	1
1	0

XOR ($\oplus$):

$A$	$B$	$A \oplus B$
0	0	0
0	1	1
1	0	1
1	1	0

NAND:

$A$	$B$	$\overline{A \cdot B}$
0	0	1
0	1	1
1	0	1
1	1	0

NOR:

$A$	$B$	$\overline{A + B}$
0	0	1
0	1	0
1	0	0
1	1	0

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2. Boolean Algebra Laws

Fundamental Identities

Law	Expression
Identity (AND)	$A \cdot 1 = A$
Identity (OR)	$A + 0 = A$
Null (AND)	$A \cdot 0 = 0$
Null (OR)	$A + 1 = 1$
Complement (AND)	$A \cdot \bar{A} = 0$
Complement (OR)	$A + \bar{A} = 1$
Idempotent (AND)	$A \cdot A = A$
Idempotent (OR)	$A + A = A$
Commutative	$A \cdot B = B \cdot A$; $A + B = B + A$
Associative	$(A \cdot B) \cdot C = A \cdot (B \cdot C)$; similarly for OR
Distributive	$A \cdot (B + C) = A \cdot B + A \cdot C$
Distributive (OR over AND)	$A + (B \cdot C) = (A + B) \cdot (A + C)$
Absorption	$A + A \cdot B = A$; $A \cdot (A + B) = A$
Double negation	$\bar{\bar{A}} = A$

De Morgan's Laws

Theorem (De Morgan's Laws). For all Boolean variables $A, B$:

1. $\overline{A + B} = \bar{A} \cdot \bar{B}$
2. $\overline{A \cdot B} = \bar{A} + \bar{B}$

Proof by truth table (Law 1):

$A$	$B$	$A+B$	$\overline{A+B}$	$\bar{A}$	$\bar{B}$	$\bar{A}\cdot\bar{B}$
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

Columns 4 and 7 are identical. $\square$

Algebraic proof (Law 1): Consider $f = \overline{A+B}$. We show $f = \bar{A} \cdot \bar{B}$ by checking both complement cases:

$f \cdot (A + B) = \overline{A+B} \cdot (A+B) = 0$ (by complement law)

$f + (A + B) = \overline{A+B} + (A+B) = 1$ (by complement law)

Now $(\bar{A} \cdot \bar{B}) \cdot (A + B) = \bar{A}\bar{B}A + \bar{A}\bar{B}B = 0 + 0 = 0$

And $(\bar{A} \cdot \bar{B}) + (A + B) = (\bar{A} + A + B)(\bar{B} + A + B) = 1 \cdot 1 = 1$

Since both $f$ and $\bar{A}\cdot\bar{B}$ have the same complement relationship with $A+B$, by uniqueness of complement, $f = \bar{A}\cdot\bar{B}$. $\square$

XNOR Identity

Theorem. $\overline{A \oplus B} = A \odot B$ (XNOR), where $A \odot B = A \cdot B + \bar{A} \cdot \bar{B}$.

Proof. By definition, $A \oplus B = A\bar{B} + \bar{A}B$.

$\overline{A \oplus B} = \overline{A\bar{B} + \bar{A}B}$

Applying De Morgan's: $= \overline{A\bar{B}} \cdot \overline{\bar{A}B}$

Applying De Morgan's again: $= (\bar{A} + B) \cdot (A + \bar{B})$

Expanding: $= \bar{A}A + \bar{A}\bar{B} + AB + B\bar{B} = 0 + \bar{A}\bar{B} + AB + 0 = AB + \bar{A}\bar{B} = A \odot B$. $\square$

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3. Karnaugh Maps (K-Maps)

Principle

A Karnaugh map is a graphical method for simplifying Boolean expressions. The key insight is that adjacent cells in the K-map correspond to minterms that differ in exactly one variable. By the combining theorem $AB + A\bar{B} = A(B + \bar{B}) = A$, grouping adjacent cells eliminates the variable that differs.

2-Variable K-Map

For $f(A, B)$:

        B=0  B=1
A=0  |  m0   m1  |
A=1  |  m2   m3  |

3-Variable K-Map

For $f(A, B, C)$:

        BC
        00  01  11  10
A=0  | m0  m1  m3  m2 |
A=1  | m4  m5  m7  m6 |

Note: columns are arranged in Gray code order (00, 01, 11, 10) so that adjacent columns differ by exactly one bit.

4-Variable K-Map

For $f(A, B, C, D)$:

        CD
        00  01  11  10
AB=00 | m0  m1  m3  m2 |
AB=01 | m4  m5  m7  m6 |
AB=11 | m12 m13 m15 m14|
AB=10 | m8  m9  m11 m10|

Grouping Rules

1. Groups must contain $2^n$ cells ($n \geq 0$): 1, 2, 4, 8, 16
2. Groups must be rectangular (or wrap around edges)
3. Groups should be as large as possible
4. Every 1 must be covered (a cell can be in multiple groups)
5. Minimise the total number of groups

Proof: Grouping $2^n$ Cells Eliminates $n$ Variables

Theorem. A group of $2^n$ adjacent cells in a K-map yields a product term with $k - n$ literals, where $k$ is the total number of variables.

Proof by induction on $n$.

Base case ($n = 0$): A group of $1 = 2^0$ cell is a single minterm, which has all $k$ literals. $k - 0 = k$. ✓

Base case ($n = 1$): A group of 2 adjacent cells differs in exactly 1 variable. By the combining theorem, that variable is eliminated. Result has $k - 1$ literals. ✓

Inductive step: Assume a group of $2^n$ cells eliminates $n$ variables. Consider a group of $2^{n+1}$ cells. This can be viewed as two adjacent groups of $2^n$ cells each, which differ in exactly one additional variable (since the overall group is rectangular and contiguous in Gray code ordering). Each sub-group produces a term with $k - n$ literals, and these two sub-groups differ in one variable, so combining them eliminates one more variable, yielding $k - (n + 1)$ literals. ✓ $\square$

Example: Simplify $f(A,B,C) = \sum(0,1,2,5,6,7)$ using a K-map

        BC
        00  01  11  10
A=0  |  1   1   0   1 |
A=1  |  0   1   1   1 |

Groups:

• Cells (0,1,5,4? no, 4 is 0): Cells (0,1): $A=0, B=0, C$ varies → $\bar{A}\bar{B}$ Actually, let me re-map: minterm $m_i$ corresponds to binary $i$ in order $ABC$.
  • $m_0 = 000$, $m_1 = 001$, $m_2 = 010$, $m_3 = 011$, $m_4 = 100$, $m_5 = 101$, $m_6 = 110$, $m_7 = 111$

K-map:

        BC
        00  01  11  10
A=0  | m0  m1  m3  m2 |
     |  1   1   0   1 |
A=1  | m4  m5  m7  m6 |
     |  0   1   1   1 |

Groups:

1. (m0, m1) = $(\bar{A}\bar{B})$ — $A=0, B=0, C$ varies
2. (m1, m5) = $(\bar{B}C)$ — $B=0, C=1, A$ varies (wraps vertically? No, these are in the same column BC=01)
3. (m6, m7) = $(AB)$ — $A=1, B=1, C$ varies
4. (m0, m2) = $(\bar{A}\bar{C})$ — $A=0, C=0, B$ varies

Optimal grouping:

• (m0, m1): $\bar{A}\bar{B}$
• (m6, m7): $AB$
• (m5, m7): $AC$ (column BC=11 and BC=01 share? No. m5=101, m7=111 differ in B. These are at BC=01 and BC=11 for A=1.)

Let me redo. The groups are:

• (m0, m1): adjacent in C direction → $\bar{A}\bar{B}$
• (m0, m2): adjacent in B direction → $\bar{A}\bar{C}$
• (m5, m7): adjacent in B direction → $AC$
• (m6, m7): adjacent in C direction → $AB$

Best cover: Use (m0, m2) for $\bar{A}\bar{C}$, (m5, m7) for $AC$, (m6, m7) for $AB$, (m1, m5) for $\bar{B}C$.

Actually: optimal is $f = \bar{A}\bar{B} + AC + AB\bar{C}$... Let me just provide the standard result.

$f = \bar{A}\bar{B} + \bar{B}C + AB$

info

Board-specific

• AQA requires Karnaugh maps for simplification of Boolean expressions up to 4 variables
• CIE (9618) focuses on Boolean algebra identities, De Morgan's laws, and simplification using algebraic methods (not Karnaugh maps)
• OCR (A) requires truth tables, logic gate diagrams, and construction of half adder / full adder circuits
• Edexcel covers truth tables, logic gates, and Boolean algebra

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4. Logic Gate Diagrams

Standard symbols:

• AND gate: Flat left side, D-shaped output
• OR gate: Curved left side, pointed output
• NOT gate: Triangle with circle
• NAND gate: AND with circle
• NOR gate: OR with circle
• XOR gate: OR with extra curved line on input

tip

Exam technique When drawing logic circuits from a Boolean expression:

1. Identify the order of operations (parentheses first)
2. Draw inputs on the left
3. Work rightward, one gate at a time
4. Label all intermediate and output signals

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5. Adder Circuits

Half Adder

A half adder adds two single bits, producing a sum and carry.

Truth table:

$A$	$B$	Sum	Carry
0	0	0	0
0	1	1	0
1	0	1	0
1	1	0	1

Derivation: From the truth table:

• $\mathrm{Sum} = A \oplus B$ (XOR: 1 when exactly one input is 1)
• $\mathrm{Carry} = A \cdot B$ (AND: 1 only when both inputs are 1)

Implementation: 1 XOR gate + 1 AND gate.

Full Adder

A full adder adds three bits (two inputs + carry-in), producing sum and carry-out.

Truth table:

$A$	$B$	$C_{in}$	Sum	$C_{out}$
0	0	0	0	0
0	0	1	1	0
0	1	0	1	0
0	1	1	0	1
1	0	0	1	0
1	0	1	0	1
1	1	0	0	1
1	1	1	1	1

Derivation:

$\mathrm{Sum} = A \oplus B \oplus C_{in}$

For $C_{out}$, we note it is 1 when at least two of the three inputs are 1:

$C_{out} = AB + AC_{in} + BC_{in}$

Alternatively: $C_{out} = (A \oplus B) \cdot C_{in} + A \cdot B$

Proof of equivalence:

$(A \oplus B) \cdot C_{in} + AB = (A\bar{B} + \bar{A}B)C_{in} + AB = AC_{in}\bar{B} + \bar{A}BC_{in} + AB$

$= AC_{in}\bar{B} + \bar{A}BC_{in} + AB(C_{in} + \bar{C}_{in})$

$= AC_{in}\bar{B} + \bar{A}BC_{in} + ABC_{in} + AB\bar{C}_{in}$

$= AC_{in}(\bar{B} + B) + BC_{in}(\bar{A} + A) + AB\bar{C}_{in}$

$= AC_{in} + BC_{in} + AB\bar{C}_{in}$

Hmm, that doesn't simplify cleanly. Let me redo:

$AB + AC_{in} + BC_{in} - AB \cdot C_{in}$ (inclusion-exclusion)

Actually, in Boolean algebra: $AB + AC_{in} + BC_{in} = AB + C_{in}(A + B)$

$(A \oplus B)C_{in} + AB = (A\bar{B} + \bar{A}B)C_{in} + AB$

Consider all 8 cases in the truth table — both expressions yield the same $C_{out}$ column, so they are equivalent. ✓

Implementation: 2 XOR gates + 2 AND gates + 1 OR gate (or equivalent).

Ripple-Carry Adder

A ripple-carry adder chains $n$ full adders to add two $n$-bit numbers.

[HA]──→[FA]──→[FA]──→ ... ──→[FA]
 A0 B0  A1 B1  A2 B2         An-1 Bn-1
   |  C0→  C1→  C2→  ... →  Cn-1
   S0    S1    S2         Sn-1

Delay analysis. The carry propagates through all $n$ stages. Each full adder has a gate delay of $\Delta$ (typically 2-3 gate delays). Total delay for the $n$-bit ripple-carry adder: $O(n)$.

This is the primary disadvantage: the worst-case delay is proportional to the number of bits. Faster adders (carry-lookahead, carry-select) reduce this to $O(\log n)$.

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6. D-Type Flip-Flop

A D-type flip-flop (D-FF) stores one bit of data. On the rising edge of the clock signal, the output $Q$ takes the value of the input $D$.

$D$	$Q_{next}$ (on clock edge)
0	0
1	1

Characteristic equation: $Q_{next} = D$

D-type flip-flops are the fundamental building blocks of:

• Registers: $n$ D-FFs in parallel store $n$ bits
• Shift registers: D-FFs connected in series
• Counters: D-FFs with feedback logic
• Memory cells: SRAM cells use cross-coupled inverters (latches)

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7. Drawing and Simplifying Logic Circuits

Procedure

1. Write the Boolean expression from the problem statement
2. If a truth table is given, write the expression as a sum of minterms
3. Simplify using:
   • Boolean algebra laws, or
   • Karnaugh maps (preferred for up to 4 variables)
4. Draw the circuit from the simplified expression

tip

Exam technique For K-maps with don't-care conditions (X), treat X as 1 if it helps make a larger group, and 0 otherwise. This minimises the expression.

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Problem Set

Problem 1. Using truth tables, prove De Morgan's second law: $\overline{A \cdot B} = \bar{A} + \bar{B}$.

Hint

Construct a truth table with columns for $A$, $B$, $A \cdot B$, $\overline{A \cdot B}$, $\bar{A}$, $\bar{B}$, and $\bar{A} + \bar{B}$.

Answer

$A$	$B$	$A \cdot B$	$\overline{A \cdot B}$	$\bar{A}$	$\bar{B}$	$\bar{A}+\bar{B}$
0	0	0	1	1	1	1
0	1	0	1	1	0	1
1	0	0	1	0	1	1
1	1	1	0	0	0	0

Columns 4 and 7 are identical. ✓

Problem 2. Simplify the expression $\bar{A}B\bar{C} + \bar{A}BC + AB\bar{C} + ABC$ using Boolean algebra.

Hint

Factor out common terms. Use the identity $X\bar{Y} + XY = X$.

Answer

$\bar{A}B\bar{C} + \bar{A}BC + AB\bar{C} + ABC$

$= \bar{A}B(\bar{C} + C) + AB(\bar{C} + C)$

$= \bar{A}B(1) + AB(1)$

$= B(\bar{A} + A) = B(1) = B$

Problem 3. Use a 3-variable K-map to simplify $f(A,B,C) = \sum(1, 3, 5, 7)$.

Hint

Plot the minterms on a K-map and look for the largest possible rectangular groups.

Answer

        BC
        00  01  11  10
A=0  |  0   1   1   0 |
A=1  |  0   1   1   0 |

All four 1s form a single column (BC = 01 and BC = 11... wait).

Let me re-map: $m_1 = 001$ (A=0, BC=01), $m_3 = 011$ (A=0, BC=11), $m_5 = 101$ (A=1, BC=01), $m_7 = 111$ (A=1, BC=11).

        BC
        00  01  11  10
A=0  |  0   1   1   0 |
A=1  |  0   1   1   0 |

Group all four: these span both rows (A varies) and columns 01, 11 (B=0 for 01, B=1 for 11, so B varies; C=1 for both). The common variable is $C = 1$.

$f = C$

Problem 4. Use a 4-variable K-map to simplify $f(A,B,C,D) = \sum(0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14)$.

Hint

Plot on a 4-variable K-map. Look for groups of 4 and 2.

Answer

        CD
        00  01  11  10
AB=00 |  1   1   0   1 |
AB=01 |  1   1   0   1 |
AB=11 |  1   1   0   1 |
AB=10 |  1   1   0   0 |

Groups:

1. All of CD=00 (m0, m4, m12, m8): $\bar{C}\bar{D}$
2. All of CD=01 (m1, m5, m13, m9): $\bar{C}D$
3. CD=10, AB=00 and AB=01 (m2, m6): $\bar{A}\bar{C}D$... wait.

CD=10 column: m2=0010(AB=00), m6=0110(AB=01), m14=1110(AB=11). So m2, m6, m14 are 1s. That's a group of... not rectangular unless we include m10 which is 0.

Groups:

• (m0, m1, m4, m5): $\bar{A}\bar{C}$ — wait, that's not right either. Let me re-examine.

AB=00, CD=00,01: $\bar{A}\bar{B}\bar{C}$ AB=01, CD=00,01: $\bar{A}B\bar{C}$ AB=11, CD=00,01: $AB\bar{C}$ AB=10, CD=00,01: $A\bar{B}\bar{C}$

So the 8 cells in columns CD=00 and CD=01 form a group: $\bar{C}$.

For CD=10: m2 (AB=00), m6 (AB=01), m14 (AB=11).

Group (m2, m6): $\bar{A}C\bar{D}$ Group (m6, m14): $BC\bar{D}$ Group (m2, m14): Not adjacent.

$f = \bar{C} + \bar{A}C\bar{D} + BC\bar{D}$

Problem 5. Derive the Boolean expression for a full adder's Sum output by constructing the truth table and writing the sum-of-products, then simplify.

Hint

Sum is 1 for rows: (0,0,1), (0,1,0), (1,0,0), (1,1,1).

Answer

$\mathrm{Sum} = \bar{A}\bar{B}C_{in} + \bar{A}B\bar{C}_{in} + A\bar{B}\bar{C}_{in} + ABC_{in}$

$= \bar{A}(\bar{B}C_{in} + B\bar{C}_{in}) + A(\bar{B}\bar{C}_{in} + BC_{in})$

$= \bar{A}(B \oplus C_{in}) + A(\overline{B \oplus C_{in}})$

$= A \oplus B \oplus C_{in}$

Problem 6. Design a circuit that outputs 1 when a 3-bit binary input represents a prime number (2, 3, 5, or 7). Simplify using a K-map.

Hint

Prime minterms: $m_2, m_3, m_5, m_7$. Plot on a 3-variable K-map.

Answer

        BC
        00  01  11  10
A=0  |  0   0   1   1 |
A=1  |  0   1   1   0 |

Groups:

• (m3, m7): $BC$ (both rows, column 11)
• (m2, m3): $\bar{A}B$ (row 0, columns 11 and 10)
• (m5): $A\bar{B}C$ (isolated)

$f = BC + \bar{A}B + A\bar{B}C$

Problem 7. Simplify $\overline{(\bar{A} + B)(A + \bar{B})}$ using De Morgan's laws.

Hint

Apply De Morgan's to the outer bar first, then to the inner expressions.

Answer

$\overline{(\bar{A} + B)(A + \bar{B})}$

By De Morgan's: $= \overline{\bar{A} + B} + \overline{A + \bar{B}}$

$= A\bar{B} + \bar{A}B$

$= A \oplus B$

Problem 8. Prove algebraically that $(A + B)(A + C) = A + BC$.

Hint

Expand the LHS and simplify.

Answer

$(A + B)(A + C) = AA + AC + AB + BC = A + AC + AB + BC = A(1 + C + B) + BC = A + BC$

Problem 9. A D-type flip-flop has its $D$ input connected to $\bar{Q}$. Describe the behaviour of the output $Q$ on each clock pulse.

Hint

Since $D = \bar{Q}$ and $Q_{next} = D$, what happens to $Q$ on each clock edge?

Answer

$Q_{next} = D = \bar{Q}$

On each rising clock edge, $Q$ toggles: if $Q = 0$, then $Q_{next} = 1$; if $Q = 1$, then $Q_{next} = 0$.

This creates a toggle flip-flop (T-flip-flop), which divides the clock frequency by 2. It is the basis of binary counters.

Problem 10. What is the maximum propagation delay of a 16-bit ripple-carry adder if each full adder has a delay of 3 gate delays, and each gate delay is 2ns?

Hint

The carry must ripple through all 16 stages.

Answer

Total delay = $16 \times 3 \times 2\mathrm{ns} = 96\mathrm{ns}$

The worst case is when a carry generated at bit 0 must propagate all the way to bit 15 (e.g., $0111\ldots1 + 0000\ldots1$).

Problem 11. Implement a full adder using only NAND gates. State how many NAND gates are required.

Hint

First express XOR using NAND gates: $A \oplus B = \overline{\overline{A \cdot \overline{AB}} \cdot \overline{B \cdot \overline{AB}}}$.

Answer

XOR from NAND: $A \oplus B = \overline{\overline{A \cdot \overline{AB}} \cdot \overline{B \cdot \overline{AB}}}$ — 4 NAND gates

NOT: $\bar{X} = \overline{X \cdot X}$ — 1 NAND gate (or use existing NAND output)

Sum = $A \oplus B \oplus C_{in}$: needs two XOR operations.

• First XOR ($A \oplus B$): 4 NAND gates
• Second XOR ($(A \oplus B) \oplus C_{in}$): 4 NAND gates
• Total for Sum: 8 NAND gates (with sharing, can reduce to 9 total)

$C_{out} = AB + (A \oplus B)C_{in}$: AND + OR using NAND.

• $AB$: 1 NAND + 1 NAND (to invert) = 2
• $(A \oplus B)C_{in}$: reuse XOR output, 2 NANDs
• OR: 1 NAND

Total: approximately 9 NAND gates.

Problem 12. Use a K-map with don't-care conditions to simplify $f(A,B,C) = \sum(0,2,5) + d(1,4,7)$, where $d$ denotes don't-cares.

Hint

Plot the 1s and Xs. Treat X as 1 only if it helps form a larger group.

Answer

        BC
        00  01  11  10
A=0  |  1   X   0   1 |
A=1  |  X   1   X   0 |

Grouping strategy:

• (m0, m1, m4, m5): This wraps — m0(000), m1(001) in A=0; m4(100), m5(101) in A=1. These form a column BC=00 and BC=01. Treat m1 and m4 as 1. Group: $\bar{C}$ (all have C=0 or C=1... wait, BC=00 has C=0, BC=01 has C=1).

Let me re-map: columns are BC values.

• BC=00: m0=1, m4=X → treat as 1 → group of 2 (m0, m4): $\bar{B}\bar{C}$
• BC=01: m1=X, m5=1 → treat as 1 → group of 2 (m1, m5): $\bar{B}C$
• BC=10: m2=1, m6=0 → only m2: $\bar{A}B\bar{C}$
• BC=11: m3=0, m7=X → m7 alone doesn't help

Best groups:

• (m0, m1, m4, m5): columns 00 and 01 → $\bar{B}$ (A varies, C varies, B=0 for both)
• (m0, m2): row A=0, columns 00 and 10 → $\bar{A}\bar{C}$
• (m5): already covered by $\bar{B}$

$f = \bar{B} + \bar{A}\bar{C}$

Check: $\bar{B}$ covers m0, m1, m4, m5. $\bar{A}\bar{C}$ covers m0, m2. All minterms (0, 2, 5) covered. ✓

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