1. Lesson Objectives Aims Be able to define problems using Boolean logic
Manipulate Boolean expressions including the use of Karnaugh maps to simplify those expressions
Use De Morgan’s Laws, distribution, association, commutation and double negation
Use logic diagrams and truth tables – half and full adders

2. Do NOT trust the book on this topic:
Note
Do NOT trust the book on this topic:
There ARE mistakes in there
It does NOT provide a good explanation of these topics.

3. Biconditional equivalence
Key terms
Symbol
Formal term
Informal term ∧
Conjunction
AND ∨
Disjunction OR
Negation
NOT →
Implication IF ↔
Biconditional equivalence
Equality ⊕
Exclusive or
XOR

4. Conjunction - AND P Q
P AND Q T F P Q
P AND Q 1

5. Disjunction - OR P Q
P OR Q T F P Q
P OR Q 1

6. Inverse (NOT) P
NOT P T F P
NOT P 1

7. Writing Boolean Expressions

11. The easiest way to simplify an expression with up to 4 inputs
Karnaugh Mapping
The easiest way to simplify an expression with up to 4 inputs
Basic gist:
Get the Boolean expression
Make truth table
Plot karnaugh map
Look for blocks (in powers of 2 (book makes a mistake here!!))
Interpret what those blocks mean

12. More rules
It does not matter which order you plot the inputs on the map
A zero will be interpreted as NOT
Each rule is conjoined by OR
You are looking for what does NOT change in a block
If you have two inputs together in a block that do not change, they will be ANDed

13. Example Simplify ¬A ∧ (B ⋁ ¬A) Truth Table (2 inputs) A B ¬A B ⋁ ¬A 11 A B ¬A
B ⋁ ¬A
¬A Λ (B ⋁ ¬A) 1 A B 1 A B 1 A B ¬A 1

14. Plot the Karnaugh Map A B 1 A B
¬A Λ (B ⋁ ¬A) 1 A B 1

15. Notes
It does not matter where you plot A and B, try it – you can reverse if you like
The order of the 1’s and 0’s in the table headings don’t matter either as long as only ONE variable changes at a time (irrelevant on 2 inputs, relevant on 3 or more)

16. Now look for groups of 1’s
Next
Now look for groups of 1’s
They MUST be in powers of 2 (you cannot select 5 or 3 for example) but you CAN have a group of 1 A B 1

17. Interpret the group ¬A ∧ (B ⋁ ¬A) ¬A A B 1 is
You are looking for the variable which does NOT change in the group.
In this example, B changes from 0 to 1, so we ignore it.
A stays the same – 0
If the variable that stays the same is a 1 then you write it down “as is”
If the variable that stays the same is a 0 then it is NOT that variable.
In this example, A stays the same, but is a Zero, so the rule for this group is ¬A
There are no more groups so the simplification of
¬A ∧ (B ⋁ ¬A) is ¬A A B 1

18. Harder Question – 3 inputs
Simplify A ∧ B ⋁ A ∧ (B ⋁ C) ⋁ B ∧ (B ⋁ C)
Hint, your Karnaugh Map will look like this:
Note the order of the column AB – it does not go 00, 01, 10, 11 as you would expect. This is because only ONE variable may change from one column to the next AB C 00 01 11 10 1

19. A B C
A Λ B
B ⋁ C
A Λ (B ⋁ C)
B Λ (B ⋁ C)
A Λ B ⋁ A Λ (B ⋁ C) ⋁ B Λ (B ⋁ C) 1

20. AB C 00 01 11 10 1 A B C A Λ B B ⋁ C A Λ (B ⋁ C) B Λ (B ⋁ C)
A Λ B ⋁ A Λ (B ⋁ C) ⋁ B Λ (B ⋁ C) 1 AB C 00 01 11 10 1

21. B A Λ C AB C 00 01 11 10 1 AB C 00 01 11 10 1

22. A ∧ B ⋁ A ∧ (B ⋁ C) ⋁ B ∧ (B ⋁ C)
= B ⋁ (A Λ C)

24. Answers

25. Now try this…

27. Those groups give us: (A Λ B Λ C) V (A Λ ¬B Λ ¬C Λ ¬D) And also: (¬A Λ C Λ ¬D) V (¬A Λ ¬C Λ ¬D)

28. (¬A Λ C Λ ¬D) V (¬A Λ ¬C Λ ¬D) This can be simplified: ¬A V ¬A = ¬A C V ¬C = 1 (So removed) ¬D V ¬D = ¬D Therefore (¬A Λ C Λ ¬D) V (¬A Λ ¬C Λ ¬D) = (¬A Λ ¬D)

30. Simplifying expressions
Rules…

31. Basic rules
Work out the brackets first – expand them
Then look for rules that cancel out
Then re-arrange if necessary to create other simplification rules from the table
Continue until reduced to simplest form

32. Exercise
Its worksheet time

33. Simplification questions

34. Answers – hello cheats

35. What do the outputs look like? Sum = XOR Carry = AND
Circuits
Half Adder:
What do the outputs look like?
Sum = XOR
Carry = AND
INPUTS                 OUTPUTS
A             B             SUM      CARRY
0              0              0              0
0              1              1              0
1              0              1              0
1              1              0              1

36. Half Adder
You will need to be able to draw this circuit from memory in the exam.

37. Why is it called a half adder?
Because it doesn’t take in to account any carry values being brought IN to the calculation

38. Full Adder Truth Table:
INPUTS                 OUTPUTS
A             B             CIN         COUT    S
0              0              0              0              0
0              0              1              0              1
0              1              0              0              1
0              1              1              1              0
1              0              0              0              1
1              0              1              1              0
1              1              0              1              0
1              1              1              1              1

39. Full adder circuit

40. S = (A ⊕ B) ⊕ Cin
Cout = (A ∧ B) ∨ (Cin ∧ (A ⊕ B) )

41. 4 bit Adder
What happens to the final Cout?
What implication does the cascaded carry have?

42. Implication Boolean implication A implies B simply meansQ
P → Q T F
Boolean implication A implies B simply means
"if A is true, then B must be true".
This implies that if A isn't true, then B can be anything.
This can also be read as (not A) or B
"either A is false, or B must be true"

43. Review/Success Criteria