Fundamentals of logic 1

FUNDAMENTALS OF LOGIC
BY
LAKSHMI R
ASST. PROFESSOR
DEPARTMENT OF ISE

PROPOSITION
• Statement or declaration which in a given context, can be either true or false,
but not both.
LAKSHMI R, ASST. PROFESSOR, DEPT. OF ISE
Propositions:
i. France borders Belgium
ii. 8 is an even number
(May not be true always!!)
i. Barcelona is in Italy
ii. 2 is rational
Not Propositions:
i. Can you pass the sheet?
ii. Listen to me.
iii. What a lovely dress!
iv. Give me a cup of tea.

TRUTH VALUE
• The truth or falsity of a proposition is called its truth value
Propositions:
i. France borders Belgium – with truth value TRUE (1)
ii. 8 is an even number – with truth value TRUE (1)
iii. Barcelona is in Italy - with truth value FALSE (0)
iv. 2 is rational - with truth value FALSE (0)

LOGICAL CONNECTIVES
• New phrases are obtained – not, or, and, if then, if and only if etc.,
• Compound propositions - The new proposition obtained by the use of logical
connectives eg: France borders Belgium and Norway borders Sweden
• Components or primitives – original propositions from which a compound
proposition is obtained.
• Simple propositions – Proposition that do not contain any logical connectives

LOGICAL CONNECTIVES
UNDERSTANDING

LOGICAL CONNECTIVES CONTD., - TRUTH TABLES
Negation (NOT) Conjunction (AND) Disjunction (OR)
Exclusive Disjunction (XOR) Conditional or Implication (if
then)
Bidirectional or Double
Implication (if and only if)
PQ AND Q  P

LOGICAL CONNECTIVES CONTD., EXAMPLES
Negation:
p: Madrid is a capital city
¬p : Madrid is not a capital city
Conjunction:
P: 2 is an irrational number
Q: 2 + 5 = 7
P ∧ Q : 2 is an irrational number and
2 + 5 = 7
Disjunction:
p: Triangles have three sides
q : Bangalore is in Karnataka
p ∨ q : Triangles have three sides or
Bangalore is in Karnataka
Exclusive Disjunction:
p: Alex eats an apple
q: Alex eats an orange
p ⊻ q : Alex eats an apple or Alex
eats an orange, but not both

LOGICAL CONNECTIVES CONTD., EXAMPLES
Conditional
p: I weigh more than 120 pounds
q : I shall enrol in an exercise class
p → q : if I weigh more than 120 pounds, then I shall enrol in an
exercise class
Bidirectional: (p → q) ∧ (q → p)
p: I shall enrol in an exercise class
q : I weigh more than 120 pounds
p ↔ q : I shall enrol in an exercise class if and only if I weigh more
than 120 pounds

PROBLEM 1:
Let s, t, and u denote the following primitive statements.
s: Zac goes out for a walk
t: The moon is out
u: It is snowing
Express the following compound propositions in words
(t ∧ ¬u) → s
¬ ∧ ∨ ⊻ → ↔
If the moon is out and it is not snowing, then Zac goes out for a walk
t → (¬u → s) If the moon is out then it is not snowing then Zac goes out for a walk
¬(s ↔ (u ∨ t))
It is not the case that Zac goes out for a walk if and only if it is snowing or
the moon is out
u ∧ s It is snowing and Zac goes out for a walk
(u ∧ ¬t) → ¬s If it is snowing and the moo is not out, then Zac does not go out for a walk
s ↔ t Zac goes out for a walk if and only the moon is out
p → q ---- if p, then q
p ↔ q ---- p if and only if q

PROBLEM 2:
Construct truth table for the following compound propositions
¬ ∧ ∨ ⊻ → ↔
p ∧ ¬q
p → ¬q
p q ¬q p → ¬q
0 0 1 1
0 1 0 1
1 0 1 1
1 1 0 0
p q ¬q p ∧ ¬q
0 0 1 0
0 1 0 0
1 0 1 1
1 1 0 0
q ∧ (¬r → p)
p q r ¬r ¬r → p q ∧ (¬r → p)
0 0 0 1 0 0
0 0 1 0 1 0
0 1 0 1 0 0
0 1 1 0 1 1
1 0 0 1 1 0
1 0 1 0 1 0
1 1 0 1 1 1
1 1 1 0 1 1
(p ∨ q) ∧ r
p ∨ (q ∨ r)
(p ∧ q) → ¬r
Try this!

Let p and q be primitive statements for which the implication p → q is false.
Determine the truth values of the following
i. p ∧ q
ii. ¬ p ∨ q
iii.q → p
iv.¬ q → ¬ p
PROBLEM 3: ¬ ∧ ∨ ⊻ → ↔
Given: p → q is false
p → q is false only when p is true and
q is false
 Truth value of p is 1 (or true)
Truth value of q is 0 (or false)
Solution:

p = 1 q = 0
i. p ∧ q = 1 ∧ 0 = 0
 p ∧ q is false
ii. ¬ p ∨ p = ¬ 1 ∨ 0 = 0 ∨ 0 = 0
 ¬ p ∨ p if false
iii. q → p = 0 → 1 = 1
 q → p is true
iv. ¬ q → ¬ p = ¬ 0 → ¬ 1 = 1 → 0 = 0
 ¬ q → ¬ p is false

Let p, q, and r be propositions having truth values 0, 0, and 1 respectively. Find
the truth values of the following
i. (p ∨ q) ∨ r
ii. (p ∧ q) ∧ r
iii. (p ∧ q) → r
iv. p → (q ∧ r)
v. p ∧ (r → q)
vi. p → (q → ¬ r)
¬ ∧ ∨ ⊻ → ↔PROBLEM 4:
Answers:
i. True
ii. False
iii. True
iv. True
v. False
vi. True

• Find the possible truth values of p, q, and r in the following cases.
PROBLEM 5:
p → (q ∨ r) is false
¬ ∧ ∨ ⊻ → ↔
Solution
⁃ p → (q ∨ r) = 0
⁃ Only if p =1 and (q ∨ r) = 0
⁃ (q ∨ r) is false only if q = 0, r = 0
p ∧ (q → r) is true
Solution
⁃ This is possible only when
⁃ p = 1 and (q → r) = 1
⁃ (q → r) = 1 in 3 cases (recall the truth table
of implication)
p q r
1 0 0
p q r
1 0 0
1 0 1
1 1 1

• If a proposition q has truth value 1, determine all truth value assignments for the
primitive propositions p, q, r, and s for which the truth value of the following
proposition is 1
PROBLEM 6: ¬ ∧ ∨ ⊻ → ↔
[ q → { (¬p ∨ r ) ∧ ¬s } ] ∧ {¬s → (¬r ∧ q ) }
[ q → { (¬p ∨ r ) ∧ ¬s } ] ∧ {¬s → (¬r ∧ q ) }
1 1
q → { (¬p ∨ r ) ∧ ¬s } = 1
---- eq(1)
¬s → (¬r ∧ q ) = 1 ---- eq(2)
Given, q = 1. Substitute this.

q → { (¬p ∨ r ) ∧ ¬s } = 1
Substituting q = 1 in eq(1)
⇒ 1 → { (¬p ∨ r ) ∧ ¬s } = 1
This is possible only if
{ (¬p ∨ r ) ∧ ¬s } = 1
⇒ ¬s = 1 and
(¬p ∨ r ) = 1 --- eq (3)
If ¬s = 1, s = 0
¬s → (¬r ∧ q ) = 1
We got, ¬s = 1
Substituting ¬s = 1 and
q = 1 in eq(2)
1 → (¬r ∧ 1 ) = 1
This is possible only if
(¬r ∧ 1 ) = 1
⇒ ¬r = 1
If ¬r = 1 , r = 0
Substituting r in eq (3),
(¬p ∨ 0 ) = 1
¬p has to be 1 to get
the above equation
right.
Therefore, p = 0
Truth values
p q r s
0 1 0 0
1 2 3

Give the conjunction and disjunction of p and q, also indicate the truth value.
PROBLEM 7: ¬ ∧ ∨ ⊻ → ↔
i)
p: 4 is a perfect square
q: 27 is a prime number
ii)
p: 5 is divisible by 2
q: 7 is a multiple of 5
Conjunction: p ∧ q ⇒ 4 is a perfect
square and 27 is a prime number
p = 1, q = 0
 p ∧ q is False
Disjunction: p ∨ q ⇒ 4 is a perfect
square or 27 is a prime number
p = 1, q = 0
 p ∨ q is true
Conjunction: p ∧ q ⇒ 5 is divisible by 2
and 7 is a multiple of 5
p = 0, q = 0
 p ∧ q is False
Disjunction: p ∨ q ⇒ 5 is divisible by 2
and 7 is a multiple of 5
p = 0, q = 0
 p ∨ q is False

From the information given in each of the following
determine the truth value required.
Solve it!!!
a. p ∧ q is false and p is true. Find the truth value of q
b. p → q is true, q is false. Find the truth value of p
c. P ↔ q is true, p is false. Find the truth value of q.
¬ ∧ ∨ ⊻ → ↔PROBLEM 8:

TAUTOLOGY, CONTRADICTION AND
CONTINGENCY
• Tautology – Always true regardless of the truth values of its components
Eg: p ∨ ¬ p
• Contradiction – Always false
Eg: p ∧ ¬ p
• Contingency – Neither tautology not contingency
Eg: ¬ p ∧ q
p ¬ p p ∨ ¬ p
0 1 1
1 0 1
p ¬ p p ∧ ¬ p
0 1 0
1 0 0
p q ¬ p ¬ p ∧ q
0 0 1 0
0 1 1 1
1 0 0 0
1 1 0 0

i. (p ⊻ q) ∨ (p ↔ q) is a tautology
ii. (p ⊻ q) ∧ (p ↔ q) is a contradiction
iii. (p ⊻ q) ∧ (p → q) is a contingency
iv. P → (p ∨ q) is a tautology
PROBLEM 9: SHOW THAT
Another way of asking the same question
Show that the truth values of the following
compound propositions are independent of the
truth values of their components.

Fundamentals of logic 1

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