Discrete Mathematics Test 1

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Discrete Mathematics Test 1

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Weekly Quiz Competition

Question 1
1 / -0

What is logically equivalent to the following statement?” I pass only if you pass”
(Note that fail is equivalent to not pass.)

A
you pass only if I pass

B
If you fail then I fail

C
if you pass then I pass

D
you fail if I fail

Solution

Let p and q be two prepositional statement
p only if q ≡ p → q
p if q ≡ q → p
Contrapositive of p → q is ¬ q → ¬ p
p → q ≡ ¬ q → ¬ p
Let p: I pass.
q: you pass.
If you not pass then I not pass ≡ ¬ q → ¬ p
If you fail then I fail ≡ ¬ q → ¬ p
Important Points:
If: p → q
then:
Converse: q → p
Inverse: ~p → ~q
Contrapositive: ~q → ~p
Confusion points:
p → q ≡ ~q → ~p

Question 2

1 / -0

Match the following rules of inference with its corresponding name.

Rules of inference	Name
A. $\;\frac{{\begin{array}{*{20}{c}}{\neg q}\\{p \to q}\end{array}}}{{\therefore \neg \;p}}$	I. Modus ponens
	II. Modus tollens
B. $\frac{{\begin{array}{*{20}{c}}{p \to q}\\{q \to r}\end{array}}}{{\therefore p \to r}}$	III. Disjunctive syllogism
	IV. Hypothetical syllogism

A ------ I B ----- III

A ----- II B ----- IV

A ----- II B ----- III

A ----- I B ----- IV

Solution

Rules of inference	Name
$\;\frac{{\begin{array}{*{20}{c}}{\neg q}\\{p \to q}\end{array}}}{{\therefore \neg \;p}}$	Modus tollens
$\frac{{\begin{array}{*{20}{c}}{p \to q}\\{q \to r}\end{array}}}{{\therefore p \to r}}$	Hypothetical syllogism

Question 3
1 / -0

Which of the following is/are equivalent to [(p V q) ∧ ¬ p → ¬ q]?

A
p V ¬ q

B
¬ q → p

C
q → p

D
¬ p V q

Solution

Since p → q = ¬ p ∨ q ≡ p̅ + q
(p V q) ∧ ¬ p → ¬ q
$≡ \overline{\left( p+q \right).\bar{p}}+\bar{q}$
$≡ \overline{p\bar{p}+\bar{p}q}+\bar{q}$
≡ p + q̅ + q̅
≡ p + q̅
≡ p ∨ ¬ q
≡ ¬ q ∨ p
≡ q → p
∴ option 1 and 3 is correct
Alternate Method:
Truth Table:
p
q
¬ p
¬ q
p V q
(p V q) ∧ ¬ p
(p V q) ∧ ¬ p → ¬ q
p V ¬ q
q → p
T
T
F
F
T
F
T
T
T
T
F
F
T
T
F
T
T
T
F
T
T
F
T
T
F
F
F
F
F
T
T
F
F
T
T
T

From truth table:
(p V q) ∧ ¬ p → ¬ q ≡ p V ¬ q
Question 4
1 / -0

A logical binary relation $, is defined as follows:
X
Y
X $ Y
FALSE
FALSE
TRUE
FALSE
TRUE
TRUE
TRUE
FALSE
TRUE
TRUE
TRUE
FALSE

Let ~ be the unary negation NOT operator, with higher precedence that $. Which of the following is equivalent to X V Y?

A
X $ ~Y

B
~X $ Y

C
~X $ ~Y

D
~ (X $ Y)

Solution

Formula:
X → Y ≡ ~X ∨ Y ≡ X’ + Y
AND ≡ ∧ ≡ .
OR ≡ ∨ ≡ +
NOT ≡ ~ ≡ ‘
Calculation:
X $ Y ≡ X’.Y’ + X’.Y + X.Y’
∴ X $ Y ≡ X’ + X.Y’ ≡ X’ + Y’
Option 3:
X $ Y ≡ X’ $ Y’
~X $ ~Y ≡ (X’)’ + (Y’)’ ≡ X + Y ≡ X ∨ Y
Question 5
1 / -0

Consider the statement
Not all good-looking people are nice.
P(x): good-looking people
N(x): Nice
Which one of the following logical formula represents the above statement?

A
∀x (P(x) ∧ ¬ N(x))

B
∃x(P(x) ∨ ¬ N(x))

C
∃x(P(x) ∧ ¬ N(x))

D
∀x (P(x) → N(x))

Solution

NOT (all good-looking people are nice.)
≡ ¬ (∀x (P(x) → N(x)))
≡ ¬ ∀x (P(x) → N(x))
P(x) → N(x) ≡ ¬ P(x) ∨ N(x)
≡ ∃x (¬ (¬ P(x) ∨ N(x)))
≡ ∃x (P(x) ∧ ¬ N(x)))
Question 6
1 / -0

If T(x) denotes is a trigonometric function, P(x) denotes x is a periodic function and C(x) denotes x is a continuous function then the statement “It is not the case that some trigonometric functions are not periodic” can be logically represented as

A
¬ ∃x[T(x) ∧ ¬ P(x)]

B
¬ ∃x[T(x) ∨ ¬ P(x)]

C
¬ ∃x[¬T(x) ∧ ¬ P(x)]

D
¬ ∃x[T(x) ∧ P(x)]

Solution

Statement: “It is not the case that some trigonometric functions are not periodic”
"some trigonometric functions are not periodic" means
There exist some trigonometric functions which are also not periodic.
∃x [T(x) ∧ ¬ P(x)]
The negation of it.
"It is not the case that some trigonometric functions are not periodic"
¬ ∃x[T(x) ∧ ¬ P(x)]
Question 7
1 / -0

Which of the following is true?
I. ¬ (P ↔ Q) ≡ P ↔ ¬ Q
II. (P ↔ Q) ≡ (P ∧ Q) ∨ (¬ P ∧ ¬ Q)
III. (P ↔ Q) ≡ ¬ P ↔ ¬ Q

A
Only I

B
Only I and II

C
Only II and III

D
I, II and III

Solution

TRUTH TABLE:
P
Q
(P ↔ Q)
¬ (P ↔ Q)
P ↔ ¬ Q
¬ P ↔ ¬ Q
(P∧Q)∨(¬P ∧ ¬Q)
F
F
T
F
F
T
T
F
T
F
T
T
F
F
T
F
F
T
T
F
F
T
T
T
F
F
T
T

I, II and III are all true.
Question 8
1 / -0

Consider the following statements:
P: Good mobile phones are not cheap
Q: Cheap mobile phones are not good
Which one of the following is/are CORRECT?

A
P implies Q

B
P implies not of Q

C
P is equivalent to Q

D
Q implies P

Solution

Concept
(A → ¬B) ⟺ (¬A ∨ ¬B)
(B → ¬A) ⟺ (¬B ∨ ¬A) ⟺ (¬A ∨ ¬B) (Disjunction is commutative)
Hence, (A → ¬B) ⟺ (B → ¬A)
Explanation:
Suppose,
A: Good mobile phones.
B: Cheap mobile phones.
Therefore, P and Q can be written as,
P: A → ¬B
Q: B → ¬A
Truth Table for P and Q as outputs,
A
B
¬A
¬B
P
Q
0
0
1
1
1
1
0
1
1
0
1
1
1
0
0
1
1
1
1
1
0
0
0
0

P and Q are equivalent. So, option_4 is correct.
Question 9
1 / -0

Consider the statements below:
“There is a country that borders both India and Nepal.”
Which of the following represents the above sentence correctly?

A
∃c Border(Country(c), India ∧ Nepal)

B
∃c Country(c) ∧ Border(c, India) ∧ Border(c, Nepal)

C
[∃c Country(c)] ⇒ [Border(c, India) ∧ Border(c, Nepal)]

D
∃c Country(c) ⇒ [ Border(c, India) ∧ Border(c, Nepal)]

Solution

c- It represents country
∃c – it means there exists a country
Border(c, India) – it means border between India and c
Border(c, Nepal) – it means border between Nepal and c
Variables are two: India and Nepal
Given statement: “There is a country that borders both India and Nepal.”
AND is true only when Border(c, India) and Border(c, Nepal) is 1.
Country must have border with both India and Nepal. So, and operation supports it.
AND is denoted by ∧ in proposition logic.
Correct option is:
∃c Country(c) ∧ Border(c, India) ∧ Border(c, Nepal)
It means there exists a country c and it borders both India and Nepal.
Question 10
1 / -0

If X ≡ A → ¬ B, Y ≡ ¬ B → ¬ C, Z ≡ A→ ¬ C then (X ∧ Y) → Z is
I. Valid
II. Satisfiable

A
Only I

B
Only II

C
Both I and II

D
Neither I nor II

Solution

Concepts:
Valid: If for all the combination of variable, expression returns true then it is valid.
Satisfiable: If there exists at least one combination of variable which return true then it is satisfiable
Derivation:
Since p → q ≡ ¬ p ∨ q ≡ p’ + q
X ≡ A → ¬ B ≡ A’ + B’
Y ≡ ¬ B → ¬ C ≡ B + C’
Z ≡ A → ¬ C ≡ A’ + C’
(X ∧ Y) → Z ≡ (X.Y)’ + Z
≡ ((A’ + B’) (B + C’))’ + A’ + C’
≡ (A’B + A’C’ + B’C’)’ + A’ + C’
By Demorgan’s Law
≡ (A + B’).(A + C).(B + C) + A’ + C’
≡ (A + B’C).(B + C) + A’ + C’
≡ AB + AC + B’C + A’ + C'
≡ AB + A’ + AC + C’ + B’C
≡ (A’ + B) + (A + C’) + B’C
≡ A’ + B + A + C’ + B’
≡ T
Therefore, it is valid and satisfiable
Tips and Tricks:
By hypothetical Syllogism
X: A → ¬ B
Y: ¬ B → ¬ C
___________
Z: A → ¬ C
∴ (X ∧ Y) → Z ≡ T

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Re-Attempt Weekly Quiz Competition

Rules of inference	Name
A. \(\;\frac{{\begin{array}{*{20}{c}}{\neg q}\\{p \to q}\end{array}}}{{\therefore \neg \;p}}\)	I. Modus ponens
	II. Modus tollens
B. \(\frac{{\begin{array}{*{20}{c}}{p \to q}\\{q \to r}\end{array}}}{{\therefore p \to r}}\)	III. Disjunctive syllogism
	IV. Hypothetical syllogism

Rules of inference	Name
\(\;\frac{{\begin{array}{*{20}{c}}{\neg q}\\{p \to q}\end{array}}}{{\therefore \neg \;p}}\)	Modus tollens
\(\frac{{\begin{array}{*{20}{c}}{p \to q}\\{q \to r}\end{array}}}{{\therefore p \to r}}\)	Hypothetical syllogism

p	q	¬ p	¬ q	p V q	(p V q) ∧ ¬ p	(p V q) ∧ ¬ p → ¬ q	p V ¬ q	q → p
T	T	F	F	T	F	T	T	T
T	F	F	T	T	F	T	T	T
F	T	T	F	T	T	F	F	F
F	F	T	T	F	F	T	T	T

X	Y	X $ Y
FALSE	FALSE	TRUE
FALSE	TRUE	TRUE
TRUE	FALSE	TRUE
TRUE	TRUE	FALSE

P	Q	(P ↔ Q)	¬ (P ↔ Q)	P ↔ ¬ Q	¬ P ↔ ¬ Q	(P∧Q)∨(¬P ∧ ¬Q)
F	F	T	F	F	T	T
F	T	F	T	T	F	F
T	F	F	T	T	F	F
T	T	T	F	F	T	T

A	B	¬A	¬B	P	Q
0	0	1	1	1	1
0	1	1	0	1	1
1	0	0	1	1	1
1	1	0	0	0	0

Discrete Mathematics Test 1

Result

Discrete Mathematics Test 1

Score

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Rank

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SHARING IS CARING

Question 1

you pass only if I pass

If you fail then I fail

if you pass then I pass

you fail if I fail

Solution

Question 2

A ------ I B ----- III

A ----- II B ----- IV

A ----- II B ----- III

A ----- I B ----- IV

Solution

Question 3

p V ¬ q

¬ q → p

q → p

¬ p V q

Solution

Question 4

X $ ~Y

~X $ Y

~X $ ~Y

~ (X $ Y)

Solution

Question 5

∀x (P(x) ∧ ¬ N(x))

∃x(P(x) ∨ ¬ N(x))

∃x(P(x) ∧ ¬ N(x))

∀x (P(x) → N(x))

Solution

Question 6

¬ ∃x[T(x) ∧ ¬ P(x)]

¬ ∃x[T(x) ∨ ¬ P(x)]

¬ ∃x[¬T(x) ∧ ¬ P(x)]

¬ ∃x[T(x) ∧ P(x)]

Solution

Question 7

Only I

Only I and II

Only II and III

I, II and III

Solution

Question 8

P implies Q

P implies not of Q

P is equivalent to Q

Q implies P

Solution

Question 9

∃c Border(Country(c), India ∧ Nepal)

∃c Country(c) ∧ Border(c, India) ∧ Border(c, Nepal)

[∃c Country(c)] ⇒ [Border(c, India) ∧ Border(c, Nepal)]

∃c Country(c) ⇒ [ Border(c, India) ∧ Border(c, Nepal)]

Solution

Question 10

Only I

Only II

Both I and II

Neither I nor II

Solution

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