## Propositional-Logic

 Question 1
Let p and q be two propositions. Consider the following two formulae in propositional logic.

Which one of the following choices is correct?
 A Both S1and S2 are tautologies. B Neither S1and S2 are tautology. C S1is not a tautology but S2is a tautology. D S1is a tautology but S2is not a tautology.
Question 1 Explanation:

A tautology is a formula which is "always true" . That is, it is true for every assignment of truth values to its simple components.

Method 1:
S1: (~p ^ (p Vq)) →q
The implication is false only for T->F condition.
Let's consider q as false, then
(~p ^ (p Vq)) will be (~p ^ (p VF)) = (~p ^ (p)) =F.
It is always F->F which is true for implication. So there are no cases that return false, thus its always True i.e. its Tautology.

S2:

q->(~p (p Vq))

The false case for implication occurs at T->F case.
Let q=T then (~p (p Vq))  = (~p (p VT))= ~p. (It can be false for p=T).
So there is a case which yields T->F = F. Thus its not Valid or Not a Tautology.

Method 2:

 Question 2

(a) Let * be a Boolean operation defined as
If C = A * B then evaluate and fill in the blanks:
(i) A * A = _______
(ii) C * A = _______
(b) Solve the following boolean equations for the values of A, B and C:

 A Theory Explanation.
 Question 3

Obtain the principal (canonical) conjunctive normal form of the propositional formula

(p ∧ q) V (¬q ∧ r)

Where ‘∧’ is logical and, ‘v’ is inclusive or and ¬ is negation.

 A Theory Explanation.
 Question 4

Which one of the following predicate formulae is NOT logically valid?
Note that W is a predicate formula without any free occurrence of x.

 A ∃x(p(x) → W) ≡ ∀x p(x) → W B ∀x(p(x) → W) ≡ ∀x p(x) → W C ∃x(p(x) ∧ W) ≡ ∃x p(x) ∧ W D ∀x(p(x) ∨ W) ≡ ∀x p(x) ∨ W
Question 4 Explanation:
Basic Rules:
~p→q ≡ ~p∨q
Demorgan laws:
~(∀x(a(x)) ≡ ∃x~a(x)
~(∃x(a(x)) ≡ ∀x~a(x)
(A) ∃x(p(x)→w) ≡ ∀x p(x)→w
LHS: ∃x(p(x)→w) ≡ ∃x(~p(x)∨w)
≡ ∃x(~p(x))∨w
Demorgan’s law:
~(∀x(a(x)) = ∃x ~ a(x)
≡ ~(∀x P(x)) ∨ w
≡ (∀x) P(x) → w ≡ RHS
It’s valid.
(B) ∀x(P(x) → w) ≡ ∀x(~P(x) ∨ w)
≡ ∀x(~P(x)) ∨ w
≡ ~(∃x P(x)) ∨ w
≡ ∃x P(x) → w
This is not equal to RHS.
(C) ∃x(P(x) ∧ w) ≡ ∃x P(x) ∧ w
‘w’ is not a term which contains x.
So the quantifier does not have any impact on ‘w’.
Thus it can be written as
∃x(P(x)) ∧ w) ≡ ∃x P(x) ∧ w
(D) ∀(x)(P(x) ∨ w) ≡ ∀x P(x) ∨ w
‘w’ is not a term which contains ‘x’.
So the quantifier does not have an impact on ‘w’.
Thus ∀(x)(P(x) ∨ w) ≡ ∀x P(x) ∨ w
 Question 5

What is the correct translation of the following statement into mathematical logic?

“Some real numbers are rational”

 A ∃x (real(x) ∨ rational(x)) B ∀x (real(x) → rational(x)) C ∃x (real(x) ∧ rational(x)) D ∃x (rational(x) → real(x))
Question 5 Explanation:

∃x (real(x) ∧ rational(x))
(A) ∃x(real(x) ∨ rational(x))
means There exists some number, which are either real or rational.
(B) ∀x (real(x)→rational(x))
If a number is real then it is rational.
(D) ∃x (rational(x)→real(x))
There exists a number such that if it is rational then it is real.
 Question 6

Consider the following logical inferences.

I1: If it rains then the cricket match will not be played.
The cricket match was played.
Inference: There was no rain.
I2: If it rains then the cricket match will not be played.
It did not rain.
Inference: The cricket match was played.

Which of the following is TRUE?

 A Both I1 and I2 are correct inferences B I1 is correct but I2 is not a correct inference C I1 is not correct but I2 is a correct inference D Both I1 and I2 are not correct inferences
Question 6 Explanation:
I1: If it rains then the cricket match will not be played.
The cricket match was played.
Let p = it rains
q = playing cricket/ match played
If (it rains) then (the match will not be played)
p ⇒ (∼q)
Inference: There was no rain. (i.e., p = F)
So for any F ⇒ (∼q) is true.
So this inference is valid.
I2: If it rains then the cricket match will not be played.
It did not rain.
p ⇒ (∼q)
Inference: The cricket match was played.
q = T
p ⇒ (∼q)
p ⇒ (∼T)
p ⇒ F
This is false for p = T, so this is not true.
 Question 7

Which of the following is false? Read ∧ as AND, ∨ as OR, ~ as NOT, → as one way implication and ↔ as two way implication.

 A ((x → y) ∧ x) → y B ((x → y) ∧ (x ∧ y)) → x C (x → (x ∨ ψ)) D ((x ∨ y) ↔ (x → y)
Question 7 Explanation:
When x = F and y = F
then option (D) will be False.
 Question 8

Which of the following propositions is a tautology?

 A (p ∨ q) → p B p ∨ (q → p) C p ∨ (p → q) D p → (p → q)
Question 8 Explanation:
 Question 9

What is the converse of the following assertion?

I stay only if you go.
 A I stay if you go B If I stay then you go C If you do not go then I do not stay D If I do not stay then you go
Question 9 Explanation:
"I stay only you go" = "If I stay then you go"
⇒ i.e., A→B
Where A = If I stay; B = you go
Converse for (A→B) is (B→A)
⇒ If you go then I stay.
 Question 10

(a) Show that the formula [(~p ∨ q) ⇒ (q ⇒ p)] is not a tautology.
(b) Let A be a tautology and B be any other formula. Prove that (A ∨ B) is a tautology.

 A Theory Explanation.
 Question 11

Let a, b, c, d be propositions. Assume that the equivalence a ↔ (b ∨ -b) and b ↔ c hold. Then the truth-value of the formula (a ∧ b) → (a ∧ c) ∨ d is always

 A True B False C Same as the truth-value of b D Same as the truth-value of d
Question 11 Explanation:
a ↔ (b ∨-b) and b ↔ c
Given ⇒ (a∧b) → (a∧c) ∨d
⇒ (a∧b) → (a∧c) ∨d (b⇔c)
⇒ T∨d
⇒ T
 Question 12

Consider two well-formed formulas in prepositional logic

F1: P ⇒ ¬P          F2: (P⇒¬P)∨(¬P⇒P)

Which of the following statements is correct?

 A F1 is satisfiable, F2 is valid B F1 unsatisfiable, F2 is satisfiable C F1 is unsatisfiable, F2 is valid D F1 and F2 are both satisfiable
Question 12 Explanation:

F1 is satisfiable; F2 is valid.
 Question 13

“If X then Y unless Z” is represented by which of the following formulas in prepositional logic? (“¬“, is negation, “∧” is conjunction, and “→” is implication)

 A (X∧¬Z)→Y B (X∧Y)→¬Z C X→(Y∧¬Z) D (X→Y)∧¬Z
Question 13 Explanation:
"If X then Y unless Z" ⇒ ¬Z → (X→Y)
⇒ Z ∨ ¬X ∨ Y
⇒ ¬X ∨ Z ∨ Y
Option A:
(X ∧ ¬Z) → Y = ¬(X ∧ ¬Z ) ∨ Y = ¬X ∨ Z ∨ Y Hence, option (A) is correct.
 Question 14

The following resolution rule is used in logic programming.

Derive clause (P ∨ Q) from clauses (P ∨ R), (Q ∨ ¬R)

Which of the following statements related to this rule is FALSE?

 A ((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q) is logically valid B (P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R)) is logically valid C (P ∨ Q) is satisfiable if and only if (P∨R) ∧ (Q∨¬R) is satisfiable D (P ∨ Q) ⇒ FALSE if and only if both P and Q are unsatisfiable
Question 14 Explanation:
(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R))

It is may be True (or) False depending on values. So this is not valid.
 Question 15

Consider the following formula a and its two interpretations I1 and I2

α: (∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Qyy]] ⇒ (∀x)[¬Px]
I1: Domain: the set of natural numbers
Px ≡ 'x is a prime number'
Qxy ≡ 'y divides x'
I2: same as I1 except that Px = 'x is a composite number'.

Which of the following statements is true?

 A I1 satisfies α, I2 does not B I2 satisfies α, I1 does not C Neither I2 nor I1 satisfies α D Both I1 and I2 satisfy α
Question 15 Explanation:
Given that:
(∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Qyy]] ⇒(∀x)[¬Px]
Qyy is always true, because y divide y, then ¬Qyy is false.
∀x[(P(x) ⇔ ∀y [Qxy ⇔ False]]
∀y [Qxy ⇔ False] can be written as ∀y[¬axy]
⇒(∀x)[P(x) ⇔ ∀y[¬Qxy]]
Here, ¬Qxy says that y doesnot divides x, which is not always be true.
For example, if x=y then it is false then ∀y[¬Qxy] is not true for all values of y.
⇒(∀x)[P(x) ⇔ False]
⇒(∀x)[¬P(x) = RHS]
LHS = RHS
⇒ Irrespective of x, whether x is prime of composite number I1 and I2 satisfies α.
 Question 16

Which of the following is a valid first order formula? (Here α and β are first order formulae with x as their only free variable)

 A ((∀x)[α] ⇒ (∀x)[β]) ⇒ (∀x)[α⇒β] B (∀x)[α] ⇒ (∃x)[α ∧ β] C ((∀x)[α ∨ β] ⇒ (∃x)[α] ⇒ (∀x)[α] D (∀x)[α ⇒ β] ⇒ ((∀x)[α] ⇒ (∀x)[β])
Question 16 Explanation:
Option D is valid.
Here, α, β are holding values of x. Then and RHS saying that α holding the value of x and β is holding value of x.
Then LHS ⇒ RHS.
 Question 17

Show that proposition C is a logical consequence of the formula

A ∧ (A →(B ∨ C)) ∧ (B → ~A)

Using truth tables.

 A Theory Explanation.
 Question 18

(a) Uses Modus ponens (A, A →|= B) or resolution to show that the following set is inconsistent:

(1) Q(x) → P(x)V ~ R(a)
(2) R(a) ~ Q(a)
(3) Q(a)
(4) ~ P(y)

Where x and y are universally quantified variables, a is a constant and P, Q, R are monadic predicates.

(b) Let S be the set of all integers and let n > 1 be a fixed integer. Define for a, b ∈ S, a R biff a-b is a multiple of n. Show that R is an equivalence relation and finds its equivalence classes for n = 5.

 A Theory Explanation.
 Question 19
Choose the correct choice(s) regarding the following propositional logic assertion S:
 A S is a contradiction. B The anecdote of S is logically equivalent to the consequent of S. C S is a tautology. D S is neither a tautology nor a contradiction.
Question 19 Explanation:
 Question 20

Consider the following first order logic formula in which R is a binary relation symbol.

∀x∀y (R(x, y)  => R(y, x))
The formula is

 A satisfiable and valid B satisfiable and so is its negation C unsatisfiable but its negation is valid D satisfiable but its negation is unsatisfiable
Question 20 Explanation:
The given relation is known to be symmetry. We have both symmetric relations possible as well as antisymmetric but neither always holds for all sets. So they both are valid but are satisfiable.
 Question 21

Which one of these first-order logic formula is valid?

 A ∀x(P(x) ⇒ Q(x)) ⇒ (∀xP(x) ⇒ ∀xQ(x)) B ∃x(P(x) ∨ Q(x)) ⇒ (∃xP(x) ⇒ ∃xQ(x)) C ∃x(P(x) ∧ Q(x)) (∃xP(x) ∧ ∃xQ(x)) D ∀x∃y P(x, y) ⇒ ∃y∀x P(x, y)
Question 21 Explanation:
LHS = for every x, if P holds then Q holds
RHS = if P(x) holds for all x, then Q(x) holds for all x
LHS ⇒ RHS (✔)
RHS ⇒ LHS (️×)
 Question 22

Which of the following first order formula is logically valid? Here α(x) is a first order formula with x as a free variable, and β is a first order formula with no free variable.

 A [β→(∃x,α(x))]→[∀x,β→α(x)] B [∃x,β→α(x)]→[β→(∀x,α(x))] C [(∃x,α(x))→β]→[∀x,α(x)→β] D [(∀x,α(x))→β]→[∀x,α(x)→β]
Question 22 Explanation:
[(∃x,α(x))→β]→[∀x,α(x)→β]
L.H.S. : If there is an x such that α(x) is true, then β is true.
R.H.S. : For all x, if α(x) true, then β is true.
Here, the given LHS and RHS are to be same as β is a formula which can be independent of x (if β is true for one x, it is true for every x, and vice-versa).
Here, LHS = RHS
So, Option C is valid.
 Question 23

Which of the following is the negation of [∀x, α →(∃y, β →(∀ u, ∃v, y))]

 A [∃x, α → (∀y, β → (∃u, ∀v, y))] B [∃x, α → (∀y, β → (∃u, ∀ v, ¬y))] C [∀x, ¬α → (∃y, ¬β → (∀u, ∃v, ¬y))] D [∃x, α ʌ (∀y, β ʌ (∃u, ∀v, ¬y))]
Question 23 Explanation:
~[∀x, α → (∃y, β → (∀u, ∃v, y))]
⇔ [∃x, [α × ~(∃y, β → (∀u, ∃v, y))]
⇔ [∃x, [α × ∀y, ~(β → (∀u, ∃v, y)]
⇔ [∃x, [α × ∀y, ~(β × ~(∀u, ∃v, y)]
⇔ [∃x, [α × ∀y(β × (∃u, ∀v, y)]
 Question 24

Which of the following well-formed formulas are equivalent?

 A P → Q B ¬Q → ¬P C ¬P ∨ Q D ¬Q → P E A, B and C.
Question 24 Explanation:
P → Q ⇔ ¬P ∨ Q
¬Q → ¬P ⇔ Q ∨ ¬P
¬P ∨ Q ⇔ ¬P ∨ Q
¬Q → P ⇔ Q ∨ P
A, B and C are equivalent.
 Question 25

Choose the correct alternatives (More than one may be correct). Indicate which of the following well-formed formulae are valid:

 A (P⇒Q) ∧ (Q⇒R) ⇒ (P⇒R) B (P⇒Q) ⇒ (¬P⇒¬Q) C (P∧(¬P∨¬Q)) ⇒ Q D (P⇒R) ∨ (Q⇒R) ⇒ ((P∨Q)⇒R)
Question 25 Explanation:
To prove any well formed formula valid or tautology try to use this analogy.
Since implication A → B is False only when A = T and B = F. So to prove any implication is valid or not try to get
TRUE → FALSE, if we succeed then it is not valid, if we not then well formed formula is valid.
So, for option (A),
Substitute P=T and R=F
RHS:
P→R becomes False.
LHS:
(P→Q) ∧ (P→R)
To get true here we need T∧T. So substitute Q=T which makes P→Q TRUE and P→R FALSE.
So, T∧F = F which makes LHS = False.
Hence, we are unable to get T→F which proves well formed formula given in option (A) is valid.
Similarly, try for (B), (C), (D). We get T → F in these options which says these well formed formula is invalid.
 Question 26

The following propositional statement is (P → (Q v R)) → ((P v Q) → R)

 A satisfiable but not valid B valid C a contradiction D None of the above
Question 26 Explanation:
(P→(Q∨R)) → ((P∨Q)→R)
If P=T; Q=T; R=T
(P→(T∨T)) → ((T∨T)→R)
(P→T) → (T→R)
(T→T) → (T→T)
T→T
T(Satisfiable)
 Question 27

Identify the correct translation into logical notation of the following assertion.

"Some boys in the class are taller than all the girls"

Note: taller(x,y) is true if x is taller than y.

 A (∃x) (boy(x) → (∀y) (girl(y) ∧ taller(x,y))) B (∃x) (boy(x) ∧ (∀y) (girl(y) ∧ taller(x,y))) C (∃x) (boy(x) → (∀y) (girl(y) → taller(x,y))) D (∃x) (boy(x) ∧ (∀y) (girl(y) → taller(x,y)))
Question 27 Explanation:
Don't confuse with '∧' and '→'
'∧' → predicts statements are always true, no matter the value of x.
'→' → predicts there is no need of left predicate to be true always, but whenever it becomes true, then right predicate must be true.
Option D:
There exists a some boys who are taller than of all girls y.
 Question 28

Which one of the following well formed formulae is a tautology?

 A ∀x ∃y R(x,y) ↔ ∃y ∀x R(x,y) B (∀x [∃y R(x,y) → S(x,y)]) → ∀x∃y S(x,y) C [∀x ∃y (P(x,y) → R(x,y)] ↔ [∀x ∃y (¬ P(x,y)∨R(x,y)] D ∀x ∀y P(x,y) → ∀x ∀y P(y,x)
Question 28 Explanation:
Since P→R = ¬P∨R
[∀x ∃y (P(x,y) → R(x,y)] ↔ [∀x ∃y (¬ P(x,y)∨R(x,y)] is a tautology.
 Question 29
Which one of the following is not equivalent to p←→q
 A A B B C C D D
 Question 30

Which one of the following well-formed formulae in predicate calculus is NOT valid?

 A (∀x p(x) ⇒ ∀x q(x)) ⇒ (∃x ¬p(x) ∨ ∀x q(x)) B (∃x p(x) ∨ ∃x q(x)) ⇒ ∃x (p(x) ∨ q(x)) C ∃x (p(x) ∧ q(x)) ⇒ (∃x p(x) ∧ ∃x q(x)) D ∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x))
Question 30 Explanation:
For the formulae to be valid there should not be implication like T → F.
But in option (D), we can generate T → F.
Hence, not valid.
 Question 31
Consider the following expressions:
(i) false
(ii) Q
(iii) true
(iv) P ∨ Q
(v) ¬Q ∨ P
The number of expressions given above that are logically implied by P ∧ (P ⇒ Q) is _________.
 A 4 B 5 C 6 D 7
Question 31 Explanation:
The expression is logically implied by P ∧ (P → Q) means
(P ∧ (P → Q))→ expression is a tautology. So we have to find
How many tautological formulas are there for the given inputs.
(P ∧ (P → Q)) → True is always tautology
(P ∧ (P → Q)) → False is not a tautology
(P ∧ (P → Q)) → Q is a tautology
(P ∧ (P → Q)) → ¬Q ∨ P is a tautology
(P ∧ (P → Q)) → P ∨ Q is a tautology
So there are 4 expressions logically implied by (P ∧ (P → Q))
 Question 32
Geetha has a conjecture about integers, which is of the form ∀x P(x) → ∃yQ(x, y) , where P is a statement about integers, and Q is a statement about pairs of integers. Which of the following (one or more) option(s) would imply Geetha’s conjecture?
 A ∃x P(x) ∧ ∀yQ(x, y) B ∀x∀yQ(x, y) C ∃y∀x P(x) ⇒ Q(x, y) D ∃x P(x) ∧ ∃yQ(x, y)
 Question 33
The statement (¬p) ⇒ (¬q) is logically equivalent to which of the statements below?
I. p ⇒ q
II. q ⇒ p
III. (¬q) ∨ p
IV. (¬p) ∨ q
 A I only B I and IV only C II only D II and III only
Question 33 Explanation:
Method 1:
Construct Truth tables:
~p ⇒ ~q

II, III are equivalent to (~p) ⇒ (~q)
Method 2:
(I) p⇒q ≡ ~p∨q
(II) q⇒p ≡ ~q∨p
(III) (~q) ∨ p ≡ ~q∨p
(IV) (~p) ∨ p ≡ ~p∨q
Also, from question:
(~p) ⇒ (~q)
≡ p∨~q
So, (II) & (III) are equivalent to the statement given in question.
 Question 34
Consider the first-order logic sentence
φ ≡ ∃s∃t∃u∀v∀w∀x∀y ψ(s,t,u,v,w,x,y)
where ψ(s,t,u,v,w,x,y) is a quantifier-free first-order logic formula using only predicate symbols, and possibly equality, but no function symbols. Suppose φ has a model with a universe containing 7 elements.
Which one of the following statements is necessarily true?
 A There exists at least one model of φ with universe of size less than or equal to 3. B There exists no model of φ with universe of size less than or equal to 3. C There exists no model of φ with universe of size greater than 7. D Every model of φ has a universe of size equal to 7.
Question 34 Explanation:
φ = ∃s∃t∃u∀v∀w∀x∀y ψ (s,t,u,v,w,x,y)
"∃" there exists quantifier decides whether a sentence belong to the model or not.
i.e., ~∃ will make it not belong to the model. (1) We have ‘7’ elements in the universe, So max. size of universe in a model = ‘7’
(2) There are three '∃' quantifiers, which makes that a model have atleast “3” elements. So, min. size of universe in model = ‘7’.
(A) is False because: (2)
(B) is true
(C) is false because of (1)
(D) is false, because these all models with size {3 to 7} not only ‘7’.
 Question 35

Consider the first order predicate formula φ:

∀x[(∀z z|x ⇒ ((z = x) ∨ (z = 1))) ⇒ ∃w (w > x) ∧ (∀z z|w ⇒ ((w = z) ∨ (z = 1)))]

Here 'a|b' denotes that 'a divides b', where a and b are integers. Consider the following sets:

S1.  {1, 2, 3, ..., 100}
S2.  Set of all positive integers
S3. Set of all integers

Which of the above sets satisfy φ?

 A S1 and S3 B S1, S2 and S3 C S2 and S3 D S1 and S2
Question 35 Explanation:
The first order logic gives the meaning that if z is a prime number then there exists another prime number in the set which is larger than it.
One of the case:
If -7 is a number which is prime (either divided by -7 or 1 only). then there exists some number like -3 which is larger than -7 also satisfy the property (either divided by -3 or 1 only).
So, S3 is correct
It's true for all integers too.
 Question 36
Let # be a binary operator defined as X # Y = X′ + Y′ where X and Y are Boolean variables. Consider the following two statements.
S1: (P # Q) # R = P # (Q # R)
S2: Q # R = R # Q
Which of the following is/are true for the Boolean variables P, Q and R?
 A Only S1 is True B Only S2 is True C Both S1 and S2 are True D Neither S1 nor S2 are True
 Question 37

P and Q are two propositions. Which of the following logical expressions are equivalent?

I. P∨∼Q
II. ∼(∼P∧Q)
III. (P∧Q)∨(P∧∼Q)∨(∼P∧∼Q)
IV. (P∧Q)∨(P∧∼Q)∨(∼P∧Q)
 A Only I and II B Only I, II and III C Only I, II and IV D All of I, II, III and IV
Question 37 Explanation:
I. P∨∼Q (✔️)
II. ∼(∼P∧Q)⇒(P∨∼Q)≡I (✔️)
III. (P×Q)∨(P×∼Q)∨(∼P×∼Q)
P∧(Q∨∼Q)∨(∼P∧∼Q)
P∨(∼P×∼Q)
(P∨∼P)×(P∨∼Q)
(P∨∼Q)≡I=II (✔️)
IV. (P×Q)∨(P∧∼Q)∨(∼P×Q)
P×(Q∨∼Q)∨(∼P∧Q)
P∨(∼P×Q)
(P∨∼P)×(P∨Q)
(P∨Q)≠I (❌)
So I≡II≡III (✔️)
 Question 38

Let fsa and pda be two predicates such that fsa(x) means x is a finite state automaton, and pda(y) means that y is a pushdown automaton. Let equivalent be another predicate such that equivalent (a, b) means a and b are equivalent. Which of the following first order logic statements represents the following:

Each finite state automaton has an equivalent pushdown automaton

 A (∀x fsa(x)) ⇒ (∃y pda(y) ∧ equivalent(x,y)) B ∼∀y(∃x fsa(x) ⇒ pda(y) ∧ equivalent(x,y)) C ∀x ∃y(fsa(x) ∧ pda(y) ∧ equivalent(x,y)) D ∀x ∃y(fsa(y)∧ pda(x) ∧ equivalent(x,y))
Question 38 Explanation:
Go through the options.
Option A:
If everything is a FSA. Then there exists an equivalent PDA for everything.
Option B:
Not for the case Y, if there exists a FSA then it can have equivalent PDA.
Option C:
Everything is a PDA and consists equivalent PDA.
Option D:
Everything is a PDA and has exist an equivalent FSA. In option A we are getting the equivalent of a and b.
 Question 39

Let P, Q and R be three atomic prepositional assertions. Let X denote (P ∨ Q) → R and Y denote (P → R) ∨ (Q → R). Which one of the following is a tautology?

 A X ≡ Y B X → Y C Y → X D ¬Y → X
Question 39 Explanation:
X: (P∨Q) → R
⇒ ∼(P∨Q) ∨ R
⇒ (∼P∧∼Q) ∨ R
⇒ (∼P∨R) × (∼Q∨R)
⇒ (P→R) ∧ (Q→R)
Option B: X→Y
[(P→R) × (Q→R)] → [(P→R) ∨ (Q→R)]
∼[(P→R) × (Q→R) ∨ (P→R) ∨ (Q→R)]
[∼(P→R) ∨ ∼(Q→R)] ∨ [(P→R) ∨ (Q→R)]
[∼(P→R) ∨ (P→R)] ∨ [∼(P→R) ∨ (Q→R)] ∨ [(Q→R) ∨ (P→R)] ∨ [∼(Q→R) ∨ (Q→R)]
T ∨ [∼(P→R) ∨ (Q→R)] ∨ [(Q→R) ∨ (P→R)] V T
T (✔️)
There are 39 questions to complete.

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