## Propositional-Logic

 Question 1

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.
Engineering-Mathematics       Propositional-Logic       Gate 2018
Question 1 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 2
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
Engineering-Mathematics       Propositional-Logic       Gate 2017 set-01
Question 2 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 3

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
Engineering-Mathematics       Propositional-Logic       GATE 2016 set-2
Question 3 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 4

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))
Engineering-Mathematics       Propositional-Logic       GATE 2016 set-2
Question 4 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 5
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)
Engineering-Mathematics       Propositional-Logic       GATE 2015 -(Set-2)
Question 5 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 6
 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
Engineering-Mathematics       Propositional-Logic       Gate 2012
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
 A ∃x (real(x) ∨ rational(x)) B ∀x (real(x) → rational(x)) C ∃x (real(x) ∧ rational(x)) D ∃x (rational(x) → real(x))
Engineering-Mathematics       Propositional-Logic       Gate 2012
Question 7 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 8
 A P(x) being true means that x is a prime number B P(x) being true means that x is a number other than 1 C P(x) is always true irrespective of the value of x D P(x) being true means that x has exactly two factors other than 1 and x
Engineering-Mathematics       Propositional-Logic       Gate 2011
Question 8 Explanation:
Statement: x is not equal to 1 and if there exists some z for all y such that product of y and z is x, then y is either the no. itselfor 1.
This is the definition of prime nos.
 Question 9
Suppose the predicate F(x, y, t) is used to represent the statement that person x can fool person y at time t. Which one of the statements below expresses best the meaning of the formula ∀x∃y∃t(¬F(x, y, t))?
 A Everyone can fool some person at some time B No one can fool everyone all the time C Everyone cannot fool some person all the time D No one can fool some person at some time
Engineering-Mathematics       Propositional-Logic       2010
Question 9 Explanation:
F(x,y,t) ⇒ Person 'x' can fool person 'y' at time 't'.
For better understanding propagate negation sign outward by applying Demorgan's law.
∀x∃y∃t(¬F(x, y, t)) ≡ ¬∃x∀y∀t(F(x,y,t))
Now converting ¬∃x∀y∀t(F(x,y,t)) to English is simple.
¬∃x∀y∀t(F(x,y,t)) ⇒ There does not exist a person who can fool everyone all the time.
Which means "No one can fool everyone all the time".
Hence, Option (B) is correct.
 Question 10
 A ∀x(P(x) → (G(x) ∧ S(x))) B ∀x((G(x) ∧ S(x)) → P(x)) C ∃x((G(x) ∧ S(x)) → P(x) D ∀x((G(x) ∨ S(x)) → P(x))
Engineering-Mathematics       Propositional-Logic       2009
Question 10 Explanation:
Interpreting the options will lead to
(A) for all ornaments, if it is precious then they should be gold and silver.
But, given statement does not says that, “ only gold and silver are precious “ . So this is wrong.
(B) For all ornaments, which contains gold and silver are precious. Which is only the shaded region in the venn diagrams. But, it misses p,r regions. So, this is wrong option.
C) Some ornaments, which are gold and silver are precious. It is false, because all gold or silver ornaments are precious.
D) For all ornaments, Any ornament which is gold or silver is precious. Which is true.
 Question 11
 A ¬Q□¬P B P□¬Q C ¬P□Q D ¬P□¬Q
Engineering-Mathematics       Propositional-Logic       2009
Question 11 Explanation:
The options are simple to draw the truth table then go with the corresponding options. P∨Q=P□️Q
So, option B is correct.
 Question 12
 A I and III B I and IV C II and III D II and IV
Engineering-Mathematics       Propositional-Logic       2009
Question 12 Explanation:
I) ¬∀x(P(x)) = ∃x(¬P(x)) [De morgan's Law]
II ) ¬∃x(P(x))= ∀x(~P(x))
III) ¬∃x(¬P(x)) = ∀x(P(x))
 Question 13
 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))
Engineering-Mathematics       Propositional-Logic       Gate-2008
Question 13 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.
So answer is option A.
 Question 14
 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
Engineering-Mathematics       Propositional-Logic       Gate-2008
Question 14 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 15
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]
Engineering-Mathematics       Propositional-Logic       Gate 2008-IT
Question 15 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 16
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))]
Engineering-Mathematics       Propositional-Logic       Gate 2008-IT
Question 16 Explanation: Question 17
Let Graph(x) be a predicate which denotes that x is a graph. Let Connected(x) be a predicate which denotes that x is connected. Which of the following first order logic sentences DOES NOT represent the statement: “Not every graph is connected”?
 A ¬∀x (Graph (x) ⇒ Connected (x)) B ¬∃x (Graph (x) ∧ ¬Connected (x)) C ¬∀x (¬Graph (x) ∨ Connected (x)) D ∀x (Graph (x) ⇒ ¬Connected (x))
Engineering-Mathematics       Propositional-Logic       Gate-2007
Question 17 Explanation:
Option (A) and (C) are same, because in option (C) the given expression is
Given expression is
¬∀x(¬Graph(x) ∨ Connected(x)
which can be rewritten as,
¬∀x(Graph(x) ⇒ Connected(x)
which is equivalent to option (A)
(∵ ¬p∨q ≡ p→q)
So, option (A) and (C) cannot be the answer.
Coming to option (B), the given expression is,
∃x (Graph (x) ∧ ¬Connected (x))
"There exist some graph which is not connected", which is equivalent in saying that "Not every graph is connected".
Coming to option (D),
For all x graph is not connected, which is not correct.
Hence, option (D) is the answer.
 Question 18
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)
Engineering-Mathematics       Propositional-Logic       Gate 2007-IT
Question 18 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 19
 A P1 is a tautology, but not P2 B P2 is a tautology, but not P1 C P1 and P2 are both tautologies D Both P1 and P2 are not tautologies
Engineering-Mathematics       Propositional-Logic       Gate-2006
Question 19 Explanation:
It’s better to draw truth table such that Both P1 and P2 are not Tautologies.
 Question 20
 A ∀x [(tiger(x) ∧ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}] B ∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) ∧ attacks(x)}] C ∀x [(tiger(x) ∨ lion(x)) → {(attacks(x) → (hungry (x)) ∨ threatened (x))}] D ∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]
Engineering-Mathematics       Propositional-Logic       Gate-2006
Question 20 Explanation:
Tigers and lions attack if they are hungry (or) threatened.
Here we have two cases.
i) If Tiger is hungry (or) threaten that will attack.
ii) If Lion is hungry (or) threaten that will attack.
If Tiger is hungry (or) threaten then both lion and tiger will not attack only Tiger will attack and vice-versa.
Then answer is
∀x[(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]
Note: Don’t confuse with the statement Tiger and Lion.
 Question 21
 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
Engineering-Mathematics       Propositional-Logic       Gate 2006-IT
Question 21 Explanation:
The given relationis 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 22
 A ∀(x) [teacher(x) → ∃ (y) [student(y) → likes (y, x)]] B ∀(x) [teacher(x) → ∃ (y) [student(y) ∧ likes (y, x)]] C ∃(y) ∀(x) [teacher(x) → [student(y) ∧ likes (y, x)]] D ∀(x) [teacher(x) ∧ ∃ (y)[student(y) → likes (y, x)]]
Engineering-Mathematics       Propositional-Logic       Gate-2005
Question 22 Explanation:
Option A: If x is a teacher, then there exist a y such that if y is a student , then y likes x.
Option B: If x is a teacher, then there exists some y, who is a student and like x. (✔️)
Option C: There exists a student who likes all teachers.
Option D: If x is a teacher and then there exists some y, if y is a student then y likes x.
 Question 23
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
Engineering-Mathematics       Propositional-Logic       Gate-2005
Question 23 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 (✔️)
 Question 24
 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)))
Engineering-Mathematics       Propositional-Logic       Gate-2004
Question 24 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 25
The following propositional statement is (P ⇒ (QÚR)) ⇒ ((P Ù Q) ⇒ R)
 A satisfiable but not valid B valid C a contradiction D None of the above
Engineering-Mathematics       Propositional-Logic       Gate-2004
Question 25 Explanation:
(P ⇒ (QÚR)) ⇒ ((P Ù Q) ⇒ R)
(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 26
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])
Engineering-Mathematics       Propositional-Logic       Gate-2003
Question 26 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 27
 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
Engineering-Mathematics       Propositional-Logic       Gate-2003
Question 27 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 28
 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
Engineering-Mathematics       Propositional-Logic       Gate-2003
Question 28 Explanation:
(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R)) It is may be True (or) False depending on values. So this is not valid.
 Question 29
“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
Engineering-Mathematics       Propositional-Logic       Gate-2002
Question 29 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 30
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
Engineering-Mathematics       Propositional-Logic       Gate-2000
Question 30 Explanation:
a ↔ (b ∨-b) and b ↔ c
Given ⇒ (a∧b) → (a∧c) ∨d
⇒ (a∧b) → (a∧c) ∨d (b⇔c)
⇒ T∨d
⇒ T
 Question 31
 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
Engineering-Mathematics       Propositional-Logic       Gate-1998
Question 31 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 32
Which of the following propositions is a tautology?
 A (p ∨ q) → p B p ∨ (q → p) C p ∨ (p → q) D p → (p → q)
Engineering-Mathematics       Propositional-Logic       Gate-1997
Question 32 Explanation: Question 33
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)
Engineering-Mathematics       Propositional-Logic       Gate-1996
Question 33 Explanation:
When x = F and y = F
then option (D) will be False.
 Question 34
 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
Engineering-Mathematics       Propositional-Logic       Gate-2001
Question 34 Explanation: F1 is satisfiable; F2 is valid.
 Question 35
 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)
Engineering-Mathematics       Propositional-Logic       Gate-1990
Question 35 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 36
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.
Engineering-Mathematics       Propositional-Logic       Gate-1989
Question 36 Explanation:
P → Q ⇔ ¬P ∨ Q
¬Q → ¬P ⇔ Q ∨ ¬P
¬P ∨ Q ⇔ ¬P ∨ Q
¬Q → P ⇔ Q ∨ P
A, B and C are equivalent.
There are 36 questions to complete.