PropositionalLogic
Question 1 
Consider the firstorder logic sentence
where ψ(s,t,u,v,w,x,y) is a quantifierfree firstorder 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?
There exists at least one model of φ with universe of size less than or equal to 3.
 
There exists no model of φ with universe of size less than or equal to 3.
 
There exists no model of φ with universe of size greater than 7.  
Every model of φ has a universe of size equal to 7.

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’.
"∃" 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
I only  
I and IV only  
II only  
II and III only 
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.
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 _________.
4  
5  
6  
7 
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))
(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 wellformed formulae in predicate calculus is NOT valid?
(∀x p(x) ⇒ ∀x q(x)) ⇒ (∃x ¬p(x) ∨ ∀x q(x))  
(∃x p(x) ∨ ∃x q(x)) ⇒ ∃x (p(x) ∨ q(x))  
∃x (p(x) ∧ q(x)) ⇒ (∃x p(x) ∧ ∃x q(x))  
∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x)) 
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.
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?
∀x ∃y R(x,y)↔ ∃y ∀x R(x,y)
 
(∀x [∃y R(x,y)→S(x,y)])→ ∀x∃y S(x,y)
 
[∀x ∃y (P(x,y)→R(x,y)]↔[∀x ∃y ( ¬ P(x,y)∨R(x,y)]  
∀x ∀y P(x,y)→ ∀x ∀y P(y,x)

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.
[∀x ∃y (P(x,y)→R(x,y)]↔[∀x ∃y ( ¬ P(x,y)∨R(x,y)] is a tautology.
Question 6 
Both I_{1} and I_{2} are correct inferences  
I_{1} is correct but I_{2} is not a correct inference  
I_{1} is not correct but I_{2} is a correct inference  
Both I_{1} and I_{2} are not correct inferences 
Question 6 Explanation:
I_{1}: 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.
I_{2}: 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.
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.
I_{2}: 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 
∃x (real(x) ∨ rational(x))  
∀x (real(x) → rational(x))  
∃x (real(x) ∧ rational(x))  
∃x (rational(x) → real(x)) 
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 
P(x) being true means that x is a prime number  
P(x) being true means that x is a number other than 1
 
P(x) is always true irrespective of the value of x
 
P(x) being true means that x has exactly two factors other than 1 and x 
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.
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))?
Everyone can fool some person at some time  
No one can fool everyone all the time  
Everyone cannot fool some person all the time
 
No one can fool some person at some time 
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.
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 
∀x(P(x) → (G(x) ∧ S(x)))
 
∀x((G(x) ∧ S(x)) → P(x))  
∃x((G(x) ∧ S(x)) → P(x)
 
∀x((G(x) ∨ S(x)) → P(x))

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.
(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 
¬Q□¬P
 
P□¬Q
 
¬P□Q
 
¬P□¬Q

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.
P∨Q=P□️Q
So, option B is correct.
Question 12 
I and III  
I and IV
 
II and III
 
II and IV

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))
II ) ¬∃x(P(x))= ∀x(~P(x))
III) ¬∃x(¬P(x)) = ∀x(P(x))
Question 13 
(∀x fsa(x)) ⇒ (∃y pda(y) ∧ equivalent(x,y))
 
∼∀y(∃x fsa(x) ⇒ pda(y) ∧ equivalent(x,y))  
∀x ∃y(fsa(x) ∧ pda(y) ∧ equivalent(x,y))
 
∀x ∃y(fsa(y)∧ pda(x) ∧ equivalent(x,y)) 
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.
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 
Only I and II
 
Only I, II and III  
Only I, II and IV
 
All of I, II, III and IV 
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 (✔️)
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.
[β→(∃x,α(x))]→[∀x,β→α(x)]  
[∃x,β→α(x)]→[β→(∀x,α(x))]  
[(∃x,α(x))→β]→[∀x,α(x)→β]  
[(∀x,α(x))→β]→[∀x,α(x)→β]

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 viceversa).
Here, LHS = RHS
So, Option C is valid.
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 viceversa).
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))]
[∃ x, α → (∀y, β → (∃u, ∀ v, y))]  
[∃ x, α → (∀y, β → (∃u, ∀ v, ¬y))]  
[∀ x, ¬α → (∃y, ¬β → (∀u, ∃ v, ¬y))]  
[∃ x, α ʌ (∀y, β ʌ (∃u, ∀ v, ¬y))] 
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”?
¬∀x (Graph (x) ⇒ Connected (x))  
¬∃x (Graph (x) ∧ ¬Connected (x))  
¬∀x (¬Graph (x) ∨ Connected (x))  
∀x (Graph (x) ⇒ ¬Connected (x))

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.
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 firstorder logic formula is valid?
∀x(P(x) ⇒ Q(x)) ⇒ (∀xP(x) ⇒ ∀xQ(x))  
∃x(P(x) ∨ Q(x)) ⇒ (∃xP(x) ⇒ ∃xQ(x))  
∃x(P(x) ∧ Q(x)) (∃xP(x) ∧ ∃xQ(x))  
∀x∃y P(x, y) ⇒ ∃y∀x P(x, y)

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 (️❌)
RHS = if P(x) holds for all x, then Q(x) holds for all x
LHS ⇒ RHS (✔)
RHS ⇒ LHS (️❌)
Question 19 
P1 is a tautology, but not P2
 
P2 is a tautology, but not P1
 
P1 and P2 are both tautologies  
Both P1 and P2 are not tautologies

Question 19 Explanation:
It’s better to draw truth table such that
Both P1 and P2 are not Tautologies.
Both P1 and P2 are not Tautologies.
Question 20 
∀x [(tiger(x) ∧ lion(x)) → {(hungry(x) ∨ threatened(x))
→ attacks(x)}]  
∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x))
∧ attacks(x)}]  
∀x [(tiger(x) ∨ lion(x)) → {(attacks(x) → (hungry (x)) ∨ threatened (x))}]  
∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]

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 viceversa.
Then answer is
∀x[(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]
Note: Don’t confuse with the statement Tiger and Lion.
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 viceversa.
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 
satisfiable and valid  
satisfiable and so is its negation  
unsatisfiable but its negation is valid  
satisfiable but its negation is unsatisfiable 
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 
∀(x) [teacher(x) → ∃ (y) [student(y) → likes (y, x)]]  
∀(x) [teacher(x) → ∃ (y) [student(y) ∧ likes (y, x)]]
 
∃(y) ∀(x) [teacher(x) → [student(y) ∧ likes (y, x)]]
 
∀(x) [teacher(x) ∧ ∃ (y)[student(y) → likes (y, x)]]

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.
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?
X ≡ Y  
X → Y
 
Y → X
 
¬Y → X

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 (✔️)
⇒ ∼(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 
(∃x) (boy(x) → (∀y) (girl(y) ∧ taller(x,y)))
 
(∃x) (boy(x) ∧ (∀y) (girl(y) ∧ taller(x,y)))
 
(∃x) (boy(x) → (∀y) (girl(y) → taller(x,y)))
 
(∃x) (boy(x) ∧ (∀y) (girl(y) → taller(x,y)))

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.
'∧' → 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)
satisfiable but not valid
 
valid
 
a contradiction
 
None of the above

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)
(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)
((∀x)[α] ⇒ (∀x)[β]) ⇒ (∀x)[α⇒β]
 
(∀x)[α] ⇒ (∃x)[α ∧ β]  
((∀x)[α ∨ β] ⇒ (∃x)[α] ⇒ (∀x)[α]
 
(∀x)[α ⇒ β] ⇒ ((∀x)[α] ⇒ (∀x)[β])

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.
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 
I_{1} satisfies α, I_{2} does not  
I_{2} satisfies α, I_{1} does not
 
Neither I_{2} nor I_{1} satisfies α
 
Both I_{1} and I_{2} satisfy α

Question 27 Explanation:
Given that:
(∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Q_{yy}]] ⇒(∀x)[¬Px]
Q_{yy} is always true, because y divide y, then ¬Q_{yy} 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 I_{1} and I_{2} satisfies α.
(∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Q_{yy}]] ⇒(∀x)[¬Px]
Q_{yy} is always true, because y divide y, then ¬Q_{yy} 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 I_{1} and I_{2} satisfies α.
Question 28 
((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q) is logically valid  
(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R)) is logically valid
 
(P ∨ Q) is satisfiable if and only if (P∨R) ∧ (Q∨¬R) is satisfiable  
(P ∨ Q) ⇒ FALSE if and only if both P and Q are unsatisfiable

Question 28 Explanation:
(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R))
It is may be True (or) False depending on values. So this is not valid.
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)
(X∧¬Z)→Y  
(X∧Y)→¬Z  
X→(Y∧¬Z)  
(X→Y)∧¬Z 
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.
⇒ 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 truthvalue of the formula (a ∧ b) → (a ∧ c) ∨ d is always
True  
False  
Same as the truthvalue of b  
Same as the truthvalue of d 
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
Given ⇒ (a∧b) → (a∧c) ∨d
⇒ (a∧b) → (a∧c) ∨d (b⇔c)
⇒ T∨d
⇒ T
Question 31 
I stay if you go  
If I stay then you go  
If you do not go then I do not stay  
If I do not stay then you go 
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.
⇒ 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?
(p ∨ q) → p  
p ∨ (q → p)  
p ∨ (p → q)  
p → (p → q) 
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.
((x → y) ∧ x) → y  
((x → y) ∧ (x ∧ y)) → x  
(x → (x ∨ ψ))  
((x ∨ y) ↔ (x → y) 
Question 33 Explanation:
When x = F and y = F
then option (D) will be False.
then option (D) will be False.
Question 34 
F1 is satisfiable, F2 is valid  
F1 unsatisfiable, F2 is satisfiable  
F1 is unsatisfiable, F2 is valid  
F1 and F2 are both satisfiable 
Question 34 Explanation:
F1 is satisfiable; F2 is valid.
Question 35 
(P ⇒ Q) ∧ (Q ⇒ R) ⇒ (P ⇒ R)  
(P ⇒ Q) ⇒ (¬P ⇒ ¬Q)  
(P ∧ (¬P ∨ ¬Q)) ⇒ Q  
(P ⇒ R) ∨ (Q ⇒ R) ⇒ ((P ∨ Q) ⇒ R) 
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.
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 wellformed formulas are equivalent?
P → Q  
¬Q → ¬P  
¬P ∨ Q  
¬Q → P  
A, B and C. 
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.
¬Q → ¬P ⇔ Q ∨ ¬P
¬P ∨ Q ⇔ ¬P ∨ Q
¬Q → P ⇔ Q ∨ P
A, B and C are equivalent.
There are 36 questions to complete.