Mathematical Reasoning

Step 1: Understanding the Concept:
A Boolean expression is a tautology if it is true for all possible truth values of its variables. For an implication to be a tautology, whenever is true, must also be true.
Step 2: Detailed Explanation:
Let's test Option (B): .
The expression is .
Truth Table:

Since the last column is all True, the expression is a tautology for these operators.
If we checked (A): . If , the LHS is but RHS is , so it's not a tautology.
Step 3: Final Answer:
The operators are and .

The correct answer is 9
There are total 9 compound propositions, out of which 6 contain ~s. So if we assign s as false, these 6 propositions will be true.
In remaining 3 compound propositions, two contain p and the third contains ~r. So if we assign p and r as true and false respectively, these 3 propositions will also be true.
Therefore, maximum number of propositions that can be true are 9.

The correct answer is (C) : (~p) ∧ q ∧ r
p ∨ q ⇒ (~ r ∨ p)
≡ ~ (p ∨ q) ∨ (~ r ∨ p)
≡ (~ p ∧ ~ q) ∨ (p ∨ ~ r)
≡ [(~ p ∨ p) ∧ (~ q ∨ p)] ∨ ~ r
≡ (~ q ∨ p) ∨ ~ r
Its negation is ~ p ∧ q ∧ r.

Let's solve the logical equivalence problem given in the question.

1. We need to determine for which value of r the expression is equivalent to .
2. Start by analyzing the logical expressions:
   • means both p and r need to be true.
   • means p is true and q is false.
   • Consider the possible scenarios:
     • If p is false, both and become false, satisfying the condition .
     • Thus,

The correct answer is (C) : ~p
Clearly r must be equal to ~ p
∵ ~ p ∨ ~ p = ~ p
and (p ∧ q) ∨ ~ p = p
∴ ~ p ⇒ p = tautology.

The correct answer is (A) :
Case-I If ∇ is same as ∧
Then (p∧q) ⇒ ((pΔq) ∧r) is equivalent to ~ (p∧q) ∨ ((pΔq) ∧r) is equivalent to (~ (p∧q) ∨ (pΔq))∧ (~ (p∧q) ∨r)
Which cannot be a tautology
For both Δ (i.e.∨ or ∧)

Case-II If ∇ is same as ∨
Then (p∨q) ⇒ ((pΔq) ∨r) is equivalent to
~(p∨q) ∨ (pΔq) ∨r which can be a tautology if Δ is also same as ∨.
Hence both Δ and ∇ are same as ∨.
Now (p∇q) Δr is equivalent to (p∨q∨r).

To determine the equivalent logical statement for , we will use logical equivalences and analyze each option.

The implication can be rewritten using the definition of implication: is equivalent to .

This means:

Simplifying involves the following steps:

1. The expression using De Morgan's Laws becomes .
2. Thus, becomes .
3. The distributive law allows us to further simplify this to:
4. We know that is a tautology and simplifies to , so the expression reduces to:
   • Which simplifies finally to
   • is equivalent to .

Therefore, the equivalent statement to is .

This matches the correct option: .

So, the correct option is (D):

The correct answer is (D) : (~ p) ∧ (~ q)
p⇔ (q⇒p)
~ (p⇔ (q⇔p))
≡p⇔ ~ (q⇒p)
≡p⇔ (q∧ ~ p)
≡ (p⇒ (q∧ ~ p)) ∧ ((q∧ ~ p) ⇒p))
≡ (~ p∨ (q∧ ~ p)) ∧ ((~ q∨p) ∨p))
≡ ((~ p∨q) ∧ ~ p) ∧ (~ q∨p)
≡ ~ p∧ (~ q∨p)
≡ (~ p∧ ~ q) ∨ (~ p∧p)
≡ (~ p∧ ~ q) ∨c
≡ (~ p∧ ~ q)

Step 1: Analyze Each Statement
(i) :

Using distributive property:

This simplifies to:


This is not always true, so it is not a tautology.
(ii) :


This simplifies to:

Hence, it is tautology
(iii) :




This is not always true, so it is not a tautology
(iv)

This is not always true, so it is not a tautology.
Conclusion
The statement is a tautology.

We will evaluate the truth values of the given statements and for different truth values of , , and . The truth table is as follows:

Analysis of Statements

For statement , the truth table shows that is not true for all cases. Specifically, it is false when .

For statement , is also not a tautology because it is not true in all cases.

Conclusion

Neither nor is a tautology.

Therefore, the correct answer is:

Let r=(∼(P∧Q))∨((∼P)∧Q);s= ((∼P)∧(∼Q))

Option (A) : ((∼P)∨Q)∧((∼Q)∨P)
is equivalent to (not of only P)∧( not of only Q )
=( Both P,Q) and (neither P nor Q)

To find the maximum value of n such that (66)! is divisible by 3ⁿ, we need to count the number of factors of 3 in (66)!, since each factor of 3 contributes to the divisibility by 3.
To count the number of factors of 3 in (66)!, we can use the formula:

= 22 + 7 + 2 + 0 = 31,
where ⌊x⌋ denotes the greatest integer less than or equal to x.
This means that (66)! is divisible by 3³¹. Therefore, the maximum value of n such that (66)! is divisible by is n = 31.
So, the correct answer is 31

The correct answer is 5.