The statement p → (q → p) is equivalent...

MathematicsTypes of Sets and Set OperationsJEE Main 2008Easy

The statement $p \to (q \to p)$ is equivalent to

(A)

p \to (p \to q)

(B)

p \to (p \lor q)

(C)

p \to (p \land q)

(D)

p \to (p \leftrightarrow q)

Answer:B

Visualized Solution (English)

Introduction to the Statement

Given statement: $p \to (q \to p)$
Objective: Find an equivalent logical expression.

The Conditional Identity

Recall the identity: $A \to B \equiv \neg A \lor B$

Simplifying the Inner Bracket

Inner part: $(q \to p) \equiv \neg q \lor p$

Expanding the Full Expression

Full expression: $p \to (\neg q \lor p) \equiv \neg p \lor (\neg q \lor p)$

Applying the Associative Law

Using Associative Law: $(\neg p \lor p) \lor \neg q$

Identifying the Tautology

Since $\neg p \lor p \equiv T$
Expression becomes $T \lor \neg q \equiv T$
The statement is a Tautology.

Testing Option 2

Option 2: $p \to (p \lor q)$
Apply identity: $\neg p \lor (p \lor q)$

Verifying the Equivalence

Rearrange: $(\neg p \lor p) \lor q$
Result: $T \lor q \equiv T$
Both are tautologies, hence equivalent.

Final Conclusion

Key Takeaway: Two statements are equivalent if they share the same truth table.
Both $p \to (q \to p)$ and $p \to (p \lor q)$ are Tautologies.
Final Answer: Option 2

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Conceptually Similar Problems

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The Boolean Expression $(p \land \sim q) \lor q \lor (\sim p \land q)$ is equivalent to:

(A)

p \lor q

(B)

p \lor \sim q

(C)

\sim p \land q

(D)

p \land q

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The negation of $\sim s \lor (\sim r \land s)$ is equivalent to :

(A)

s \lor (r \lor \sim s)

(B)

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\sim (Q \leftrightarrow (P \land \sim R))

(B)

\sim Q \leftrightarrow\sim P \land R

(C)

\sim (P \land \sim R) \leftrightarrow Q

(D)

\sim P \land (Q \leftrightarrow\sim R)

Answer:A

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Consider Statement-1 : $(p \land \sim q) \land (\sim p \land q)$ is a fallacy. Statement-2 : $(p \to q) \leftrightarrow (\sim q \to\sim p)$ is a tautology.

(A)

Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

(B)

Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-I.

(C)

Statement-1 is true; Statement-2 is false.

(D)

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The statement $\sim (p \leftrightarrow\sim q)$ is:

(A)

a tautology

(B)

a fallacy

(C)

eqivalent to

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(D)

equivalent to

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Statement-1: $\sim (p \leftrightarrow\sim q)$ is equivalent to $p \leftrightarrow q$ . Statement-2: $\sim (p \leftrightarrow\sim q)$ is a tautology

(A)

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

(B)

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(C)

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(D)

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Let p be the statement "x is an irrational number", q be the statement "y is a transcendental number", and r be the statement "x is a rational number if y is a transcendental number". Statement-1 : r is equivalent to either q or p Statement-2 : r is equivalent to ~(p <-> q).

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STATEMENT-1: $(p^{2} - q) (b^{2} - a c) \geq 0$ and STATEMENT-2: $b \neq = p a$ or $c \neq = q a$

(A)

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

(B)

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

(C)

STATEMENT-1 is True, STATEMENT-2 is False

(D)

STATEMENT-1 is False, STATEMENT-2 is True

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If $a \times b = b \times c = c \times a$ then $a + b + c =$

(A)

a b c

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- 1

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0

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For two given events $A$ and $B, P (A \cap B)$ is

* Multiple Correct Options

(A)

not less than

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(B)

not greater than

P (A) + P (B)

(C)

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(D)

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The statement p→(q→p) is equivalent to

Visualized Solution (English)

Introduction to the Statement

The Conditional Identity

Simplifying the Inner Bracket

Expanding the Full Expression

Applying the Associative Law

Identifying the Tautology

Testing Option 2

Verifying the Equivalence

Final Conclusion

Conceptually Similar Problems

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