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Match the following :

$\begin{array}{clcl}  & \textbf{List-I} && \textbf{List-II} \\ \text{(a)}&(p \rightarrow q) \Leftrightarrow (\neg q \rightarrow \neg p) & \text{(i)} & \text{Contrapositive} \\ \text{(b)}& [(p \wedge q) \rightarrow r] \Leftrightarrow [p \rightarrow (q \rightarrow r)]& \text{(ii)} & \text{Exportation law} \\ \text{(c)} & (p \rightarrow q) \Leftrightarrow [(p \wedge \neg q) \rightarrow o] & \text{(iii)} & \text{Reduction as absurdum} \\ \text{(d)} & (p \leftrightarrow q) \Leftrightarrow [(p \rightarrow q) \wedge (q \rightarrow p)]& \text{(iv)} & \text{Equivalence} \\ \end{array}$

$\textbf{Codes :}$

  1. $\text{(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)}$
  2. $\text{(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)}$
  3. $\text{(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)}$
  4. $\text{(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)}$
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2 Answers

Best answer
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(a)The contrapositive of a conditional statement is formed by negating both the hypothesis and the conclusion, and then interchanging the resulting negations. In other words, the contrapositive negates and switches the parts of the sentence. It does BOTH the jobs of the INVERSE and the CONVERSE.

Ex:-(p→q)⇔(¬q→¬p)

(b)Exportation is a valid rule of replacement in propositional logic. The rule allows conditional statements having conjunctive antecedents to be replaced by statements having conditional consequents and vice versa inlogical proofs. It is the rule that:

((P ∧ Q ) $\rightarrow$ R)  (p$\rightarrow$(Q$\rightarrow$R))

 

(C)

 

(d)Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true.It is logically equivalent to (p → q) ∧ (q → p).

List-I List-II
a. $(p \rightarrow q) \Leftrightarrow (\neg q \rightarrow \neg p)$ i. Contrapositive
b. $[(p \wedge q) \rightarrow r] \Leftrightarrow [p \rightarrow (q \rightarrow r)]$ ii. Exportation law
c. $(p \rightarrow q) \Leftrightarrow [(p \wedge \neg q) \rightarrow o]$ iii. Reductio as absurdum
d. $(p \leftrightarrow q) \Leftrightarrow [(p \rightarrow q) \wedge (q \rightarrow p)]$ iv. Equivalence

Answer:-  a-i, b-ii, c-iii, d-iv

 

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a.contra positive

d.equivalence

By looking only this two options we get Ans as A.
Answer:

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