First slide

Introduction to mathematical reasoning

Question

Statement-I: $~ (A \Leftrightarrow\sim B)$ is equivalent to $A \Leftrightarrow B$
Statement-2: $A \lor (~ (A \land ~ B))$ is a tautology.

Moderate

A

STATEMENT- I is True, STATEMENT-2 is True; S TATEMENT-2 is a correct explanation for STATEMENT- I
B

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

STATEMENT-I is True, STATEMENT-2 is False
D

STATEMENT-I is False, STATEMENT-2 is True

Solution

We have
$\begin{array}{l} A \lor [~ (A \land ~ B)] \\ \equiv A \lor [~ A \lor B] \\ \equiv (A \lor ~ A) \lor B \equiv T \lor B \equiv T \end{array}$
Thus, Statement-2 is true.
We have
$\begin{array}{l} \equiv ~ [A \Leftrightarrow ~ B] \\ =\sim [(A \to\sim B) \land (~ B \to A)] \\ \equiv\sim [(~ A \lor ~ B) \land (B \lor A)] \end{array}$
$\begin{array}{l} \equiv\sim [~ (A \land B) \land (A \lor B)] \\ \equiv [(A \land B) \lor (~ A \land ~ B)] \\ \equiv [(A \land B) \lor (~ A)] \land [(A \land B) \lor (~ B)] \\ \equiv [(A \lor ~ A) \lor (B \lor - A)] \land [(A \lor ~ B) \land (B \lor - B)] \\ \equiv [T \land (~ A \lor B)] \land [(A \lor ~ B) \land T] \\ \equiv (A \to B) \land (B \to A) \equiv A \Leftrightarrow B \end{array}$

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