Question 13 (1 point) Select all that are (equivalent to)...

Question 13 (1 point) Select all that are (equivalent to) negations of $P \implies Q$. a) $P \land (\neg Q)$ b) P but not Q c) $(\neg P) \lor Q$ d) $\neg (P \implies Q)$ e) $P \lor (\neg Q)$

Transcript

00:01 Hi, in the given question we have to simplify the expression negate of p and q or negate p and negate q or p and negate q or p and q.

00:22 We have to simplify this expression.

00:24 If we multiply inside the bracket by the negate we know that negate to negate is always positive.

00:31 Therefore we get p and q.

00:35 If we multiply by negate to the or we get and then we have negate to negate is always positive.

00:44 Therefore we get p into negate of and is or into q.

00:50 Similarly for this or we get the negate of or is and and inside the bracket we get p or q.

01:00 So the answer of given expression is p and q or p or q and or q and p or q and which is a required solution.

01:18 Next we have to state that whether the given expression that is p and q or negate p or p and negate q is is tautology contradiction or what? we must know that when the result is always positive or when the result is always true, then it is tautology.

01:58 And when the result is always false, then the given expression is contradiction.

02:17 So let's simplify the given expression as by using the truth table.

02:22 We know that the 2 treble of p, q.

02:31 We have the given values are p and q.

02:36 Next we will find the value of negation of p.

02:41 Then we have to find the value of negation of q.

02:46 Let's find the value of p and negation of q.

02:51 Then we will find our expression that is p and negation of q...

Question

Question 13 (1 point) Select all that are (equivalent to) negations of $P \implies Q$. a) $P \land (\neg Q)$ b) P but not Q c) $(\neg P) \lor Q$ d) $\neg (P \implies Q)$ e) $P \lor (\neg Q)$

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