Mathematical Reasoning

Using DeMorgan's law,

$ \sim(p \wedge q)=\sim p \vee \sim q $

Thus, our expression becomes

$ (\sim p \vee \sim q) \vee(p \wedge \sim q) \vee(\sim p \wedge \sim q) $

Rearrange and group similar terms

$ \sim p \vee(\sim q \vee(p \wedge \sim q) \vee(\sim p \wedge \sim q)) $

Since, $\sim q \vee(p \wedge \sim q) \vee(\sim p \wedge \sim q)$ simplifies to $\sim q \vee p$, we get

$ \sim p \vee q $

The Boolean expression simplifies to $\sim p \vee q$.

The Boolean expression $ \sim(p \vee q) \vee (\sim p \wedge q) $ is simplified as follows:

Start with the given expression:

$ \sim(p \vee q) \vee (\sim p \wedge q) $

Apply De Morgan's Law:

$ (\sim p \wedge \sim q) \vee (\sim p \wedge q) $

Use the Distributive Law:

$ \sim p \wedge (\sim q \vee q) $

By the Complement Law, $(\sim q \vee q)$ is a tautology ($t$):

$ \sim p \wedge t $

Finally, apply the Identity Law:

$ \sim p $

Therefore, the expression simplifies to $ \sim p $.

So, options (a) and (d) are not true.

$\sim$ (p $\to$ q) $\equiv$ p $\wedge$ $\sim$ q, so option (c) is not true.

Now, p $\to$ q $\sim$ p $\vee$ q

$\sim$ q $\to$ $\sim$ p $\equiv$ [$\sim$ ($\sim$ q) $\vee$ $\sim$ p]

$\equiv$ q $\vee$ $\sim$ p $\equiv$ p $\vee$ q $\Rightarrow$ p $\to$ q $\equiv$ ($\sim$ q $\to$ $\sim$ p)