Show that ~ (p → q) and p ∧~q are logically equivalent. (Hint: you can use a truth table to prove it or you apply De Morgan law to show the ~(p → q) is p ∧~q.

Here the truth table for $\sim (p\to q)$ given below:

Table 1

Again truth table for $(p\land \sim q)$ given below:

Table 2

Here we see from Table 1 and Table 2 , the truth table for $\sim (p\to q)$ and $(p\land \sim q)$ are equal.

Therefore $\sim(p\to q) \equiv (p\land \sim q)$