Question

Use Theorem 2.1.1 to verify the logical equivalence. Give a reason for each step. -(pv –q) v(-p^q) = ~p

Answer 1

Answer:

The statement $\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q)$ is equivalent to $\lnot p$, $\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv \lnot p$

Step-by-step explanation:

We need to prove that the following statement $\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q)$ is equivalent to $\lnot p$ with the use of Theorem 2.1.1.

So

$\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv$

$\equiv (\lnot p \land \lnot(\lnot q))\lor(\lnot p \land \lnot q)$ by De Morgan's law.

$\equiv (\lnot p \land q)\lor(\lnot p \land \lnot q)$ by the Double negative law

$\equiv \lnot p \land (q \lor \lnot q)$ by the Distributive law

$\equiv \lnot p \land t$ by the Negation law

$\equiv \lnot p$ by Universal bound law

Therefore $\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv \lnot p$