Question

Without using the truth table show that $\sim (p\vee q)\vee (\sim p\wedge q)\equiv \sim p$.

Answer 1

First use De Morgan law to simplify $\sim (p\vee q)$ and apply distributive law to simplify the result obtained earlier. Then finally use negation law to find the result.
Let$p$, $q$ and $r$ be proportions, then De Morgan law states that negation of a conjunction is the disjunction of negations which can be written symbolically as $\sim (p\vee q)\equiv \sim p\wedge \sim q$. Distributive law is symbolically written as$p\wedge (q\vee r)\equiv (p\wedge q)\vee (p\wedge r)$ which means disjunction will distribute over conjunction and Negation law is defined as $\sim q\vee q\equiv True$ that means conjunction of $q$ and $\sim q$will always be true.

Complete step by step answer:
We are given in question that we have to show $\sim (p\vee q)\vee (\sim p\wedge q)\equiv r$ without using the truth table. We will start from left hand side (L.H.S) of expression that is,
$\sim (p\vee q)\vee (\sim p\wedge q)$
Now we will be using De Morgan law $\sim (p\vee q)\equiv \sim p\wedge \sim q$ to simplify $\sim (p\vee q)$ then the statement becomes,
$\equiv (\sim p\wedge \sim q)\vee (\sim p\wedge q)$
Now use Distributive law $p\wedge (q\vee r)\equiv (p\wedge q)\vee (p\wedge r)$ to simplify $(\sim p\wedge \sim q)\vee (\sim p\wedge q)$. So after applying distributive law we get,
$\equiv \sim p\wedge (\sim q\vee q)$
Now use Negation law $\sim q\vee q\equiv True$ to the above expression we get,
$\equiv \sim p\wedge True$
$\equiv \sim p$ $[\because \sim p\wedge True\equiv \sim p]$
As disjunction of negation of $p$and true value will always be negation of $p$.
This expression is equal to the right hand side (R.H.S) of the expression given in the question.
Therefore, L.H.S=R.H.S
Hence, we proved that$\sim (p\vee q)\vee (\sim p\wedge q)\equiv \sim p$.

Note:
We should take care while simplifying $(\sim p\wedge \sim q)\vee (\sim p\wedge q)$ using Distributive property $a\wedge (b\vee c)\equiv (a\wedge b)\vee (a\wedge c)$. Here, $a=\sim p$ , $b=\sim q$ and $c=q$ so $(\sim p\wedge \sim q)\vee (\sim p\wedge q)$ will get simplified as $\sim p\wedge (\sim q\vee q)$. $\sim p\wedge True\equiv \sim p$ because if we take $p$ to be true value then negation of $p$will be false so disjunction of true and false will be false. Similarly if we take $p$ to be false value then negation of $p$will be true so disjunction of true and false will be false that implies whatever be the value of negation of $p$will be the result of expression $\sim p\wedge True$. Hence, $\sim p\wedge True\equiv \sim p$is proved.