Question

The negation of $p \to ( \sim p \vee q)$ is
A)$p \vee (p \vee \sim q)$
B) $p \to \sim (p \vee q)$
C) $p \to q$
D) $p \wedge \sim q$

Answer 1

Hint : This is the type of logical equivalence. In these types of questions, we basically took two statements and with the help of those statements if the statement is logically equivalent then both in the given statements are resulting in the value being true.
In this question, we make a truth table $p \to ( \sim p \vee q)$ and find the negation of it. After finding the negation, we used to solve each option by using a truth table. The correct answer will match the truth table of $p \to ( \sim p \vee q)$.

Formula used : Here we use two logical equivalence. The properties are shown below;
1. \[p \to q \Leftrightarrow \sim p \vee q\]
2. $ \sim (p \vee q) \equiv \sim p \wedge \sim q$ (De-Morgan’s theorem)

Complete step-by-step solution.
Let us make the truth table of $p \to ( \sim p \vee q)$ , we get;
Before making the truth table we can write $p \to ( \sim p \vee q)$ as;
$p \to ( \sim p \vee q) \leftrightarrow \sim p \vee ( \sim p \vee q)$
Now truth table for $ \sim \left( { \sim p \vee ( \sim p \vee q)} \right)$ we get;

p	q	$ \sim p$	$ \sim p \vee q$	$ \sim p \vee ( \sim p \vee q)$	$ \sim \left( { \sim p \vee ( \sim p \vee q)} \right)$
T	T	F	T	T	F
T	F	F	F	F	T
F	T	T	T	T	F
F	F	T	T	T	F

Here, $ \sim p$ means “Not” of $p$.
$ \vee $ means “OR”
$ \sim $means negation.
Now, let us take option A) and see whether it matches with $ \sim \left( { \sim p \vee ( \sim p \vee q)} \right)$.
The truth table of Option A) is written as;

p	q	$ \sim q$	$(p \vee \sim q)$	$p \vee (p \vee \sim q)$
T	T	F	T	T
T	F	T	T	T
F	T	F	F	F
F	F	T	T	T

Now, we can see that it does not make with the solution of $ \sim p \vee ( \sim p \vee q)$ , therefore it is not the correct option.
Let us take Option B), we have;
$p \to \sim (p \vee q)$
According to the De Morgan theorem, we have;
$ \sim (p \vee q) \equiv \sim p \wedge \sim q$
Therefore,
$p \to \sim p \wedge \sim q$
By using $p \to q$ logical equivalence we get;
$p \to \sim p \wedge \sim q \Leftrightarrow \sim p \vee ( \sim p \wedge \sim q)$
Now, let us make the logical table for $ \sim p \vee ( \sim p \wedge \sim q)$ , we get;

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

Here, $ \wedge $ stands for “AND”
Hence, option B) does not belong to the $ \sim \left( { \sim p \vee ( \sim p \vee q)} \right)$. Therefore, it is not the correct option.
Now, let us solve the truth table for Option C), we get;
$p \to q$
As we know, the logical equivalence of $p \to q$ is,
$p \to q \Leftrightarrow \sim p \vee q$
The truth table is written as;

p	q	$ \sim p$	$( \sim p \vee q)$
T	T	F	T
T	F	F	F
F	T	T	T
F	F	T	T

Here, we can see that it is not matching with the solution of $ \sim \left( { \sim p \vee ( \sim p \vee q)} \right)$, Hence option c) is incorrect.
Let us take the option D), we have;
$p \wedge \sim q$
The truth table is written as;

p	q	$ \sim q$	$p \wedge \sim q$
T	T	F	F
T	F	T	T
F	T	F	F
F	F	T	F

Therefore, we can see that option D) matches with $ \sim \left( { \sim p \vee ( \sim p \vee q)} \right)$ or $p \to ( \sim p \vee q)$

Hence, option D) is correct

Note : The two logical expressions are said to be logical equivalence if both will have the same truth values in all the cases. Here we use three types of propositions based on the truth values. The three types of propositions are Tautology, Contradiction, and Contingency.