Question

The expression $\sim \left( \sim p\to q \right)$ is logically equivalent to

& A.\sim p\hat{\ }\sim q \\\ & B.p\hat{\ }q \\\ & C.\sim p\hat{\ }q \\\ & D.p\hat{\ }\sim q \\\ \end{aligned}$$

Answer 1

To solve this question we will first make truth table of $\sim p,\sim \left( \sim p\to q \right),\sim q,p\to q\text{ and }\sim \text{p}\to \text{q}$ and then to check if given expression is equivalent to which of the option given, we will match truth tables of all.
If all the values of truth tables are equal then they are equivalent. Also $\sim p$ denotes "negation of p" and $p\to q$ denotes p implies q.

Complete step by step answer:
Two statements are logically equivalent if their truth tables have the same value at each entry.
We have, if p and q are any statements, then $\sim p\Rightarrow \sim q$ denotes negation of p or negation of q. Its truth table is given as:

p	$\sim p$
T	F
F	T

Or

q	$\sim q$
T	F
F	T

Again if p and q are two statements then $p\to q$ denotes p implies q.
Its truth table is given as:

p	q	$p\to q$
T	F	F
T	T	T
F	T	T
F	F	T

Here, T denotes True and F denotes False.
Now finally we will make a truth table of $\sim \left( \sim p\to q \right)$ then make a truth table of all options and see which one matches, then it is equivalent.

p	q	$\sim p$	$\sim q$	$\sim \left( \sim p\to q \right)$
T	F	F	T	F
T	T	F	T	F
F	T	T	T	F
F	F	T	F	T

Hence, we have obtained truth table of $\sim \left( \sim p\to q \right)$
Consider option A $\sim p\hat{\ }\sim q$
Let us first define what is $p\hat{\ }q$
When p and q are two statements then $p\hat{\ }q$ is p and q whose truth table is as below:

p	q	$p\hat{\ }q$
T	F	F
T	T	T
F	T	F
F	F	F

Then the truth table of $\sim p\hat{\ }\sim q$ is

p	q	$\sim p$	$\sim q$	$\sim p\hat{\ }\sim q$
T	F	F	T	F
T	T	F	T	F
F	T	T	T	F
F	F	T	F	T

Clearly, the truth table of $\sim p\hat{\ }\sim q\text{ and }\sim \left( \sim p\to q \right)$ matches. So option A is correct.
Consider option B $p\hat{\ }q$
Truth table of $p\hat{\ }q$ is

p	q	$p\hat{\ }q$
T	F	F
T	T	T
F	T	F
F	F	F

Clearly, the truth table of $p\hat{\ }q\text{ and }\sim \left( \sim p\to q \right)$ differs. So option B is wrong.
Consider option C $\sim p\hat{\ }q$
Truth table of $\sim p\hat{\ }q$ is

p	q	$\sim p$	$\sim p\hat{\ }q$
T	F	F	F
T	T	F	F
F	T	T	T
F	F	T	F

Again the truth table of $\sim p\hat{\ }q\text{ and }\sim \left( \sim p\to q \right)$ doesn't match. So, option C is wrong.
Consider option D $p\hat{\ }\sim q$
Truth table of $p\hat{\ }\sim q$ is

p	q	$\sim q$	$p\hat{\ }\sim q$
T	F	T	T
T	T	F	F
F	T	F	F
F	F	T	F

Again the truth table of $p\hat{\ }\sim q\text{ and }\sim \left( \sim p\to q \right)$ doesn't match so option D is wrong.
Therefore, $\sim \left( \sim p\to q \right)$ is logically equivalent to $\sim p\hat{\ }\sim q$

So, the correct answer is “Option A”.

Note: The possibility of confusion here in this question can be at option C. Observing truth tables of option C we get we have 3 false (F) and 1 true (T) which is same as truth table of $\sim \left( \sim p\to q \right)$ As truth table of $\sim \left( \sim p\to q \right)$ also has 3 false F and 1 true T. Even then option C is wrong, this is so because the values of the truth table differs at position of T or F.